All_Solutions/HW_04_LinearAlgebra_Due_10_30_Solutions.ipynb

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   "cell_type": "markdown",
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    "## EE 502 P: Analytical Methods for Electrical Engineering\n",
    "    \n",
    "# Homework 4: Linear Algebra\n",
    "## Due 30 October, 2019 at 6:00 PM\n",
    "### <span style=\"color: red\">YOUR NAME HERE</span>\n",
    "\n",
    "Copyright &copy; 2019, University of Washington\n",
    "\n",
    "<hr>\n",
    "\n",
    "**Instructions**: Use this notebook as a template. Answer all questions using well formatted Markdown with embedded LaTeX equations, executable Jupyter cells, or both. Submit your homework solutions as an `.ipynb` file via Canvas.\n",
    "\n",
    "<span style=\"background: yellow; padding: 6px; border: 1pt solid black\">\n",
    "Although you may discuss the homework with others, you must turn in your own, original work.\n",
    "</span>\n",
    "\n",
    "**Things to remember:**\n",
    "- Use complete sentences. Equations should appear in text as grammatical elements.\n",
    "- Comment your code.\n",
    "- Label your axes. Title your plots. Use legends where appropriate. \n",
    "- Before submitting a notebook, choose Kernel -> Restart and Run All to make sure your notebook runs when the cells are evaluated in order. \n",
    "\n",
    "Note : Late homework will be accepted up to one week after the due date and will be worth 50% of its full credit score. "
   ]
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    "### 0. Warmup (Do not turn in)\n",
    "\n",
    "- Make sure you get download, read, and run the notebook for lecture 4. Work through the notebook cell by cell and see what happens when you change the expressions, and make up some of your own.\n",
    "- The material covered in class is intended to be an introductory subset the general subject of Linear Algebra, for example covered in the following online books:\n",
    "    - [Beezer](http://linear.ups.edu/html/fcla.html). Does not cover matrix exponentials, Caley-Hamilton, SVD, or PCA.\n",
    "    - [Hefferon](http://joshua.smcvt.edu/linearalgebra/book.pdf). Also does not cover the above.\n",
    "- The more advanced ideas are covered online as well. For example:\n",
    "    - Matrix Exponentials are typically not covered until a course on linear different equations. But the [Wikipedia page](https://en.wikipedia.org/wiki/Matrix_exponential) is pretty good.\n",
    "    - [Caley-Hamilton](https://brilliant.org/wiki/cayley-hamilton-theorem/)\n",
    "    - [SVD](https://www.cs.cmu.edu/~venkatg/teaching/CStheory-infoage/book-chapter-4.pdf)\n",
    "    - [PCA](https://medium.com/@jonathan_hui/machine-learning-singular-value-decomposition-svd-principal-component-analysis-pca-1d45e885e491)\n",
    "- If you want the **best** textbooks in linear algebra, get\n",
    "    - [Strang](https://www.amazon.com/Introduction-Linear-Algebra-Gilbert-Strang/dp/0980232775). Introductory. See also [Strang's Online Lectures](https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/).\n",
    "    - [Horn and Johnson](http://www.cse.zju.edu.cn/eclass/attachments/2015-10/01-1446086008-145421.pdf)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
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   "source": [
    "import math\n",
    "import scipy # Has linear algebra\n",
    "import scipy.ndimage\n",
    "import numpy as np\n",
    "import sympy as sm\n",
    "sm.init_printing(use_latex='mathjax')\n",
    "import matplotlib.pyplot as plt\n",
    "import matplotlib.patches as pch\n",
    "%matplotlib inline\n",
    "\n",
    "import pandas as pd\n",
    "import seaborn as sns\n",
    "sns.set()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2019-10-08T16:48:02.889303Z",
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   "source": [
    "### 1. Linear independence  (10pts)\n",
    "\n",
    "Which of the following sets of vectors are linearly independent? Why?\n",
    "\n",
    "a) $\\left (\n",
    "\\begin{array}{c}\n",
    "1 \\\\\n",
    "2 \\\\\n",
    "3 \\\\\n",
    "4\n",
    "\\end{array}\n",
    "\\right )$ and $\\left (\n",
    "\\begin{array}{c}\n",
    "5 \\\\\n",
    "6 \\\\\n",
    "7 \\\\\n",
    "8\n",
    "\\end{array}\n",
    "\\right )$\n",
    "\n",
    "b) $\\left (\n",
    "\\begin{array}{c}\n",
    "1 \\\\\n",
    "1 \\\\\n",
    "0 \\\\\n",
    "0\n",
    "\\end{array}\n",
    "\\right )$, $\\left (\n",
    "\\begin{array}{c}\n",
    "0 \\\\\n",
    "0 \\\\\n",
    "1 \\\\\n",
    "1\n",
    "\\end{array}\n",
    "\\right )$, $\\left (\n",
    "\\begin{array}{c}\n",
    "0 \\\\\n",
    "1 \\\\\n",
    "1 \\\\\n",
    "0\n",
    "\\end{array}\n",
    "\\right )$ and $\\left (\n",
    "\\begin{array}{c}\n",
    "1 \\\\\n",
    "0 \\\\\n",
    "0 \\\\\n",
    "1\n",
    "\\end{array}\n",
    "\\right )$\n",
    "\n",
    "c) $\\left (\n",
    "\\begin{array}{c}\n",
    "1 \\\\\n",
    "1 \\\\\n",
    "0 \\\\\n",
    "0\n",
    "\\end{array}\n",
    "\\right )$, $\\left (\n",
    "\\begin{array}{c}\n",
    "0 \\\\\n",
    "0 \\\\\n",
    "1 \\\\\n",
    "1\n",
    "\\end{array}\n",
    "\\right )$, $\\left (\n",
    "\\begin{array}{c}\n",
    "0 \\\\\n",
    "1 \\\\\n",
    "1 \\\\\n",
    "0\n",
    "\\end{array}\n",
    "\\right )$ and $\\left (\n",
    "\\begin{array}{c}\n",
    "0 \\\\\n",
    "1 \\\\\n",
    "0 \\\\\n",
    "1\n",
    "\\end{array}\n",
    "\\right )$\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    10/3pts for explaining why they are linearly independent\n",
    "    10/3pts (b) for explaining why they are NOT linearly independent\n",
    "    10/3pts (c) for explaining why they are linearly independent"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.a)\n",
    "\n",
    "Yes, it is clearly independent, but let's say some more about why. The row reduced echelon form of the matrix of these vectors stacked horizontally has two leading ones. (i.e. For this set of vectors to be linearly independent, the number of leading ones has to equal the number of vectors.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The rank of A is  2 and the column dimension is 2\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}1 & 0\\\\0 & 1\\\\0 & 0\\\\0 & 0\\end{matrix}\\right], \\quad \\left ( 0, \\quad 1\\right )\\right )$$"
      ],
      "text/plain": [
       "⎛⎡1  0⎤        ⎞\n",
       "⎜⎢    ⎥        ⎟\n",
       "⎜⎢0  1⎥        ⎟\n",
       "⎜⎢    ⎥, (0, 1)⎟\n",
       "⎜⎢0  0⎥        ⎟\n",
       "⎜⎢    ⎥        ⎟\n",
       "⎝⎣0  0⎦        ⎠"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "'''\n",
    "rref returns a tuple of two elements. The first is the reduced row echelon form, \n",
    "and the second is a tuple of indices of the pivot columns.\n",
    "'''\n",
    "x = np.array([[1],[2],[3],[4]])\n",
    "y = np.array([[5],[6],[7],[8]])\n",
    "A = np.hstack([x,y])\n",
    "\n",
    "A = sm.Matrix(A)\n",
    "print('The rank of A is ', A.rank(),'and the column dimension is', np.shape(A)[1])\n",
    "A.rref()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.b)\n",
    "\n",
    "No, it is linearly dependent. Observe that \n",
    "$$d = a + b - c $$\n",
    "\n",
    "$$\n",
    "\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ 1 \\end{pmatrix}  = \n",
    "\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\\\ 0 \\end{pmatrix} +\n",
    "\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\\\ 1 \\end{pmatrix} - \n",
    "\\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\\\ 0 \\end{pmatrix} \n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The rank of A is 3 \n",
      "The column  dimension is 4 \n",
      "det(A)= 0\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}1 & 0 & 0 & 1\\\\0 & 1 & 0 & 1\\\\0 & 0 & 1 & -1\\\\0 & 0 & 0 & 0\\end{matrix}\\right], \\quad \\left ( 0, \\quad 1, \\quad 2\\right )\\right )$$"
      ],
      "text/plain": [
       "⎛⎡1  0  0  1 ⎤           ⎞\n",
       "⎜⎢           ⎥           ⎟\n",
       "⎜⎢0  1  0  1 ⎥           ⎟\n",
       "⎜⎢           ⎥, (0, 1, 2)⎟\n",
       "⎜⎢0  0  1  -1⎥           ⎟\n",
       "⎜⎢           ⎥           ⎟\n",
       "⎝⎣0  0  0  0 ⎦           ⎠"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a = np.array([[1],[1],[0],[0]])\n",
    "b = np.array([[0],[0],[1],[1]])\n",
    "c = np.array([[0],[1],[1],[0]])\n",
    "d = np.array([[1],[0],[0],[1]]) \n",
    "\n",
    "A = np.hstack([a,b,c,d])\n",
    "A = sm.Matrix(A)\n",
    "\n",
    "print('The rank of A is', A.rank(), '\\nThe column  dimension is', np.shape(A)[1], \\\n",
    "      '\\ndet(A)=', A.det())\n",
    "A.rref()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.c)\n",
    "\n",
    "Yes. We'll use another test to show this. Recall that a set of $n$ vectors of length $n$ is linearly independent if the matrix with these vectors as columns has a non-zero determinant. We show this in python below."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$-2$$"
      ],
      "text/plain": [
       "-2"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a = np.array([[1],[1],[0],[0]])\n",
    "b = np.array([[0],[0],[1],[1]])\n",
    "c = np.array([[0],[1],[1],[0]])\n",
    "d = np.array([[0],[1],[0],[1]])\n",
