diff --git a/lectures/linear_equations.md b/lectures/linear_equations.md index 4b9ccb0f..fec1bf7e 100644 --- a/lectures/linear_equations.md +++ b/lectures/linear_equations.md @@ -11,152 +11,144 @@ kernelspec: name: python3 --- -# Linear Equations and Matrix Algebra +# 线性方程和矩阵代数 ```{index} single: Linear Equations and Matrix Algebra ``` -## Overview +## 概述 -Many problems in economics and finance require solving linear equations. +经济学和金融学中的许多问题都需要解线性方程。 -In this lecture we discuss linear equations and their applications. +在本讲座中，我们将讨论线性方程及其应用。 -To illustrate the importance of linear equations, we begin with a two good -model of supply and demand. +为了说明线性方程的重要性，我们从一个两种商品的供需模型开始。 -The two good case is so simple that solutions can be calculated by hand. +两种商品的情况非常简单，可以手动计算解。 -But often we need to consider markets containing many goods. +但我们经常需要考虑包含多种商品的市场。 -In the multiple goods case we face large systems of linear equations, with many equations -and unknowns. +在多种商品的情况下，我们面对的是大型线性方程组，有许多方程和未知数。 -To handle such systems we need two things: +为了处理这样的系统，我们需要两样东西： -* matrix algebra (and the knowledge of how to use it) plus -* computer code to apply matrix algebra to the problems of interest. +* 矩阵代数（以及如何使用它的知识）以及 +* 将矩阵代数应用于感兴趣问题的计算机代码。 -This lecture covers these steps. +本讲座涵盖了这些步骤。 -We will use the following packages: +我们将使用以下的库： ```{code-cell} ipython3 import numpy as np import matplotlib.pyplot as plt ``` +## 两种商品的例子 -## A two good example +在本节中，我们将讨论一个简单的两种商品例子，并通过以下两种方法解决： -In this section we discuss a simple two good example and solve it by +1. 纸笔计算 +2. 矩阵代数 -1. pencil and paper -2. matrix algebra +正如我们将看到的，第二种方法更具普遍性。 -The second method is more general, as we will see. +### 纸笔计算方法 +假设我们有两种相关的商品，比如： -### Pencil and paper methods +* 丙烷和乙醇，或 +* 大米和小麦等。 -Suppose that we have two related goods, such as +为了简化问题，我们将它们标记为商品0和商品1。 -* propane and ethanol, and -* rice and wheat, etc. - -To keep things simple, we label them as good 0 and good 1. - -The demand for each good depends on the price of both goods: +每种商品的需求取决于两种商品的价格： ```{math} :label: two_eq_demand + \begin{aligned} q_0^d = 100 - 10 p_0 - 5 p_1 \\ q_1^d = 50 - p_0 - 10 p_1 \end{aligned} ``` -(We are assuming demand decreases when the price of either good goes up, but -other cases are also possible.) +（我们假设当任一商品的价格上涨时需求会下降，但其他情况也是可能的。） +让我们假设供给由以下方程给出： -Let's suppose that supply is given by ```{math} :label: two_eq_supply + \begin{aligned} q_0^s = 10 p_0 + 5 p_1 \\ q_1^s = 5 p_0 + 10 p_1 \end{aligned} ``` -Equilibrium holds when supply equals demand ($q_0^s = q_0^d$ and $q_1^s = q_1^d$). +当供给等于需求时（$q_0^s = q_0^d$ 和 $q_1^s = q_1^d$），市场达到均衡。 -This yields the linear system +这产生了以下线性系统： ```{math} :label: two_equilibrium + \begin{aligned} 100 - 10 p_0 - 5 p_1 = 10 p_0 + 5 p_1 \\ 50 - p_0 - 10 p_1 = 5 p_0 + 10 p_1 \end{aligned} ``` -We can solve this with pencil and paper to get +我们可以用纸笔计算得到： $$ - p_0 = 4.41 \quad \text{and} \quad p_1 = 1.18. + p_0 = 4.41 \quad \text{和} \quad p_1 = 1.18. $$ -Inserting these results into either {eq}`two_eq_demand` or {eq}`two_eq_supply` yields the -equilibrium quantities +将这些结果代入{eq}`two_eq_demand`或{eq}`two_eq_supply`中，可得均衡数量： $$ - q_0 = 50 \quad \text{and} \quad q_1 = 33.82. + q_0 = 50 \quad \text{和} \quad q_1 = 33.82. $$ +### 展望未来 -### Looking forward - -Pencil and paper methods are easy in the two good case. - -But what if there are many goods? +在两种商品的情况下，纸笔计算方法很容易。 -For such problems we need matrix algebra. +但如果有很多种商品呢？ -Before solving problems with matrix algebra, let's first recall the -basics of vectors and matrices, in both theory and computation. +对于这样的问题，我们需要矩阵代数。 +在用矩阵代数解决问题之前，让我们先回顾一下向量和矩阵的基础知识，包括理论和计算。 -## {index}`Vectors ` +## {index}`向量 ` ```{index} single: Linear Algebra; Vectors ``` -A **vector** of length $n$ is just a sequence (or array, or tuple) of $n$ numbers, which we write as $x = (x_1, \ldots, x_n)$ or $x = \begin{bmatrix}x_1, \ldots, x_n\end{bmatrix}$. +一个长度为$n$的**向量**就是一个由$n$个数字组成的序列（或数组，或元组），我们将其写作$x = (x_1, \ldots, x_n)$或$x = \begin{bmatrix}x_1, \ldots, x_n\end{bmatrix}$。 -We can write these sequences either horizontally or vertically. +我们可以将这些序列横向或纵向写出。 -But when we use matrix operations, our default assumption is that vectors are -column vectors. +但当我们使用矩阵运算时，我们默认假设向量是列向量。 -The set of all $n$-vectors is denoted by $\mathbb R^n$. +所有$n$维向量的集合用$\mathbb R^n$表示。 ```{prf:example} :label: le_ex_dim -* $\mathbb R^2$ is the plane --- the set of pairs $(x_1, x_2)$. -* $\mathbb R^3$ is 3 dimensional space --- the set of vectors $(x_1, x_2, x_3)$. +* $\mathbb R^2$是平面 --- 即所有$(x_1, x_2)$对的集合。 +* $\mathbb R^3$是三维空间 --- 即所有$(x_1, x_2, x_3)$向量的集合。 +向量通常在视觉上表示为从原点到某点的箭头。 ``` -Often vectors are represented visually as arrows from the origin to the point. - -Here's a visualization. +这里是一个可视化示例。 ```{code-cell} ipython3 :tags: [hide-input] fig, ax = plt.subplots() -# Set the axes through the origin +# 通过原点建立坐标轴 for spine in ['left', 'bottom']: ax.spines[spine].set_position('zero') for spine in ['right', 'top']: @@ -175,17 +167,16 @@ for v in vecs: plt.show() ``` -### Vector operations +### 向量运算 ```{index} single: Vectors; Operations ``` -Sometimes we want to modify vectors. +有时我们需要修改向量。 -The two most common operators on vectors are addition and scalar -multiplication, which we now describe. +对向量最常见的两种运算是加法和标量乘法，我们现在来描述这两种运算。 -When we add two vectors, we add them element-by-element. +当我们对两个向量进行加法运算时，我们是逐元素相加。 ```{prf:example} :label: le_ex_add @@ -213,7 +204,7 @@ $$ $$ ``` -In general, +一般来说， $$ x + y = @@ -237,13 +228,13 @@ x + y = \end{bmatrix}. $$ -We can visualise vector addition in $\mathbb{R}^2$ as follows. +我们可以在$\mathbb{R}^2$中将向量加法可视化如下。 ```{code-cell} ipython3 :tags: [hide-input] fig, ax = plt.subplots() -# Set the axes through the origin +# 通过原点建立坐标轴 for spine in ['left', 'bottom']: ax.spines[spine].set_position('zero') for spine in ['right', 'top']: @@ -275,7 +266,7 @@ for i, v in enumerate(vecs): plt.show() ``` -Scalar multiplication is an operation that multiplies a vector $x$ with a scalar elementwise. +标量乘法是一种将向量 $x$ 与一个标量进行元素级别相乘的运算。 ```{prf:example} :label: le_ex_mul @@ -299,7 +290,7 @@ $$ $$ ``` -More generally, it takes a number $\gamma$ and a vector $x$ and produces +更一般地，它取一个数 $\gamma$ 和一个向量 $x$，得到 $$ \gamma x := @@ -311,13 +302,13 @@ $$ \end{bmatrix}. $$ -Scalar multiplication is illustrated in the next figure. +标量乘法在下图中进行了说明。 ```{code-cell} ipython3 :tags: [hide-input] fig, ax = plt.subplots() -# Set the axes through the origin +# 通过原点建立坐标轴 for spine in ['left', 'bottom']: ax.spines[spine].set_position('zero') for spine in ['right', 'top']: @@ -346,26 +337,25 @@ for s in scalars: plt.show() ``` -In Python, a vector can be represented as a list or tuple, -such as `x = [2, 4, 6]` or `x = (2, 4, 6)`. +在Python中，向量可以用列表或元组表示， +例如 `x = [2, 4, 6]` 或 `x = (2, 4, 6]`。 -However, it is more common to represent vectors with -[NumPy arrays](https://python-programming.quantecon.org/numpy.html#numpy-arrays). +然而，更常见的是用 +[NumPy数组](https://python-programming.quantecon.org/numpy.html#numpy-arrays)来表示向量。 -One advantage of NumPy arrays is that scalar multiplication and addition have -very natural syntax. +NumPy数组的一个优点是标量乘法和加法具有非常自然的语法。 ```{code-cell} ipython3 -x = np.ones(3) # Vector of three ones -y = np.array((2, 4, 6)) # Converts tuple (2, 4, 6) into a NumPy array -x + y # Add (element-by-element) +x = np.ones(3) # 三个元素为一的向量 +y = np.array((2, 4, 6)) # 将 (2, 4, 6) 转换为 NumPy 数组 +x + y # 每个元素相加 ``` ```{code-cell} -4 * x # Scalar multiply +4 * x # 标量乘法 ``` -### Inner product and norm +### 内积和范数 ```{index} single: Vectors; Inner Product ``` @@ -373,7 +363,7 @@ x + y # Add (element-by-element) ```{index} single: Vectors; Norm ``` -The **inner product** of vectors $x,y \in \mathbb R^n$ is defined as +向量 $x,y \in \mathbb R^n$ 的**内积**定义为 $$ x^\top y = @@ -390,50 +380,48 @@ x^\top y = := \sum_{i=1}^n x_i y_i. $$ -The **norm** of a vector $x$ represents its "length" (i.e., its distance from -the zero vector) and is defined as +向量 $x$ 的**范数**表示其"长度"（即，其与零向量的距离），定义为 $$ \| x \| := \sqrt{x^\top x} := \left( \sum_{i=1}^n x_i^2 \right)^{1/2}. $$ -The expression $\| x - y\|$ can be thought of as the "distance" between $x$ and $y$. +表达式 $\| x - y\|$ 可以被理解为 $x$ 和 $y$ 之间的"距离"。 -The inner product and norm can be computed as follows +内积和范数可以按以下方式计算 ```{code-cell} ipython3 -np.sum(x*y) # Inner product of x and y +np.sum(x*y) # x和y的内积 ``` ```{code-cell} ipython3 -x @ y # Another way to compute the inner product +x @ y # 另外一种计算内积的办法 ``` ```{code-cell} ipython3 -np.sqrt(np.sum(x**2)) # Norm of x, method one +np.sqrt(np.sum(x**2)) # x的范数，方法一 ``` ```{code-cell} ipython3 -np.linalg.norm(x) # Norm of x, method two +np.linalg.norm(x) # x的范数，方法二 ``` -## Matrix operations +## 矩阵运算 ```{index} single: Matrix; Operations ``` -When we discussed linear price systems, we mentioned using matrix algebra. +当我们讨论线性价格系统时，我们提到了使用矩阵代数。 -Matrix algebra is similar to algebra for numbers. +矩阵代数类似于数字代数。 -Let's review some details. +让我们回顾一些细节。 -### Addition and scalar multiplication +### 加法和标量乘法 -Just as was the case for vectors, we can add, subtract and scalar multiply -matrices. +就像向量一样，我们可以对矩阵进行加法、减法和标量乘法。 -Scalar multiplication and addition are generalizations of the vector case: +标量乘法和加法是向量情况的推广： ```{prf:example} :label: le_ex_asm @@ -452,7 +440,7 @@ $$ $$ ``` -In general for a number $\gamma$ and any matrix $A$, +一般来说，对于任意数 $\gamma$ 和任意矩阵 $A$， $$ \gamma A = @@ -472,8 +460,7 @@ $$ ```{prf:example} :label: le_ex_ma -Consider this example of matrix addition, - +考虑这个矩阵加法的例子， $$ \begin{bmatrix} 1 & 5 \\ @@ -492,7 +479,7 @@ $$ $$ ``` -In general, +一般来说， $$ A + B = @@ -513,27 +500,22 @@ A + B = \end{bmatrix}. $$ -In the latter case, the matrices must have the same shape in order for the -definition to make sense. +在后一种情况下，矩阵必须具有相同的形状才能使定义有意义。 -### Matrix multiplication +### 矩阵乘法 -We also have a convention for *multiplying* two matrices. +我们还有一个*相乘*两个矩阵的约定。 -The rule for matrix multiplication generalizes the idea of inner products -discussed above. +矩阵乘法的规则推广了上面讨论的内积的概念。 -If $A$ and $B$ are two matrices, then their product $A B$ is formed by taking -as its $i,j$-th element the inner product of the $i$-th row of $A$ and the -$j$-th column of $B$. +如果 $A$ 和 $B$ 是两个矩阵，那么它们的乘积 $A B$ 的形成是通过取 $A$ 的第 $i$ 行和 $B$ 的第 $j$ 列的内积作为其第 $i,j$ 个元素。 -If $A$ is $n \times k$ and $B$ is $j \times m$, then to multiply $A$ and $B$ -we require $k = j$, and the resulting matrix $A B$ is $n \times m$. +如果 $A$ 是 $n \times k$ 的，$B$ 是 $j \times m$ 的，那么要相乘 $A$ 和 $B$，我们需要 $k = j$，而得到的矩阵 $A B$ 是 $n \times m$ 的。 ```{prf:example} :label: le_ex_2dmul -Here's an example of a $2 \times 2$ matrix multiplied by a $2 \times 1$ vector. +这里是一个 $2 \times 2$ 矩阵乘以 $2 \times 1$ 向量的例子。 $$ Ax = @@ -553,10 +535,9 @@ Ax = $$ ``` -As an important special case, consider multiplying $n \times k$ -matrix $A$ and $k \times 1$ column vector $x$. +作为一个重要的特殊情况，考虑将 $n \times k$ 矩阵 $A$ 和 $k \times 1$ 列向量 $x$ 相乘。 -According to the preceding rule, this gives us an $n \times 1$ column vector. +根据前面的规则，这给我们一个 $n \times 1$ 列向量。 ```{math} :label: la_atx @@ -585,7 +566,7 @@ A x = \end{bmatrix}}_{n \times 1} ``` -Here is a simple illustration of multiplication of two matrices. +下面展示了两个矩阵的乘法。 $$ AB = @@ -603,17 +584,16 @@ AB = \end{bmatrix} $$ -There are many tutorials to help you further visualize this operation, such as - -* [this one](http://www.mathsisfun.com/algebra/matrix-multiplying.html), or -* the discussion on the [Wikipedia page](https://en.wikipedia.org/wiki/Matrix_multiplication). +有许多教程可以帮助你进一步可视化这个操作，例如 +* [这个教程](http://www.mathsisfun.com/algebra/matrix-multiplying.html)，或者 +* [维基百科页面](https://en.wikipedia.org/wiki/Matrix_multiplication)上的讨论。 ```{note} -Unlike number products, $A B$ and $B A$ are not generally the same thing. +与数字乘积不同，$A B$ 和 $B A$ 通常不是同一件事。 ``` -One important special case is the [identity matrix](https://en.wikipedia.org/wiki/Identity_matrix), which has ones on the principal diagonal and zero elsewhere: +一个重要的特殊情况是[单位矩阵](https://en.wikipedia.org/wiki/Identity_matrix)，它在主对角线上有 1，其他地方都是 0： $$ I = @@ -624,21 +604,19 @@ $$ \end{bmatrix} $$ -It is a useful exercise to check the following: - -* if $A$ is $n \times k$ and $I$ is the $k \times k$ identity matrix, then $AI = A$, and -* if $I$ is the $n \times n$ identity matrix, then $IA = A$. +验证以下内容是一个有用的练习： +* 如果 $A$ 是 $n \times k$ 矩阵，$I$ 是 $k \times k$ 单位矩阵，那么 $AI = A$，并且 +* 如果 $I$ 是 $n \times n$ 单位矩阵，那么 $IA = A$。 -### Matrices in NumPy +### NumPy中的矩阵 ```{index} single: Matrix; Numpy ``` +NumPy 数组也被用作矩阵，并且对所有标准矩阵运算都有快速、高效的函数和方法。 -NumPy arrays are also used as matrices, and have fast, efficient functions and methods for all the standard matrix operations. - -You can create them manually from tuples of tuples (or lists of lists) as follows +你可以通过以下方式从元组的元组（或列表的列表）手动创建它们 ```{code-cell} ipython3 A = ((1, 2), @@ -657,20 +635,18 @@ type(A) A.shape ``` -The `shape` attribute is a tuple giving the number of rows and columns --- -see [here](https://python-programming.quantecon.org/numpy.html#shape-and-dimension) -for more discussion. +`shape` 属性是一个给出行数和列数的元组 --- +更多讨论请参见[这里](https://python-programming.quantecon.org/numpy.html#shape-and-dimension)。 -To get the transpose of `A`, use `A.transpose()` or, more simply, `A.T`. +要获得 `A` 的转置，使用 `A.transpose()` 或更简单地使用 `A.T`。 -There are many convenient functions for creating common matrices (matrices of zeros, -ones, etc.) --- see [here](https://python-programming.quantecon.org/numpy.html#creating-arrays). +有许多方便的函数用于创建常见矩阵（零矩阵、单位矩阵等） --- 请参见[这里](https://python-programming.quantecon.org/numpy.html#creating-arrays)。 -Since operations are performed elementwise by default, scalar multiplication and addition have very natural syntax. +由于默认情况下操作是按元素执行的，标量乘法和加法具有非常自然的语法。 ```{code-cell} ipython3 -A = np.identity(3) # 3 x 3 identity matrix -B = np.ones((3, 3)) # 3 x 3 matrix of ones +A = np.identity(3) # 3 x 3 单位矩阵 +B = np.ones((3, 3)) # 3 x 3 元素为一的矩阵 2 * A ``` @@ -678,25 +654,23 @@ B = np.ones((3, 3)) # 3 x 3 matrix of ones A + B ``` -To multiply matrices we use the `@` symbol. +我们用 `@` 来进行矩阵乘法。 ```{note} -In particular, `A @ B` is matrix multiplication, whereas `A * B` is element-by-element multiplication. +其中 `A @ B` 是矩阵乘法, 但是 `A * B`是每个元素之间的运算。 ``` -### Two good model in matrix form - -We can now revisit the two good model and solve {eq}`two_equilibrium` -numerically via matrix algebra. +### 矩阵形式的两种商品模型 -This involves some extra steps but the method is widely applicable --- as we -will see when we include more goods. +我们现在可以重新审视两种商品模型，并通过矩阵代数数值求解 {eq}`two_equilibrium` 方程。 -First we rewrite {eq}`two_eq_demand` as +这涉及一些额外的步骤，但这种方法广泛适用 --- 正如我们在包含更多商品时将看到的那样。 +首先，我们将 {eq}`two_eq_demand` 重写为 ```{math} :label: two_eq_demand_mat + q^d = D p + h \quad \text{where} \quad q^d = @@ -718,14 +692,15 @@ First we rewrite {eq}`two_eq_demand` as \end{bmatrix}. ``` -Recall that $p \in \mathbb{R}^{2}$ is the price of two goods. +回想一下，$p \in \mathbb{R}^{2}$ 是两种商品的价格。 -(Please check that $