    "\n",
    "A = np.hstack([a,b,c,d])\n",
    "\n",
    "#Now check in sympy\n",
    "A = sm.Matrix(A)\n",
    "A.det()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "colab_type": "text",
    "id": "lV39KJK-KOdi"
   },
   "source": [
    "### 2. Orthogonality (10pts)\n",
    "\n",
    "a) Find two orthogonal vectors that are also orthogonal to the vector\n",
    "$v = \\left (\n",
    "\\begin{array}{c}\n",
    "1 \\\\\n",
    "-1 \\\\\n",
    "3 \\\\\n",
    "\\end{array}\n",
    "\\right ).$\n",
    "\n",
    "b) Argue that the resulting set of three vectors form a basis for for $\\mathbb{R}^3$.\n",
    "\n",
    "c) Express the vector $x = \\left (\n",
    "\\begin{array}{c}\n",
    "1 \\\\\n",
    "2 \\\\\n",
    "3\n",
    "\\end{array}\n",
    "\\right )$ as a linear combination of your three vectors. \n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    1pt for finding a vector w that is orthogonal to v\n",
    "    1pt for finding a vector u that is orthogonal to v\n",
    "    3pt if u,v,w are all mutually orthogonal\n",
    "    3pts (b) for arguing that the set forms a basis (mutually linearly independent, and there are 3 vectors)\n",
    "    2pts (c) for finding the coefficients a1, a2, a3 that represent x in the new basis\n",
    "    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2.a)\n",
    "\n",
    "We're looking for all vectors $w$ that satisfy $v \\cdot w = 0 $ for $x,y,z\\in \\mathbb{R}:$\n",
    "$$0 = \n",
    "    \\begin{pmatrix} 1 & -1 & 3 \\end{pmatrix} \\cdot\n",
    "    \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \n",
    "$$\n",
    "\n",
    "Geometrically, this means vectors in the plane $$x-y+3z = 0.$$\n",
    "\n",
    "Notice that we can reexpress one of these values in terms of the others. For instance\n",
    "\n",
    "$$y = (x+3z)$$\n",
    "so\n",
    "$$\\begin{pmatrix} x \\\\ (x+3z) \\\\ z \\end{pmatrix} = \n",
    "x\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} + \n",
    "z\\begin{pmatrix} 0 \\\\ 3 \\\\ 1 \\end{pmatrix}.$$\n",
    "Therefore, any vector in $$\\text{span}\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 3 \\\\ 1 \\end{pmatrix} \\}$$ will work.\n",
    "\n",
    "We'll just choose $x=z=1$ to get\n",
    "\n",
    "$$w = \\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\end{pmatrix}.$$\n",
    "\n",
    "Next, we need to find a vector $u$ that is orthogonal to both $v,w$. To do this we could procede as follows. From $(x ,y ,z) \\cdot w = 0$ we see that\n",
    "\n",
    "$$0 = x + 4y + z \\implies x = -4y - z$$ and if we set $$x - y +3z = 0 = x + 4y +z$$\n",
    "\n",
    "$$\\implies 0 = 5y - 2z.$$ So let's just pick $y=2$, and $z=5$. Then $x = -4(2) - (5) = -13.$\n",
    "\n",
    "Thus,\n",
    "$$u = \\begin{pmatrix} -13 \\\\ 2 \\\\ 5 \\end{pmatrix}.$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(array([[0]]), array([[0]]), array([[0]]))"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "v = np.array([[1],[-1],[3]])\n",
    "w = np.array([[1],[4],[1]])\n",
    "u = np.array([[-13],[2],[5]]) \n",
    "\n",
    "#Check if orthogonal\n",
    "v.transpose()@w, v.transpose()@u, w.transpose()@u"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**ALTERNATIVELY**\n",
    "\n",
    "We shall use Gram-Schmidt orthogonalization to determine the final vector. As we showed above, any vector in \n",
    "\n",
    "$$\\text{span}\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 3 \\\\ 1 \\end{pmatrix} \\}$$ \n",
    "\n",
    "will be orthogonal to $v$. Let's denote that first vector as $w_1$ and the second vector as $w_2$. To get $u$ we use the following formula:\n",
    "\n",
    "$$u = w_2 - proj_{w_1}(w_2)$$\n",
    "\n",
    "$$= w_2 - \\frac{w_2 \\cdot w_1}{w_1 \\cdot w_1} w_1$$\n",
    "\n",
    "$$ = \\begin{pmatrix} 0 \\\\ 3 \\\\ 1 \\end{pmatrix} - \n",
    "\\frac{\n",
    "\\begin{pmatrix} 0 & 3 & 1 \\end{pmatrix} \n",
    "\\cdot \n",
    "\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}}\n",
    "{\\begin{pmatrix} 1 & 1 & 0 \\end{pmatrix} \n",
    "\\cdot\n",
    "\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}\n",
    "}\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$$\n",
    "\n",
    "$$ = \\begin{pmatrix} 0 \\\\ 3 \\\\ 1 \\end{pmatrix} - \n",
    "\\frac{3}\n",
    "{2}\n",
    "\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$$\n",
    "\n",
    "$$\n",
    "u = \\begin{pmatrix} -\\frac{3}{2} \\\\ \\frac{3}{2} \\\\ 1 \\end{pmatrix}.$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(array([[0]]), array([[0.]]), array([[0.]]))"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "w1 = np.array([[1],[1],[0]]) # 1st vector in spanning set and the 1st orthogonal vector to v we choose\n",
    "\n",
    "w2 = np.array([[0],[3],[1]]) # 2nd vector in spanning set and is NOT orthogonal to u\n",
    "\n",
    "u_GS = w2 - ((w2.transpose() @ w1)[0] / (w1.transpose() @ w1)[0]) * w1\n",
    "\n",
    "v.transpose()@w1, v.transpose()@u_GS, w1.transpose()@u_GS"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2.b)\n",
    "\n",
    "Recall the definition of a basis: \n",
    "\n",
    "A linearly independent set of vectors $\\{v_1,...,v_n\\}$ is a **basis** for a vector space $V$ if $span(v_1,...,v_n) = V$. (i.e. if $v_1,...,v_n$ are mutually orthogonal and $dim(V) = n$). \n",
    "\n",
    "Thus, $\\{u,v,w\\}$ in any order form a basis for $\\mathbb{R}^3$ because they are mutually orthogonal, and there are enough vectors--i.e. we need 3 of such vectors to cover all of $\\mathbb{R}^3$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2.c) \n",
    "\n",
    "Notice that \n",
    "$$x = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} = \n",
    "a_1\\begin{pmatrix} -\\frac{3}{2} \\\\ \\frac{3}{2} \\\\ 1 \\end{pmatrix} + \n",
    "a_2\\begin{pmatrix} 1 \\\\ -1 \\\\ 3 \\end{pmatrix} +\n",
    "a_3\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$$\n",
    "\n",
    "can be reexpressed as $x = Az$\n",
    "\n",
    "$$\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} = \n",
    "\\begin{pmatrix} \n",
    "    -\\frac{3}{2} & 1 & 1 \\\\ \n",
    "     \\frac{3}{2} & -1 & 1 \\\\\n",
    "     1 & 3 & 0\n",
    "\\end{pmatrix}\n",
    "\\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\end{pmatrix}.\n",
    "$$\n",
    "\n",
    "We can solve for the coefficients quickly by running this linear system through a solver."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "a1= 0.8181818181818182 \n",
      "a2= 0.7272727272727273 \n",
      "a3= 1.5\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "array([[1.],\n",
       "       [2.],\n",
       "       [3.]])"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x = np.array([[1],[2],[3]])\n",
    "A = np.hstack([u_GS, v, w1])\n",
    "a1, a2, a3 = np.linalg.solve(A,x) #numerically solve for the coeffecients\n",
    "print('\\na1=', a1[0], '\\na2=', a2[0],'\\na3=', a3[0])\n",
    "\n",
    "a1*u_GS + a2*v + a3*w1 #verify that these add up to (1,2,3)^T"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**More explicitly**\n",
    "\n",
    "We can write out \n",
    "\n",
    "$$x = \\begin{pmatrix} 1 \\\\ 2 \\\\3 \\end{pmatrix} = a_1 u + a_2 v + a_3 w$$\n",
    "\n",
    "and use orthogonality to determine concise formulas for the coefficients. Thus,\n",
    "\n",
    "$$a_1 = \\frac{x \\cdot u}{u \\cdot u} = \\frac{9}{11}$$\n",
    "\n",
    "$$a_2 = \\frac{x \\cdot v}{v \\cdot v} =  \\frac{8}{11}$$\n",
    "\n",
    "$$a_3 = \\frac{x \\cdot w}{w \\cdot w} = \\frac{3}{2}.$$ And our new representation of $x$ in basis $B$ is \n",
    "\n",
    "$$x_{[B]} = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\end{pmatrix}.$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "colab_type": "text",
    "id": "0ooZr0_qKOdi"
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   "source": [
    "### 3. Magnitude (10pts)\n",
    "\n",
    "Recall the set of rotation matrices defined by\n",
    "\n",
    "$$\n",
    "R(\\theta) = \n",
    "\\begin{pmatrix}\n",
    "\\cos \\theta & \\sin \\theta \\\\\n",
    "-\\sin \\theta & \\cos \\theta\n",
    "\\end{pmatrix}.\n",
    "$$\n",
    "\n",
    "for each $\\theta \\in \\mathbb{R}$. Show that if $x \\in \\mathbb{R}^2$ that the magnitude of $x$ is equal to the magnitude of $R(\\theta)x$ for all $\\theta$. Thus, rotations do not affect magnitudes. \n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    3pts for using the definition of magnitude on Rx\n",
    "    4pts for doing the elementwise multiplication OR showing that R is orthogonal\n",
    "    3pts for a convicing argument that ||Rx|| = ||x||.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We begin with the definition of magnitude. Let $\\theta \\in \\mathbb{R},$ and $x\\in \\mathbb{R}^2$. Then we have\n",
    "$$\\begin{align}\n",
    "||R(\\theta)x|| &= \\sqrt{(R(\\theta)x)^T(R(\\theta)x)} \\\\\n",
    "&= \\sqrt{x^TR(\\theta)^TR(\\theta) x} \\\\\n",
    "&= \\sqrt{x^Tx} \\\\\n",
    "&= ||x||\n",
    "\\end{align} \n",
    "$$\n",
    "because\n",
    "$$\n",
    "\\begin{align}\n",
    "R(\\theta)^TR(\\theta) &= \n",
    "\\begin{bmatrix} \\sin^2(\\theta)+\\cos^2(\\theta) & 0 \\\\ 0 & \\sin^2(\\theta)+\\cos^2(\\theta)\\end{bmatrix} \\\\ \n",
    "&= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\text{via trig identities.}\n",
    "\\end{align}\n",
    "$$\n",
    "\n",
    "Thus, rotations do not affect magnitudes."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left[\\begin{matrix}\\sin^{2}{\\left (\\theta \\right )} + \\cos^{2}{\\left (\\theta \\right )} & 0\\\\0 & \\sin^{2}{\\left (\\theta \\right )} + \\cos^{2}{\\left (\\theta \\right )}\\end{matrix}\\right]$$"
      ],
      "text/plain": [
       "⎡   2         2                      ⎤\n",
       "⎢sin (θ) + cos (θ)          0        ⎥\n",
       "⎢                                    ⎥\n",
       "⎢                      2         2   ⎥\n",
       "⎣        0          sin (θ) + cos (θ)⎦"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#In sympy\n",
    "o = sm.symbols('theta')\n",
    "R = sm.Matrix([[sm.cos(o), sm.sin(o)],[-sm.sin(o), sm.cos(o)]])\n",
    "R.transpose()@R "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "colab_type": "text",
    "id": "dUa_52mhKOdj"
   },
   "source": [
    "### 4. Inverses (10pts)\n",
    "\n",
    "Find the inverse of each of the following matrices, or explain why no inverse exists.\n",
    "\n",
    "$\n",
    "A = \\begin{pmatrix}\n",
    "0 & 1 & 1 \\\\\n",
    "1 & 0 & 1 \\\\\n",
    "1 & 1 & 0\n",
    "\\end{pmatrix}\n",
    "$\n",
    "\n",
    "B = $\n",
    "\\begin{pmatrix}\n",
    "0 & 1 & 1 \\\\\n",
    "1 & 0 & 1 \\\\\n",
    "1 & -1 & 0\n",
    "\\end{pmatrix}\n",
    "$\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    5pts for numerically finding A^-1 or providing a derivation that leads to the correct answer\n",
    "    5pts for explaining why B does not have an inverse"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "4.a)\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The determinant is 2.0\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "array([[-0.5,  0.5,  0.5],\n",
       "       [ 0.5, -0.5,  0.5],\n",
       "       [ 0.5,  0.5, -0.5]])"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = np.array([[0,1,1],[1,0,1],[1,1,0]])\n",
    "detA = np.linalg.det(A)\n",
    "print('The determinant is',detA)\n",
    "np.linalg.inv(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "More explicitly, we can use the Cayley Hamilton Theorem to find this inverse. First, let's find the characteristic polynomial.\n",
    "\n",
    "$$\n",
    "det(sI - A) = \\begin{vmatrix}\n",
    "s & -1 & -1 \\\\\n",
    "-1 & s & -1 \\\\\n",
    "-1 & -1 & s\n",
    "\\end{vmatrix}\n",
    "$$\n",
    "\n",
    "$$\n",
    "= s \\begin{vmatrix}\n",
    "s & -1\\\\\n",
    "-1 & s\\\\\n",
    "\\end{vmatrix}\n",
    "-\n",
    "(-1) \\begin{vmatrix}\n",
    "-1 & -1\\\\\n",
    "-1 & s\\\\\n",
    "\\end{vmatrix}\n",
    "+\n",
    "(-1) \\begin{vmatrix}\n",
    "-1 & s\\\\\n",
    "-1 & -1\\\\\n",
    "\\end{vmatrix}\n",
    "$$\n",
    "\n",
    "$$ = s^3 - 3s - 2.$$\n",
    "\n",
    "Now, using C.H. we can say\n",
    "\n",
    "$$p(A) = A^3 - 3A -2I = 0$$\n",
    "\n",
    "and multiplyin both sides by $A^{-1}$ implies\n",
    "\n",
    "$$\\begin{align}\n",
    "A^{-1} &= \\frac{1}{2} (A^2 - 3I)\\\\\n",
    "&= \n",
    "\\frac{1}{2}\n",
    "\\begin{bmatrix}\n",
    "-1 & 1 & 1 \\\\\n",
    "1 & -1 & 1 \\\\\n",
    "1 & 1 & -1\n",
    "\\end{bmatrix}.\n",
    "\\end{align}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\operatorname{PurePoly}{\\left( \\lambda^{3} - 3 \\lambda - 2, \\lambda, domain=\\mathbb{Z} \\right)}, \\quad \\left[\\begin{matrix}- \\frac{1}{2} & \\frac{1}{2} & \\frac{1}{2}\\\\\\frac{1}{2} & - \\frac{1}{2} & \\frac{1}{2}\\\\\\frac{1}{2} & \\frac{1}{2} & - \\frac{1}{2}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{1}{2} & \\frac{1}{2} & \\frac{1}{2}\\\\\\frac{1}{2} & - \\frac{1}{2} & \\frac{1}{2}\\\\\\frac{1}{2} & \\frac{1}{2} & - \\frac{1}{2}\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛                                                            ⎡-1/2  1/2   1/2 \n",
       "⎜                                                            ⎢                \n",
       "⎜PurePoly(_lambda**3 - 3*_lambda - 2, _lambda, domain='ZZ'), ⎢1/2   -1/2  1/2 \n",
       "⎜                                                            ⎢                \n",
       "⎝                                                            ⎣1/2   1/2   -1/2\n",
       "\n",
       "⎤  ⎡-1/2  1/2   1/2 ⎤⎞\n",
       "⎥  ⎢                ⎥⎟\n",
       "⎥, ⎢1/2   -1/2  1/2 ⎥⎟\n",
       "⎥  ⎢                ⎥⎟\n",
       "⎦  ⎣1/2   1/2   -1/2⎦⎠"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#Let's check in sympy\n",
    "A = sm.Matrix([[0,1,1],[1,0,1],[1,1,0]])\n",
    "my_Ainv = sm.Rational(1,2) * (A**2 - 3*sm.eye(3))\n",
    "A.charpoly(), A.inv(), my_Ainv"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "4.b)\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The determinant is 0.0 therefore B is a singular matrix, and has no inverse.\n"
     ]
    }
   ],
   "source": [
    "B = np.array([[0,1,1],[1,0,-1],[1,1,0]])\n",
    "detB = np.linalg.det(B)\n",
    "try:\n",
    "    np.linalg.inv(B)\n",
    "except:\n",
    "    print('The determinant is',detB,'therefore B is a singular matrix, and has no inverse.')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "colab_type": "text",
    "id": "g2-gAVVGKOdl"
   },
   "source": [
    "### 5. Commutativity (10pts)\n",
    "\n",
    "In three dimensions, there are three different kinds of rotations, one for rotation about each axis. These are defined as follows:\n",
    "\n",
    "$$\n",
    "R_x(\\theta) = \n",
    "\\begin{pmatrix}\n",
    "1 & 0 & 0 \\\\\n",
    "0 & \\cos \\theta  & \\sin \\theta  \\\\\n",
    "0 & -\\sin \\theta & \\cos \\theta \n",
    "\\end{pmatrix}.\n",
    "$$\n",
    "\n",
    "$$\n",
    "R_z(\\theta) = \n",
    "\\begin{pmatrix}\n",
    "\\cos \\theta  & 0 & \\sin \\theta \\\\\n",
    "0 & 1 & 0 \\\\\n",
    "-\\sin \\theta & 0 & \\cos \\theta \n",
    "\\end{pmatrix}.\n",
    "$$\n",
    "\n",
    "$$\n",
    "R_z(\\theta) = \n",
    "\\begin{pmatrix}\n",
    "\\cos \\theta  & \\sin \\theta & 0 \\\\\n",
    "-\\sin \\theta & \\cos \\theta & 0 \\\\\n",
    "0 & 0 & 1\n",
    "\\end{pmatrix}.\n",
    "$$\n",
    "\n",
    "Are these matrices pairwise commutative in general? That is: Is it the case that\n",
    "\n",
    "$$\n",
    "R_x(\\theta_1)R_y(\\theta_2) = R_y(\\theta_2)R_x(\\theta_1)\n",
    "$$\n",
    "\n",
    "and similarly for $x,z$ and $y,z$? Explain your answer.\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    2.5pts for finding Rx1 @ Ry2 and Ry2 @ Rx1\n",
    "    2.5pts for finding Rx1 @ Rz2 and Rz2 @ Rx1\n",
    "    2.5pts for finding Ry1 @ Rz2 and Rz2 @ Ry1\n",
    "    2.5pts for a convincing explaination (e.g. could be physically or mathematically motivated)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "5)\n",
    "\n",
    "In general, rotation matrices **do not** commute. We use sympy to show the calculations below and verify by inspection."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}\\cos{\\left (\\theta_{2} \\right )} & 0 & \\sin{\\left (\\theta_{2} \\right )}\\\\- \\sin{\\left (\\theta_{1} \\right )} \\sin{\\left (\\theta_{2} \\right )} & \\cos{\\left (\\theta_{1} \\right )} & \\sin{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )}\\\\- \\sin{\\left (\\theta_{2} \\right )} \\cos{\\left (\\theta_{1} \\right )} & - \\sin{\\left (\\theta_{1} \\right )} & \\cos{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}\\cos{\\left (\\theta_{2} \\right )} & - \\sin{\\left (\\theta_{1} \\right )} \\sin{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{2} \\right )} \\cos{\\left (\\theta_{1} \\right )}\\\\0 & \\cos{\\left (\\theta_{1} \\right )} & \\sin{\\left (\\theta_{1} \\right )}\\\\- \\sin{\\left (\\theta_{2} \\right )} & - \\sin{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )} & \\cos{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )}\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡    cos(θ₂)          0          sin(θ₂)    ⎤  ⎡cos(θ₂)   -sin(θ₁)⋅sin(θ₂)  s\n",
       "⎜⎢                                           ⎥  ⎢                             \n",
       "⎜⎢-sin(θ₁)⋅sin(θ₂)  cos(θ₁)   sin(θ₁)⋅cos(θ₂)⎥, ⎢   0          cos(θ₁)        \n",
       "⎜⎢                                           ⎥  ⎢                             \n",
       "⎝⎣-sin(θ₂)⋅cos(θ₁)  -sin(θ₁)  cos(θ₁)⋅cos(θ₂)⎦  ⎣-sin(θ₂)  -sin(θ₁)⋅cos(θ₂)  c\n",
       "\n",
       "in(θ₂)⋅cos(θ₁)⎤⎞\n",
       "              ⎥⎟\n",
       "   sin(θ₁)    ⎥⎟\n",
       "              ⎥⎟\n",
       "os(θ₁)⋅cos(θ₂)⎦⎠"
      ]
     },
     "execution_count": 56,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "o1, o2 = sm.symbols('theta1 theta2')\n",
    "\n",
    "Rx1 = sm.Matrix([[1, 0, 0],[0, sm.cos(o1), sm.sin(o1)],[0, -sm.sin(o1), sm.cos(o1)]])\n",
    "Ry1 = sm.Matrix([[sm.cos(o1), 0, sm.sin(o1)],[0, 1, 0],[-sm.sin(o1), 0, sm.cos(o1)]])\n",
    "\n",
    "Ry2 = sm.Matrix([[sm.cos(o2), 0, sm.sin(o2)],[0, 1, 0],[-sm.sin(o2), 0, sm.cos(o2)]])\n",
    "Rz2 = sm.Matrix([[sm.cos(o2), sm.sin(o2), 0],[-sm.sin(o2), sm.cos(o2), 0],[0, 0, 1]])\n",
    "\n",
    "#Let's check the first pair\n",
    "Rx1@Ry2, Ry2@Rx1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}\\cos{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{2} \\right )} & 0\\\\- \\sin{\\left (\\theta_{2} \\right )} \\cos{\\left (\\theta_{1} \\right )} & \\cos{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{1} \\right )}\\\\\\sin{\\left (\\theta_{1} \\right )} \\sin{\\left (\\theta_{2} \\right )} & - \\sin{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )} & \\cos{\\left (\\theta_{1} \\right )}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}\\cos{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{2} \\right )} \\cos{\\left (\\theta_{1} \\right )} & \\sin{\\left (\\theta_{1} \\right )} \\sin{\\left (\\theta_{2} \\right )}\\\\- \\sin{\\left (\\theta_{2} \\right )} & \\cos{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )}\\\\0 & - \\sin{\\left (\\theta_{1} \\right )} & \\cos{\\left (\\theta_{1} \\right )}\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡    cos(θ₂)           sin(θ₂)          0   ⎤  ⎡cos(θ₂)   sin(θ₂)⋅cos(θ₁)  si\n",
       "⎜⎢                                           ⎥  ⎢                             \n",
       "⎜⎢-sin(θ₂)⋅cos(θ₁)  cos(θ₁)⋅cos(θ₂)   sin(θ₁)⎥, ⎢-sin(θ₂)  cos(θ₁)⋅cos(θ₂)  si\n",
       "⎜⎢                                           ⎥  ⎢                             \n",
       "⎝⎣sin(θ₁)⋅sin(θ₂)   -sin(θ₁)⋅cos(θ₂)  cos(θ₁)⎦  ⎣   0         -sin(θ₁)        \n",
       "\n",
       "n(θ₁)⋅sin(θ₂)⎤⎞\n",
       "             ⎥⎟\n",
       "n(θ₁)⋅cos(θ₂)⎥⎟\n",
       "             ⎥⎟\n",
       "  cos(θ₁)    ⎦⎠"
      ]
     },
     "execution_count": 57,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#Let's check the 2nd pair\n",
    "Rx1@Rz2, Rz2@Rx1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 58,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}\\cos{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{2} \\right )} \\cos{\\left (\\theta_{1} \\right )} & \\sin{\\left (\\theta_{1} \\right )}\\\\- \\sin{\\left (\\theta_{2} \\right )} & \\cos{\\left (\\theta_{2} \\right )} & 0\\\\- \\sin{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )} & - \\sin{\\left (\\theta_{1} \\right )} \\sin{\\left (\\theta_{2} \\right )} & \\cos{\\left (\\theta_{1} \\right )}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}\\cos{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{2} \\right )} & \\sin{\\left (\\theta_{1} \\right )} \\cos{\\left (\\theta_{2} \\right )}\\\\- \\sin{\\left (\\theta_{2} \\right )} \\cos{\\left (\\theta_{1} \\right )} & \\cos{\\left (\\theta_{2} \\right )} & - \\sin{\\left (\\theta_{1} \\right )} \\sin{\\left (\\theta_{2} \\right )}\\\\- \\sin{\\left (\\theta_{1} \\right )} & 0 & \\cos{\\left (\\theta_{1} \\right )}\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡cos(θ₁)⋅cos(θ₂)   sin(θ₂)⋅cos(θ₁)   sin(θ₁)⎤  ⎡cos(θ₁)⋅cos(θ₂)   sin(θ₂)  si\n",
       "⎜⎢                                           ⎥  ⎢                             \n",
       "⎜⎢    -sin(θ₂)          cos(θ₂)          0   ⎥, ⎢-sin(θ₂)⋅cos(θ₁)  cos(θ₂)  -s\n",
       "⎜⎢                                           ⎥  ⎢                             \n",
       "⎝⎣-sin(θ₁)⋅cos(θ₂)  -sin(θ₁)⋅sin(θ₂)  cos(θ₁)⎦  ⎣    -sin(θ₁)         0       \n",
       "\n",
       "n(θ₁)⋅cos(θ₂) ⎤⎞\n",
       "              ⎥⎟\n",
       "in(θ₁)⋅sin(θ₂)⎥⎟\n",
       "              ⎥⎟\n",
       "  cos(θ₁)     ⎦⎠"
      ]
     },
     "execution_count": 58,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#Let's check the 3rd pair \n",
    "Ry1@Rz2, Rz2@Ry1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "One explaination:\n",
    "\n",
    "The multiplication of rotation matrices corresponds to a composition of the rotations, applied from left to right. Compositions of rotations characterized by different axes of rotation don't share eigenvectors. Thus, they don't preserve each others' eigenspaces, and cannot commute. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 64,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left [ \\left ( 1, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}1\\\\0\\\\0\\end{matrix}\\right]\\right ]\\right ), \\quad \\left ( - \\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1} + \\cos{\\left (\\theta_{1} \\right )}, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}0\\\\- \\frac{\\sin{\\left (\\theta_{1} \\right )}}{\\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1}}\\\\1\\end{matrix}\\right]\\right ]\\right ), \\quad \\left ( \\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1} + \\cos{\\left (\\theta_{1} \\right )}, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}0\\\\\\frac{\\sin{\\left (\\theta_{1} \\right )}}{\\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1}}\\\\1\\end{matrix}\\right]\\right ]\\right )\\right ]$$"
      ],
      "text/plain": [
       "⎡               ⎛                                  ⎡⎡        0        ⎤⎤⎞  ⎛  \n",
       "⎢               ⎜                                  ⎢⎢                 ⎥⎥⎟  ⎜  \n",
       "⎢⎛      ⎡⎡1⎤⎤⎞  ⎜     ______________               ⎢⎢    -sin(θ₁)     ⎥⎥⎟  ⎜  \n",
       "⎢⎜      ⎢⎢ ⎥⎥⎟  ⎜    ╱    2                        ⎢⎢─────────────────⎥⎥⎟  ⎜  \n",
       "⎢⎜1, 1, ⎢⎢0⎥⎥⎟, ⎜- ╲╱  cos (θ₁) - 1  + cos(θ₁), 1, ⎢⎢   ______________⎥⎥⎟, ⎜╲╱\n",
       "⎢⎜      ⎢⎢ ⎥⎥⎟  ⎜                                  ⎢⎢  ╱    2         ⎥⎥⎟  ⎜  \n",
       "⎢⎝      ⎣⎣0⎦⎦⎠  ⎜                                  ⎢⎢╲╱  cos (θ₁) - 1 ⎥⎥⎟  ⎜  \n",
       "⎢               ⎜                                  ⎢⎢                 ⎥⎥⎟  ⎜  \n",
       "⎣               ⎝                                  ⎣⎣        1        ⎦⎦⎠  ⎝  \n",
       "\n",
       "                              ⎡⎡        0        ⎤⎤⎞⎤\n",
       "                              ⎢⎢                 ⎥⎥⎟⎥\n",
       " ______________               ⎢⎢     sin(θ₁)     ⎥⎥⎟⎥\n",
       "╱    2                        ⎢⎢─────────────────⎥⎥⎟⎥\n",
       "  cos (θ₁) - 1  + cos(θ₁), 1, ⎢⎢   ______________⎥⎥⎟⎥\n",
       "                              ⎢⎢  ╱    2         ⎥⎥⎟⎥\n",
       "                              ⎢⎢╲╱  cos (θ₁) - 1 ⎥⎥⎟⎥\n",
       "                              ⎢⎢                 ⎥⎥⎟⎥\n",
       "                              ⎣⎣        1        ⎦⎦⎠⎦"
      ]
     },
     "execution_count": 64,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Rx1.eigenvects()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left [ \\left ( 1, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}0\\\\1\\\\0\\end{matrix}\\right]\\right ]\\right ), \\quad \\left ( - \\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1} + \\cos{\\left (\\theta_{1} \\right )}, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}- \\frac{\\sin{\\left (\\theta_{1} \\right )}}{\\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1}}\\\\0\\\\1\\end{matrix}\\right]\\right ]\\right ), \\quad \\left ( \\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1} + \\cos{\\left (\\theta_{1} \\right )}, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}\\frac{\\sin{\\left (\\theta_{1} \\right )}}{\\sqrt{\\cos^{2}{\\left (\\theta_{1} \\right )} - 1}}\\\\0\\\\1\\end{matrix}\\right]\\right ]\\right )\\right ]$$"
      ],
      "text/plain": [
       "⎡               ⎛                                  ⎡⎡    -sin(θ₁)     ⎤⎤⎞  ⎛  \n",
       "⎢               ⎜                                  ⎢⎢─────────────────⎥⎥⎟  ⎜  \n",
       "⎢⎛      ⎡⎡0⎤⎤⎞  ⎜     ______________               ⎢⎢   ______________⎥⎥⎟  ⎜  \n",
       "⎢⎜      ⎢⎢ ⎥⎥⎟  ⎜    ╱    2                        ⎢⎢  ╱    2         ⎥⎥⎟  ⎜  \n",
       "⎢⎜1, 1, ⎢⎢1⎥⎥⎟, ⎜- ╲╱  cos (θ₁) - 1  + cos(θ₁), 1, ⎢⎢╲╱  cos (θ₁) - 1 ⎥⎥⎟, ⎜╲╱\n",
       "⎢⎜      ⎢⎢ ⎥⎥⎟  ⎜                                  ⎢⎢                 ⎥⎥⎟  ⎜  \n",
       "⎢⎝      ⎣⎣0⎦⎦⎠  ⎜                                  ⎢⎢        0        ⎥⎥⎟  ⎜  \n",
       "⎢               ⎜                                  ⎢⎢                 ⎥⎥⎟  ⎜  \n",
       "⎣               ⎝                                  ⎣⎣        1        ⎦⎦⎠  ⎝  \n",
       "\n",
       "                              ⎡⎡     sin(θ₁)     ⎤⎤⎞⎤\n",
       "                              ⎢⎢─────────────────⎥⎥⎟⎥\n",
       " ______________               ⎢⎢   ______________⎥⎥⎟⎥\n",
       "╱    2                        ⎢⎢  ╱    2         ⎥⎥⎟⎥\n",
       "  cos (θ₁) - 1  + cos(θ₁), 1, ⎢⎢╲╱  cos (θ₁) - 1 ⎥⎥⎟⎥\n",
       "                              ⎢⎢                 ⎥⎥⎟⎥\n",
       "                              ⎢⎢        0        ⎥⎥⎟⎥\n",
       "                              ⎢⎢                 ⎥⎥⎟⎥\n",
       "                              ⎣⎣        1        ⎦⎦⎠⎦"
      ]
     },
     "execution_count": 49,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ry1.eigenvects()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left [ \\left ( 1, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}0\\\\0\\\\1\\end{matrix}\\right]\\right ]\\right ), \\quad \\left ( - \\sqrt{\\cos^{2}{\\left (\\theta_{2} \\right )} - 1} + \\cos{\\left (\\theta_{2} \\right )}, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}- \\frac{\\sin{\\left (\\theta_{2} \\right )}}{\\sqrt{\\cos^{2}{\\left (\\theta_{2} \\right )} - 1}}\\\\1\\\\0\\end{matrix}\\right]\\right ]\\right ), \\quad \\left ( \\sqrt{\\cos^{2}{\\left (\\theta_{2} \\right )} - 1} + \\cos{\\left (\\theta_{2} \\right )}, \\quad 1, \\quad \\left [ \\left[\\begin{matrix}\\frac{\\sin{\\left (\\theta_{2} \\right )}}{\\sqrt{\\cos^{2}{\\left (\\theta_{2} \\right )} - 1}}\\\\1\\\\0\\end{matrix}\\right]\\right ]\\right )\\right ]$$"
      ],
      "text/plain": [
       "⎡               ⎛                                  ⎡⎡    -sin(θ₂)     ⎤⎤⎞  ⎛  \n",
       "⎢               ⎜                                  ⎢⎢─────────────────⎥⎥⎟  ⎜  \n",
       "⎢⎛      ⎡⎡0⎤⎤⎞  ⎜     ______________               ⎢⎢   ______________⎥⎥⎟  ⎜  \n",
       "⎢⎜      ⎢⎢ ⎥⎥⎟  ⎜    ╱    2                        ⎢⎢  ╱    2         ⎥⎥⎟  ⎜  \n",
       "⎢⎜1, 1, ⎢⎢0⎥⎥⎟, ⎜- ╲╱  cos (θ₂) - 1  + cos(θ₂), 1, ⎢⎢╲╱  cos (θ₂) - 1 ⎥⎥⎟, ⎜╲╱\n",
       "⎢⎜      ⎢⎢ ⎥⎥⎟  ⎜                                  ⎢⎢                 ⎥⎥⎟  ⎜  \n",
       "⎢⎝      ⎣⎣1⎦⎦⎠  ⎜                                  ⎢⎢        1        ⎥⎥⎟  ⎜  \n",
       "⎢               ⎜                                  ⎢⎢                 ⎥⎥⎟  ⎜  \n",
       "⎣               ⎝                                  ⎣⎣        0        ⎦⎦⎠  ⎝  \n",
       "\n",
       "                              ⎡⎡     sin(θ₂)     ⎤⎤⎞⎤\n",
       "                              ⎢⎢─────────────────⎥⎥⎟⎥\n",
       " ______________               ⎢⎢   ______________⎥⎥⎟⎥\n",
       "╱    2                        ⎢⎢  ╱    2         ⎥⎥⎟⎥\n",
       "  cos (θ₂) - 1  + cos(θ₂), 1, ⎢⎢╲╱  cos (θ₂) - 1 ⎥⎥⎟⎥\n",
       "                              ⎢⎢                 ⎥⎥⎟⎥\n",
       "                              ⎢⎢        1        ⎥⎥⎟⎥\n",
       "                              ⎢⎢                 ⎥⎥⎟⎥\n",
       "                              ⎣⎣        0        ⎦⎦⎠⎦"
      ]
     },
     "execution_count": 50,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Rz2.eigenvects()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "colab_type": "text",
    "id": "pPw5Epy9KOdm"
   },
   "source": [
    "### 6. Orthogonal Matrices (10pts)\n",
    "\n",
    "The set of reflection matrices is defined by\n",
    "\n",
    "$$\n",
    "S(\\theta) = \n",
    "\\begin{pmatrix}\n",
    "\\cos 2\\theta & \\sin 2 \\theta \\\\\n",
    "\\sin 2\\theta & -\\cos 2 \\theta\n",
    "\\end{pmatrix}.\n",
    "$$\n",
    "\n",
    "a) Show how $S(\\frac{\\pi}{4})$ transforms the vector $(1\\;-1)^T$. \n",
    "\n",
    "b) Show that $S(\\theta)$ does not change any vectors that happen to lie on the axis of reflection.\n",
    "\n",
    "c) Show that these matrices are orthogonal for all $\\theta$. \n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    2pts for showing how the vector is transformed\n",
    "    2pts (b) for choosing v(theta) = [cos(theta) sin(theta)]^T\n",
    "    1pt (b) for multiplying by S(theta)\n",
    "    2pts (b) for simplifying the result to equal v(theta)\n",
    "    3pts (c) for for a convincing argument to show that these matrices are orthogonal for all theta\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}1\\\\-1\\end{matrix}\\right], \\quad \\left[\\begin{matrix}0 & 1\\\\1 & 0\\end{matrix}\\right], \\quad \\left[\\begin{matrix}-1\\\\1\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡1 ⎤  ⎡0  1⎤  ⎡-1⎤⎞\n",
       "⎜⎢  ⎥, ⎢    ⎥, ⎢  ⎥⎟\n",
       "⎝⎣-1⎦  ⎣1  0⎦  ⎣1 ⎦⎠"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#6.a) First let's see what's happening in python\n",
    "\n",
    "S = lambda arg: sm.Matrix([[sm.cos(2*arg), sm.sin(2*arg)],[sm.sin(2*arg), -sm.cos(2*arg)]])\n",
    "v = sm.Matrix([[1],[-1]])\n",
    "v, S(sm.pi / 4), S(sm.pi / 4)@v"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now, more formally, let's apply $S(\\frac{\\pi}{4})$ to $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ to see what happens:\n",
    "\n",
    "$$\n",
    "\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\n",
    "\\begin{pmatrix} x \\\\ y \\end{pmatrix}\n",
    "= \n",
    "(x)\\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} +\n",
    "(y)\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}\n",
    " = \\begin{pmatrix} y \\\\ x \\end{pmatrix}$$\n",
    " \n",
    "So that means $x\\leftarrow y$ and $y \\leftarrow x$ (i.e. they swap positions). Graphically, this indicates a reflection about the line $L$ represented by the equation $y=x$ (i.e. $\\theta = \\frac{\\pi}{4}$)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [],
   "source": [
    "xs = np.linspace(-2,2,100)\n",
    "ys = xs\n",
    "\n",
    "S45 = np.array([[0,1],[1,0]]) # Reflection matrix\n",
    "p = np.array([[1],[-1]]) # Given point in R^2\n",
    "p2 = np.array([[1],[1]]) # Point on the axis of reflection\n",
    "Spx, Spy = S45@p\n",
    "Spx2, Spy2 = S45@p2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x2a2633dd8d0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "fig, ax = plt.subplots(1,1, figsize=(6,6))\n",
    "ax.plot(xs, ys, c='k')\n",
    "ax.scatter(p[0], p[1])\n",
    "ax.scatter(Spx, Spy)\n",
    "ax.scatter(p2[0], p2[1], alpha=0.4)\n",
    "ax.scatter(Spx2, Spy2, alpha=0.5)\n",
    "ax.grid('on')\n",
    "ax.legend(['Axis of Reflection','Point1','Reflected Point1','Point2','Reflected Point2'])\n",
    "ax.set_aspect('equal')\n",
    "\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "6.b)\n",
    "\n",
    "To show that $S(\\theta)$  does not change any vectors that happen to lie on the axis of reflection, we consider the transformation on the vector $v(\\theta) = \\begin{pmatrix} \\cos{(\\theta)} & \\sin{(\\theta)} \\end{pmatrix}^T$, i.e. a vector on the axis of rotation, shown in `sympy` below."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left[\\begin{matrix}\\sin{\\left (\\theta \\right )} \\sin{\\left (2 \\theta \\right )} + \\cos{\\left (\\theta \\right )} \\cos{\\left (2 \\theta \\right )}\\\\- \\sin{\\left (\\theta \\right )} \\cos{\\left (2 \\theta \\right )} + \\sin{\\left (2 \\theta \\right )} \\cos{\\left (\\theta \\right )}\\end{matrix}\\right]$$"
      ],
      "text/plain": [
       "⎡sin(θ)⋅sin(2⋅θ) + cos(θ)⋅cos(2⋅θ) ⎤\n",
       "⎢                                  ⎥\n",
       "⎣-sin(θ)⋅cos(2⋅θ) + sin(2⋅θ)⋅cos(θ)⎦"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "o = sm.symbols('theta')\n",
    "S_theta = sm.Matrix([[sm.cos(2*o), sm.sin(2*o)],[sm.sin(2*o), -sm.cos(2*o)]])\n",
    "vec = sm.Matrix([[sm.cos(o)],[sm.sin(o)]])\n",
    "S_theta @ vec"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Next, we make use of the product identities and even and odd properties of trig functions:\n",
    "\n",
    "$$\\sin{(\\theta)}\\sin{(2\\theta)} = \\frac{1}{2}(\\cos{(\\theta)} - \\cos{(3\\theta)}) $$\n",
    "\n",
    "$$\\cos{(\\theta)}\\cos{(2\\theta)} = \\frac{1}{2}(\\cos{(3\\theta)} + \\cos{(\\theta)}) $$\n",
    "\n",
    "$$\\sin{(\\theta)}\\cos{(2\\theta)} = \\frac{1}{2}(\\sin{(3\\theta)} - \\sin{(\\theta)}) $$\n",
    "\n",
    "$$\\sin{(2\\theta)}\\cos{(\\theta)} = \\frac{1}{2}(\\sin{(3\\theta)} + \\sin{(\\theta)}) $$\n",
    "\n",
    "So we see\n",
    "\n",
    "$$\n",
    "S(\\theta)v(\\theta) = \n",
    "    \\begin{pmatrix}\n",
    "        \\cos 2\\theta & \\sin 2 \\theta \\\\\n",
    "        \\sin 2\\theta & -\\cos 2 \\theta\n",
    "    \\end{pmatrix}\n",
    "    \\begin{pmatrix} \\cos{(\\theta)} \\\\ \\sin{(\\theta)} \\end{pmatrix}\n",
    "$$    \n",
    "\n",
    "$$\n",
    "= \n",
    "    \\begin{pmatrix}\n",
    "        \\sin{(\\theta)}\\sin{(2\\theta)} + \\cos{(\\theta)}\\cos{(2 \\theta)} \\\\\n",
    "        -\\sin{(\\theta)}\\cos{(2\\theta)} + \\sin{(2\\theta)}\\cos{(\\theta)}\n",
    "    \\end{pmatrix}\n",
    "$$\n",
    "\n",
    "$$\n",
    "= \n",
    "    \\begin{pmatrix}\n",
    "        \\frac{1}{2} (\\cos{(\\theta)} - \\cos{(3\\theta)}) + \\frac{1}{2} (\\cos{(3\\theta)} + \\cos{(\\theta)} )\\\\\n",
    "        \\frac{1}{2} (- \\sin{(3\\theta)} + \\sin{(\\theta)} ) + \\frac{1}{2} ( \\sin{(3\\theta)} + \\sin{(\\theta)} )\n",
    "    \\end{pmatrix}\n",
    "$$\n",
    "\n",
    "$$\n",
    "= \\begin{pmatrix} \\cos{(\\theta)} \\\\ \\sin{(\\theta)} \\end{pmatrix}\n",
    "$$\n",
    "\n",
    "$$=v(\\theta).$$\n",
    "\n",
    "Thus, $S(\\theta)$  does not change any vectors that happen to lie on the axis of reflection."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "6.c)\n",
    "\n",
    "To show that these matrices are orthogonal for all $\\theta$ we try to verify a definition:  **a matrix is orthogonal if its transpose is equal to its inverse.** Note that $1 = \\sin^2{(x)} + \\cos^2{(x)}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left[\\begin{matrix}\\sin^{2}{\\left (2 \\theta \\right )} + \\cos^{2}{\\left (2 \\theta \\right )} & 0\\\\0 & \\sin^{2}{\\left (2 \\theta \\right )} + \\cos^{2}{\\left (2 \\theta \\right )}\\end{matrix}\\right]$$"
      ],
      "text/plain": [
       "⎡   2           2                            ⎤\n",
       "⎢sin (2⋅θ) + cos (2⋅θ)            0          ⎥\n",
       "⎢                                            ⎥\n",
       "⎢                          2           2     ⎥\n",
       "⎣          0            sin (2⋅θ) + cos (2⋅θ)⎦"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "o = sm.symbols('theta')\n",
    "S_theta = sm.Matrix([[sm.cos(2*o), sm.sin(2*o)],[sm.sin(2*o), -sm.cos(2*o)]])\n",
    "S_theta.transpose() @ S_theta"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Thus, regardless of the $\\theta$ we choose, $S(\\theta)$ is orthogonal."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2019-10-23T18:01:28.779050Z",
     "start_time": "2019-10-23T18:01:28.773729Z"
    },
    "colab_type": "text",
    "id": "aNIRp4-uKOdm"
   },
   "source": [
    "### 7. Similarity (10pts)\n",
    "\n",
    "Show that the following two matrices are similar.\n",
    "\n",
    "$\n",
    "A = \\begin{pmatrix}\n",
    "2 & 4 & 6 \\\\\n",
    "0 & 2 & 0 \\\\\n",
    "1 & 1 & 0\n",
    "\\end{pmatrix}\n",
    "$\n",
    "\n",
    "B = $\n",
    "\\begin{pmatrix}\n",
    "1 & -1 & 2 \\\\\n",
    "4 & 2 & 1 \\\\\n",
    "5 & 1 & 1\n",
    "\\end{pmatrix}\n",
    "$\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    3pt for finding the characteristic polynomial of A, B and showing they are the same\n",
    "    4pt for showing that the eigenvalues are the same\n",
    "    3pts for showing that the determinants are the same\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "7)\n",
    "\n",
    "Remember from lecture the definition and properties of similarity:\n",
    "\n",
    "**Def:** $X$ and $Y$ are **similar** if there exists an invertible matrix $Q$ such that $Y = Q X Q^{-1}$. \n",
    "\n",
    "**Some (of many) properties (i.e. necessary conditions):**\n",
    "\n",
    "- Similar matrices have the same determinant. $|B| = |Q A Q^{-1}| = |Q||Q^{-1}||A| = |A|$. \n",
    "- Similar matrices have the same characteristic polynomial. \n",
    "- Similar matrices have the same eigenvalues.\n",
    "- $\\dots$\n",
    "\n",
    "We will use the **necessary and sufficient condition** that:\n",
    "\n",
    "$A$ and $B$ are similar $\\iff$ they have the same Jordan Normal Form."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}2 & 0 & 0\\\\0 & 1 + \\sqrt{7} & 0\\\\0 & 0 & - \\sqrt{7} + 1\\end{matrix}\\right], \\quad \\left[\\begin{matrix}2 & 0 & 0\\\\0 & 1 + \\sqrt{7} & 0\\\\0 & 0 & - \\sqrt{7} + 1\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡2    0        0   ⎤  ⎡2    0        0   ⎤⎞\n",
       "⎜⎢                  ⎥  ⎢                  ⎥⎟\n",
       "⎜⎢0  1 + √7     0   ⎥, ⎢0  1 + √7     0   ⎥⎟\n",
       "⎜⎢                  ⎥  ⎢                  ⎥⎟\n",
       "⎝⎣0    0     -√7 + 1⎦  ⎣0    0     -√7 + 1⎦⎠"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Pa, Ja = A.jordan_form()\n",
    "Pb, Jb = B.jordan_form()\n",
    "Ja, Jb"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Below, we illustrate that the neccesary conditions mentioned in the properties hold."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2 == 2 is True\n",
      "-sqrt(7) + 1 == -sqrt(7) + 1 is True\n",
      "1 + sqrt(7) == 1 + sqrt(7) is True\n",
      "\n",
      "Are the determinants of A and B equal? Answer: True\n"
     ]
    }
   ],
   "source": [
    "A = sm.Matrix([[2,4,6],[0,2,0],[1,1,0]])\n",
    "B = sm.Matrix([[1,-1,2],[4,2,1],[5,1,1]])\n",
    "\n",
    "for a,b in zip(A.eigenvals(), B.eigenvals()):\n",
    "    print(a, '==', b, 'is', a == b)\n",
    "    \n",
    "print('\\nAre the determinants of A and B equal? Answer:', A.det() == B.det())"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( s \\left(s - 2\\right)^{2} - 6 s + 12, \\quad s \\left(s - 2\\right)^{2} - 6 s + 12\\right )$$"
      ],
      "text/plain": [
       "⎛         2                      2           ⎞\n",
       "⎝s⋅(s - 2)  - 6⋅s + 12, s⋅(s - 2)  - 6⋅s + 12⎠"
      ]
     },
     "execution_count": 21,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# Now we check if A and B have the same characteristic equations\n",
    "s = sm.symbols('s')\n",
    "\n",
    "sI = s * sm.eye(3)\n",
    "charEqA = (sI - A).det()\n",
    "charEqB = (sI - A).det()\n",
    "\n",
    "charEqA, charEqB"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "colab_type": "text",
    "id": "uL7YVvnYKOdn"
   },
   "source": [
    "### 8. Diagonalization (10pts)\n",
    "\n",
    "Diagonalize the matrices in problem 7 (you can use `sympy`) and show they have the same diagonal form.\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    5pts for finding the diagonalization of A\n",
    "    5pts for finding the diagonalization of B"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "8)\n",
    "\n",
    "Because `diagonalize` returns a tuple $(P,D)$, where $D$ is diagonal and $M=PDP^{−1}$, we can implement this easily in python. Notice that while $P_A,P_B$ are different, $D_A,D_B$ are the same!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}7 & 1 + \\sqrt{7} & - \\sqrt{7} + 1\\\\-3 & 0 & 0\\\\2 & 1 & 1\\end{matrix}\\right], \\quad \\left[\\begin{matrix}2 & 0 & 0\\\\0 & 1 + \\sqrt{7} & 0\\\\0 & 0 & - \\sqrt{7} + 1\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡7   1 + √7  -√7 + 1⎤  ⎡2    0        0   ⎤⎞\n",
       "⎜⎢                   ⎥  ⎢                  ⎥⎟\n",
       "⎜⎢-3    0        0   ⎥, ⎢0  1 + √7     0   ⎥⎟\n",
       "⎜⎢                   ⎥  ⎢                  ⎥⎟\n",
       "⎝⎣2     1        1   ⎦  ⎣0    0     -√7 + 1⎦⎠"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "P_A, D_A = A.diagonalize()\n",
    "P_A, D_A"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}-1 & - \\frac{1}{6} + \\frac{\\sqrt{7}}{6} & - \\frac{\\sqrt{7}}{6} - \\frac{1}{6}\\\\9 & \\frac{\\sqrt{7}}{6} + \\frac{5}{6} & - \\frac{\\sqrt{7}}{6} + \\frac{5}{6}\\\\4 & 1 & 1\\end{matrix}\\right], \\quad \\left[\\begin{matrix}2 & 0 & 0\\\\0 & 1 + \\sqrt{7} & 0\\\\0 & 0 & - \\sqrt{7} + 1\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡      1   √7    √7   1⎤                      ⎞\n",
       "⎜⎢-1  - ─ + ──  - ── - ─⎥                      ⎟\n",
       "⎜⎢      6   6     6    6⎥  ⎡2    0        0   ⎤⎟\n",
       "⎜⎢                      ⎥  ⎢                  ⎥⎟\n",
       "⎜⎢     √7   5     √7   5⎥, ⎢0  1 + √7     0   ⎥⎟\n",
       "⎜⎢9    ── + ─   - ── + ─⎥  ⎢                  ⎥⎟\n",
       "⎜⎢     6    6     6    6⎥  ⎣0    0     -√7 + 1⎦⎟\n",
       "⎜⎢                      ⎥                      ⎟\n",
       "⎝⎣4      1         1    ⎦                      ⎠"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "P_B, D_B = B.diagonalize()\n",
    "P_B, D_B"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2019-10-23T18:34:47.222384Z",
     "start_time": "2019-10-23T18:34:47.217657Z"
    },
    "colab_type": "text",
    "id": "xy2IGtImKOdo"
   },
   "source": [
    "### 9. Cayley Hamilton (10pts)\n",
    "\n",
    "a) Use the Cayley Hamilton Theorem to come up with an expression for $A^n$ in terms of $n$ when \n",
    "\n",
    "$$\n",
    "A = \\begin{pmatrix}\n",
    "-\\frac{1}{2} & 1  \\\\\n",
    "0 & \\frac{1}{4}\n",
    "\\end{pmatrix} .\n",
    "$$\n",
    "\n",
    "b) Define $x_{k+1} = A x_k$ and argue that no matter what value $x_0$ is, $x_{k+1}$ converges to $(0 \\; 0)^T$. \n",
    "\n",
    "c) Choose $x_0 = (-2,3)^T$ and plot $x_k$ for $k = 0$ to $10$. Plot the two components of of $x_k$ as two separate trajectories overlaid on the same plot.\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    1pt for finding the characteristic polynomial of A\n",
    "    1pt for using the CH formula to find p(A)=0\n",
    "    2pts for finding an expression for A^n in terms of n\n",
    "    1pts (b) for finding an explicit formula for x_k+1\n",
    "    3pts (b) for for a convincing argument to show that x_k+1 converges to zero regardless of x0\n",
    "    1pt (c) for plotting the x component\n",
    "    1pt (c) for plotting the y component"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}1 & 4\\\\0 & 3\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{1}{2} & 0\\\\0 & \\frac{1}{4}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}1 & - \\frac{4}{3}\\\\0 & \\frac{1}{3}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{1}{2} & 1\\\\0 & \\frac{1}{4}\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡1  4⎤  ⎡-1/2   0 ⎤  ⎡1  -4/3⎤  ⎡-1/2   1 ⎤⎞\n",
       "⎜⎢    ⎥, ⎢         ⎥, ⎢       ⎥, ⎢         ⎥⎟\n",
       "⎝⎣0  3⎦  ⎣ 0    1/4⎦  ⎣0  1/3 ⎦  ⎣ 0    1/4⎦⎠"
      ]
     },
     "execution_count": 24,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = sm.Matrix([[sm.Rational(-1,2), 1],[0, sm.Rational(1,4)]]) \n",
    "Q, L = A.diagonalize()\n",
    "Q, L, Q.inv(), Q @ L @ Q.inv()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "9.a)\n",
    "\n",
    "Recall the **Caley-Hamilton Theorem**: A matrix solves its own characteristic polynomial (i.e. $p(A) = 0$).\n",
    "\n",
    "The characteristic polynomial of $A$ is $p(s) = (s−\\frac{1}{4})(s+ \\frac{1}{2}) = s^2 + \\frac{s}{4} - \\frac{1}{8}.$\n",
    "\n",
    "Hence \n",
    "\n",
    "$$  A^2 = -\\frac{A}{4} + \\frac{I}{8} $$\n",
    "\n",
    "$$  A^n = -\\frac{A^{n-1}}{4} + \\frac{A^{n-2}}{8}. $$\n",
    "\n",
    "We could more specifically express this as:\n",
    "\n",
    "$$ = - \\frac{1}{4} Q \\begin{pmatrix} (\\frac{-1}{2})^{n-1} & 0 \\\\ 0 & (\\frac{1}{4})^{n-1} \\end{pmatrix}Q^{-1}  + \\frac{1}{8} Q \\begin{pmatrix} (\\frac{-1}{2})^{n-2} & 0 \\\\ 0 & (\\frac{1}{4})^{n-2} \\end{pmatrix} Q^{-1} $$\n",
    "\n",
    "where $Q, Q^{-1}$ are defined in the code above."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "9.b)\n",
    "\n",
    "Let's begin by ennumerating terms of $x_{k+1} = A x_k$ and find an explicit form: \n",
    "\n",
    "$$\\begin{align}\n",
    "x_{1} &= Ax_0 \\\\\n",
    "x_{2} &= Ax_1 = A(Ax_0) = A^2x_0 \\\\\n",
    "x_{3} &= Ax_2 = A(A^2x_0) = A^3x_0\\\\\n",
    "& \\vdots \\\\\n",
    "x_{n} &= A^nx_0. \\\\\n",
    "\\end{align}$$\n",
    "\n",
    "Now, let's take the limit as $k\\rightarrow \\infty:$\n",
    "\n",
    "$$\\lim \\limits_{k\\rightarrow \\infty} x_{k+1} = \\lim \\limits_{k\\rightarrow \\infty}  A^{k+1}x_0$$\n",
    "\n",
    "Because the eigenvalues of $A$ are $\\frac{-1}{2}, \\frac{1}{4}$ (i.e. distinct and nonzero) we can diagonalize it into a matrix of eigenvectors, $Q$, and a diagonal matrix of eigenvalues, $\\Lambda:$ \n",
    "\n",
    "$$A = Q\\Lambda Q^{-1}.$$ \n",
    "\n",
    "Furthermore, because $A^2 = Q\\Lambda (Q^{-1}Q)\\Lambda Q^{-1} = Q\\Lambda^2 Q^{-1} \\implies A^n = Q\\Lambda^n Q^{-1}.$\n",
    "\n",
    "Thus,\n",
    "\n",
    "$$\\begin{align}\n",
    "\\lim \\limits_{k\\rightarrow \\infty} x_{k+1} &= x_0 \\lim \\limits_{k\\rightarrow \\infty}  Q\\Lambda^k Q^{-1} \\\\\n",
    "&= x_0 \\lim \\limits_{k\\rightarrow \\infty}  Q \n",
    "\\begin{pmatrix} (\\frac{-1}{2})^k & 0 \\\\ 0 & (\\frac{1}{4})^k \\end{pmatrix}\n",
    "Q^{-1} \\\\\n",
    "&= x_0 Q \n",
    "\\begin{pmatrix} 0 & 0 \\\\ 0 & 0 \\end{pmatrix}\n",
    "Q^{-1} \\\\\n",
    "&= 0.\n",
    "\\end{align}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x2a2636ddc18>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$$\\left [ \\left[\\begin{matrix}-2\\\\3\\end{matrix}\\right], \\quad \\left[\\begin{matrix}4\\\\\\frac{3}{4}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{5}{4}\\\\\\frac{3}{16}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}\\frac{13}{16}\\\\\\frac{3}{64}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{23}{64}\\\\\\frac{3}{256}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}\\frac{49}{256}\\\\\\frac{3}{1024}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{95}{1024}\\\\\\frac{3}{4096}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}\\frac{193}{4096}\\\\\\frac{3}{16384}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{383}{16384}\\\\\\frac{3}{65536}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}\\frac{769}{65536}\\\\\\frac{3}{262144}\\end{matrix}\\right], \\quad \\left[\\begin{matrix}- \\frac{1535}{262144}\\\\\\frac{3}{1048576}\\end{matrix}\\right]\\right ]$$"
      ],
      "text/plain": [
       "⎡                     ⎡ 13 ⎤  ⎡-23  ⎤  ⎡  49  ⎤  ⎡ -95  ⎤  ⎡ 193   ⎤  ⎡ -383  \n",
       "⎢⎡-2⎤  ⎡ 4 ⎤  ⎡-5/4⎤  ⎢ ── ⎥  ⎢──── ⎥  ⎢ ───  ⎥  ⎢ ──── ⎥  ⎢ ────  ⎥  ⎢ ───── \n",
       "⎢⎢  ⎥, ⎢   ⎥, ⎢    ⎥, ⎢ 16 ⎥, ⎢ 64  ⎥, ⎢ 256  ⎥, ⎢ 1024 ⎥, ⎢ 4096  ⎥, ⎢ 16384 \n",
       "⎢⎣3 ⎦  ⎣3/4⎦  ⎣3/16⎦  ⎢    ⎥  ⎢     ⎥  ⎢      ⎥  ⎢      ⎥  ⎢       ⎥  ⎢       \n",
       "⎣                     ⎣3/64⎦  ⎣3/256⎦  ⎣3/1024⎦  ⎣3/4096⎦  ⎣3/16384⎦  ⎣3/65536\n",
       "\n",
       "⎤  ⎡  769   ⎤  ⎡ -1535   ⎤⎤\n",
       "⎥  ⎢ ─────  ⎥  ⎢ ──────  ⎥⎥\n",
       "⎥, ⎢ 65536  ⎥, ⎢ 262144  ⎥⎥\n",
       "⎥  ⎢        ⎥  ⎢         ⎥⎥\n",
       "⎦  ⎣3/262144⎦  ⎣3/1048576⎦⎦"
      ]
     },
     "execution_count": 25,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# 9.c)\n",
    "\n",
    "k = np.arange(0,11)\n",
    "x0 = np.array([[-2],[3]])\n",
    "xks = [A**i @ x0 for i in k]\n",
    "xk_1 = [element[0] for element in xks] # Make an array of the first component of the vectors\n",
    "xk_2 = [element[1] for element in xks] # Make an array of the second component of the vectors\n",
    "\n",
    "plt.scatter(k, xk_1)\n",
    "plt.scatter(k, xk_2)\n",
    "plt.grid('on')\n",
    "plt.title(r'Trajectories of $x_{k+1} = A x_k$')\n",
    "plt.xlabel('k')\n",
    "plt.ylabel(r'$x_k$')\n",
    "plt.legend(['x trajectory','y trajectory'])\n",
    "plt.show()\n",
    "\n",
    "xks"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2019-10-23T18:47:28.833529Z",
     "start_time": "2019-10-23T18:47:28.828866Z"
    },
    "colab_type": "text",
    "id": "6LCG5pZ9KOdo"
   },
   "source": [
    "### 10. Matrix Exponential Properties (10pts)\n",
    "---\n",
    "\n",
    "Recall that for matrices $A$ and $B$ that it is not necessarily the case that $AB = BA$ (i.e. that $A$ and $B$ commute). Show that \n",
    "\n",
    "a) If $AB=BA$ then $e^Ae^B = e^Be^A$ using the definition of the matrix exponential as a series.\n",
    "\n",
    "b) Find an example in 2D where $A$ and $B$ do not commute and show that $e^Ae^B \\neq e^Be^A$.\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    2pts for stating the definitions of e^A, e^B\n",
    "    2pts for using the assumption AB=BA \n",
    "    2pts for for a convincing argument to show that e^Ae^B = e^Be^A if AB=BA\n",
    "    2pts (b) for finding two matrices that do not commute\n",
    "    2pts (b) for showing that e^Ae^B does not equal e^Be^A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "10.a)\n",
    "\n",
    "Recall from lecture, when $A \\in \\mathbb{R}^{n\\times n}$, then the we define the **matrix exponential** $e^{A}$ by\n",
    "$$\n",
    "e^{A} \\triangleq I + A + \\frac{A^2}{2!} + \\frac{A^3}{3!} + \\dots\n",
    "$$\n",
    "Now assume $AB=BA$. By using the definition we see\n",
    "\n",
    "$$\n",
    "%step 1\n",
    "e^{A}e^{B} = (I + A + \\frac{A^2}{2!} + \\frac{A^3}{3!} + \\dots)(I + B + \\frac{B^2}{2!} + \\frac{B^3}{3!} + \\dots),\n",
    "$$\n",
    "\n",
    "and we can group the terms to use in a FOIL expansion\n",
    "\n",
    "$$\n",
    "%step 2: group terms\n",
    "e^{A}e^{B} = [I + (A + \\frac{A^2}{2!} + \\frac{A^3}{3!} + \\dots)][I + (B + \\frac{B^2}{2!} + \\frac{B^3}{3!} + \\dots)].\n",
    "$$\n",
    "\n",
    "Next, the FOIL expansion gives us\n",
    "\n",
    "$$\n",
    "%step 3 FOIL expansion\n",
    "e^{A}e^{B} = [I + (B + \\frac{B^2}{2!} + \\frac{B^3}{3!} + \\dots) + (A + \\frac{A^2}{2!} + \\frac{A^3}{3!} + \\dots) + (A + \\frac{A^2}{2!} + \\frac{A^3}{3!} + \\dots)(B + \\frac{B^2}{2!} + \\frac{B^3}{3!} + \\dots)].\n",
    "$$\n",
    "\n",
    "We'll use FOIL expansion again on the last term, use the fact that $AB=BA$, and regroup some of the terms as\n",
    "\n",
    "$$\n",
    "%step 4: group expanded terms\n",
    "e^{A}e^{B} = [I + (A+B) + \\frac{1}{2!}(A^2 + B^2 + 2AB) + (\\frac{1}{3!}(A^3 + B^3) + \\frac{1}{2!}(A^2B + AB^2)) + \\dots]\n",
    "$$\n",
    "\n",
    "reexpressing them as the following:\n",
    "\n",
    "$$\n",
    "%step 5: group expanded terms\n",
    "e^{A}e^{B} = [I + (A+B) + \\frac{1}{2!}(A^2 + B^2 + 2AB) + \\frac{1}{3!}(A^3 + 3A^2B + 3AB^2 + B^3) + \\dots].\n",
    "$$\n",
    "\n",
    "Lastly, we will collect the terms as\n",
    "\n",
    "$$\n",
    "%step 6: collect terms\n",
    "e^{A}e^{B} = [I + (A+B) + \\frac{1}{2!}(A+B)^2 + \\frac{1}{3!}(A+B)^3 + \\dots].\n",
    "$$\n",
    "\n",
    "This process can be repeated to show that:\n",
    "\n",
    "$$\n",
    "%step 6: by definintion of the infinte series\n",
    "e^{A}e^{B} = e^{A+B}.\n",
    "$$\n",
    "\n",
    "And since $A+B = B+A \\implies e^{A+B} = e^{B+A}$, we can conclude that $e^Ae^B = e^Be^A$. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "10.b)\n",
    "\n",
    "We'll use two matrices from physics that are famous, the Pauli spin matrices:\n",
    "\n",
    "$\n",
    "\\sigma_x = \n",
    "\\begin{pmatrix}\n",
    "0 & 1  \\\\\n",
    "1 & 0\n",
    "\\end{pmatrix} \n",
    "$\n",
    "and\n",
    "$\n",
    "\\sigma_z = \n",
    "\\begin{pmatrix}\n",
    "1 & 0  \\\\\n",
    "0 & -1\n",
    "\\end{pmatrix}, \n",
    "$\n",
    "where:\n",
    "$\n",
    "\\sigma_x\\sigma_z = \n",
    "\\begin{pmatrix}\n",
    "0 & -1  \\\\\n",
    "1 & 0\n",
    "\\end{pmatrix} \n",
    "$\n",
    "and\n",
    "$\n",
    "\\sigma_z\\sigma_x = \n",
    "\\begin{pmatrix}\n",
    "0 & 1  \\\\\n",
    "-1 & 0\n",
    "\\end{pmatrix} \n",
    "$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Expaning like before we see\n",
    "$$\n",
    "%step 1 FOIL expansion\n",
    "e^{\\sigma_x}e^{\\sigma_z} = [I + (\\sigma_z + \\frac{\\sigma_z^2}{2!} + \\frac{\\sigma_z^3}{3!} + \\dots) + (\\sigma_x + \\frac{\\sigma_x^2}{2!} + \\frac{\\sigma_x^3}{3!} + \\dots) + (\\sigma_x + \\frac{\\sigma_x^2}{2!} + \\frac{\\sigma_x^3}{3!} + \\dots)(\\sigma_z + \\frac{\\sigma_z^2}{2!} + \\frac{\\sigma_z^3}{3!} + \\dots)]\n",
    "$$\n",
    "\n",
    "$$\n",
    "%step 2: group expanded terms\n",
    "e^{\\sigma_x}e^{\\sigma_z} = [I + (\\sigma_z+\\sigma_x) + \\frac{1}{2!}(\\sigma_x^2 + \\sigma_z^2 + \\sigma_x\\sigma_z + \\sigma_z\\sigma_x) + (\\frac{1}{3!}(\\sigma_x^3 + \\sigma_z^3) + \\frac{1}{2!}(\\sigma_x^2\\sigma_z + \\sigma_x\\sigma_z^2)) + \\dots]\n",
    "$$\n",
    "\n",
    "Notice that $2\\sigma_x\\sigma_z  = \n",
    "\\begin{pmatrix}\n",
    "0 & -2  \\\\\n",
    "2 & 0\n",
    "\\end{pmatrix} $\n",
    "but $\\sigma_x\\sigma_z + \\sigma_z\\sigma_x = \\begin{pmatrix}\n",
    "0 & 0  \\\\\n",
    "0 & 0\n",
    "\\end{pmatrix} $. So we can't say $e^{\\sigma_x}e^{\\sigma_z} = e^{\\sigma_x + \\sigma_z}$.\n",
    "\n",
    "Similarly, compare\n",
    "$$\n",
    "%step 1 FOIL expansion\n",
    "e^{\\sigma_z}e^{\\sigma_x} = [I + (\\sigma_x + \\frac{\\sigma_x^2}{2!} + \\frac{\\sigma_x^3}{3!} + \\dots) + (\\sigma_z + \\frac{\\sigma_z^2}{2!} + \\frac{\\sigma_z^3}{3!} + \\dots) + (\\sigma_z + \\frac{\\sigma_z^2}{2!} + \\frac{\\sigma_z^3}{3!} + \\dots)(\\sigma_x + \\frac{\\sigma_x^2}{2!} + \\frac{\\sigma_x^3}{3!} + \\dots)]\n",
    "$$\n",
    "\n",
    "$$\n",
    "%step 2: group expanded terms\n",
    "e^{\\sigma_z}e^{\\sigma_x} = [I + (\\sigma_x+\\sigma_z) + \\frac{1}{2!}(\\sigma_z^2 + \\sigma_x^2 + \\sigma_z\\sigma_x + \\sigma_x\\sigma_z) + (\\frac{1}{3!}(\\sigma_z^3 + \\sigma_x^3) + \\frac{1}{2!}(\\sigma_z^2\\sigma_x + \\sigma_z\\sigma_x^2)) + \\dots]\n",
    "$$\n",
    "\n",
    "Notice that $2\\sigma_z\\sigma_x  = \n",
    "\\begin{pmatrix}\n",
    "0 & 2  \\\\\n",
    "-2 & 0\n",
    "\\end{pmatrix} $\n",
    "but $\\sigma_z\\sigma_x + \\sigma_x\\sigma_z = \\begin{pmatrix}\n",
    "0 & 0  \\\\\n",
    "0 & 0\n",
    "\\end{pmatrix} $.  So we can't say $e^{\\sigma_x + \\sigma_z} = e^{\\sigma_z + \\sigma_x} = e^{\\sigma_z}e^{\\sigma_x}$.\n",
    "\n",
    "Therefore, $e^{\\sigma_x}e^{\\sigma_z} \\neq e^{\\sigma_z}e^{\\sigma_x}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "False\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$$\\left ( \\left[\\begin{matrix}e \\left(\\frac{1}{2 e} + \\frac{e}{2}\\right) & \\frac{1}{e} \\left(- \\frac{1}{2 e} + \\frac{e}{2}\\right)\\\\e \\left(- \\frac{1}{2 e} + \\frac{e}{2}\\right) & \\frac{1}{e} \\left(\\frac{1}{2 e} + \\frac{e}{2}\\right)\\end{matrix}\\right], \\quad \\left[\\begin{matrix}e \\left(\\frac{1}{2 e} + \\frac{e}{2}\\right) & e \\left(- \\frac{1}{2 e} + \\frac{e}{2}\\right)\\\\\\frac{1}{e} \\left(- \\frac{1}{2 e} + \\frac{e}{2}\\right) & \\frac{1}{e} \\left(\\frac{1}{2 e} + \\frac{e}{2}\\right)\\end{matrix}\\right]\\right )$$"
      ],
      "text/plain": [
       "⎛⎡   ⎛ -1    ⎞   ⎛   -1    ⎞    ⎤  ⎡    ⎛ -1    ⎞      ⎛   -1    ⎞⎤⎞\n",
       "⎜⎢   ⎜ℯ     ℯ⎟   ⎜  ℯ     ℯ⎟  -1⎥  ⎢    ⎜ℯ     ℯ⎟      ⎜  ℯ     ℯ⎟⎥⎟\n",
       "⎜⎢ ℯ⋅⎜─── + ─⎟   ⎜- ─── + ─⎟⋅ℯ  ⎥  ⎢  ℯ⋅⎜─── + ─⎟    ℯ⋅⎜- ─── + ─⎟⎥⎟\n",
       "⎜⎢   ⎝ 2    2⎠   ⎝   2    2⎠    ⎥  ⎢    ⎝ 2    2⎠      ⎝   2    2⎠⎥⎟\n",
       "⎜⎢                              ⎥, ⎢                              ⎥⎟\n",
       "⎜⎢  ⎛   -1    ⎞   ⎛ -1    ⎞     ⎥  ⎢⎛   -1    ⎞      ⎛ -1    ⎞    ⎥⎟\n",
       "⎜⎢  ⎜  ℯ     ℯ⎟   ⎜ℯ     ℯ⎟  -1 ⎥  ⎢⎜  ℯ     ℯ⎟  -1  ⎜ℯ     ℯ⎟  -1⎥⎟\n",
       "⎜⎢ℯ⋅⎜- ─── + ─⎟   ⎜─── + ─⎟⋅ℯ   ⎥  ⎢⎜- ─── + ─⎟⋅ℯ    ⎜─── + ─⎟⋅ℯ  ⎥⎟\n",
       "⎝⎣  ⎝   2    2⎠   ⎝ 2    2⎠     ⎦  ⎣⎝   2    2⎠      ⎝ 2    2⎠    ⎦⎠"
      ]
     },
     "execution_count": 26,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# Let's use sympy to verfiy our reasoning\n",
    "sx = sm.Matrix([[0,1],[1,0]])\n",
    "sz = sm.Matrix([[1,0],[0,-1]])\n",
    "\n",
    "LHS = sm.exp(sx)*sm.exp(sz) \n",
    "RHS = sm.exp(sz)*sm.exp(sx)\n",
    "print(LHS == RHS)\n",
    "\n",
    "LHS, RHS"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "colab_type": "text",
    "id": "Su4WKSW6KOdp"
   },
   "source": [
    "### 11. Senators Revisited (10pts)\n",
    "\n",
    "Repeat the clustering of senators by voting habit for the years 1999, 1979 and 1959. Plot them together with the 2019 plot in a grid of plots. Which years seem the most divided? \n",
    "\n",
    "Note you will need to get the data at [https://voteview.com/data](https://voteview.com/data). Choose \"Member's Votes\", \"Senate Only\", the desired year, and CSV file. \n",
    "\n",
    "I had to edit the CSV file (in a text editor or ExCEL) to remove the heading in the first row before loading the file.\n",
    "\n",
    "**Extra credit:** Color the each dot in the plots by whether the senator is a republican (red), democract (blue), or independent (green). This information is not in the data files above, so you'll need to find it elsewhere.\n",
    "[https://voteview.com/articles/data_help_members]\n",
    "\n",
    "$\\textbf{Grading breakdown:}$\n",
    "    \n",
    "    1pt for importing {S085, S095, S105, S115} OR {S086, S096, S106, S116}\n",
    "    2pts for arguing which years seem to be the most divided\n",
    "    2pts for constructing the voting matrices: A\n",
    "    2pts for constructing the voting habit matrices: T\n",
    "    2pts for plotting them together through subplot\n",
    "    1pts for labeling everything\n",
    "    +4pts for doing the extra credit"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Looking at the graphs, we can conclude that 2019 and 1999 are clearly more divided, because the clusters are more distinct and further apart in the partisan component."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "scrolled": true
   },
   "outputs": [],
   "source": [
    "# rollnumber = issue id\n",
    "# icpsr = Integer 1-99999. This is an ID code which identifies the member in question.\n",
    "# cast_code = vote outcome\n",
    "\n",
    "my_dataS085 = np.genfromtxt('S085_votes_data.csv', delimiter=',', dtype=np.int8)\n",
    "my_dataS095 = np.genfromtxt('S095_votes_data.csv', delimiter=',', dtype=np.int8)\n",
    "my_dataS105 = np.genfromtxt('S105_votes_data.csv', delimiter=',', dtype=np.int8)\n",
    "my_dataS115 = np.genfromtxt('S115_votes_data.csv', delimiter=',', dtype=np.int8)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "def senate_member_votes(data):\n",
    "    '''\n",
    "    Assumes: data is a .csv file of Senate voting data from:   https://voteview.com/data\n",
    "    with the first row deleted. Its columns are:\n",
    "    1) congress  \n",
    "    2) chamber = Senate\n",
    "    3) rollnumber = issue id\n",
    "    4) icpsr = Integer 1-99999. This is an ID code which identifies the member in question.\n",
    "    5) cast_code = vote outcome\n",
    "    6) prob\n",
    "    Returns: Matrix A is nxm where Aij is 1 if member i voted 'yes'; -1 if member i voted 'no', and 0 if they didn't vote\n",
    "    '''\n",
    "    issues = data[:,2] \n",
    "    members = data[:,3]\n",
    "    votes = data[:,4]\n",
    "    \n",
    "    member_ids = np.unique(members)\n",
    "    members = [np.where(member_ids == m)[0][0] for m in members]\n",
    "\n",
    "    yn = [ 0, 1, 1, 1, -1, -1, -1, 0, 0, 0 ]\n",
    "    votes = np.array([yn[v] for v in votes])\n",
    "\n",
    "    A = np.zeros((len(member_ids), max(issues)+1))\n",
    "    for member, issue, vote in zip(members, issues, votes):\n",
    "        A[int(member),int(issue)] = vote\n",
    "        \n",
    "    return A"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "colab": {},
    "colab_type": "code",
    "id": "a8PaocsSKOdq"
   },
   "outputs": [],
   "source": [
    "#Make Senate voting matrices for each set of data\n",
    "A_S085 = senate_member_votes(my_dataS085)\n",
    "A_S095 = senate_member_votes(my_dataS095)\n",
    "A_S105 = senate_member_votes(my_dataS105)\n",
    "A_S115 = senate_member_votes(my_dataS115)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [],
   "source": [
    "def calc_PCA(A):\n",
    "    '''\n",
    "    Assumes: A is nxm where Aij is 1 if member i voted 'yes'; -1 if member i voted 'no', and 0 if they didn't vote\n",
    "    Returns: A matrix T, where the  i th row of  T  sums up Senator i's voting habits in two numbers.\n",
    "    '''\n",
    "    X = A - np.mean(A) # Center the data at zero\n",
    "    U, s, VT = scipy.linalg.svd(X) # Singular-value decomposition\n",
    "    Sigma = np.zeros((A.shape[0], A.shape[1])) # create m x n Sigma matrix\n",
    "    Sigma[:A.shape[0], :A.shape[0]] = np.diag(s) # populate Sigma with n x n diagonal matrix\n",
    "    \n",
    "    # select\n",
    "    n_elements = 2\n",
    "    Sigma = Sigma[:, :n_elements]\n",
    "    VT = VT[:n_elements, :]\n",
    "\n",
    "    # reconstruct T\n",
    "    T = U.dot(Sigma)\n",
    "    return T"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "# The  ith row of  T  sums up Senator i's voting habits in two numbers.\n",
    "T_S085 = calc_PCA(A_S085)\n",
    "T_S095 = calc_PCA(A_S095)\n",
    "T_S105 = calc_PCA(A_S105)\n",
    "T_S115 = calc_PCA(A_S115)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x21028990320>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "fig, ax = plt.subplots(nrows=2, ncols=2, sharex=True, sharey=True, figsize=(10,10))\n",
    "\n",
    "ax[0,0].scatter(T_S085[:,0], T_S085[:,1])\n",
    "ax[0,0].set_title(\"Senators clustered by\\nvoting habits in S085\")\n",
    "ax[0,0].grid('on')\n",
    "ax[0,0].set_ylabel(\"Bipartisan component\")\n",
    "\n",
    "ax[0,1].scatter(T_S095[:,0], T_S095[:,1])\n",
    "ax[0,1].set_title(\"Senators clustered by\\nvoting habits in S095\")\n",
    "ax[0,1].grid('on')\n",
    "\n",
    "ax[1,0].scatter(T_S105[:,0], T_S105[:,1])\n",
    "ax[1,0].set_title(\"Senators clustered by\\nvoting habits in S105\")\n",
    "ax[1,0].grid('on')\n",
    "ax[1,0].set_xlabel(\"Partisan component\")\n",
    "ax[1,0].set_ylabel(\"Bipartisan component\")\n",
    "\n",
    "\n",
    "ax[1,1].scatter(T_S115[:,0], T_S115[:,1])\n",
    "ax[1,1].set_title(\"Senators clustered by\\nvoting habits in S115\")\n",
    "ax[1,1].grid('on')\n",
    "ax[1,1].set_xlabel(\"Partisan component\")\n",
    "\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "colab": {
   "name": "HW_04_LinearAlgebra_Due_10_30.ipynb",
   "provenance": []
  },
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.7.4"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 1
}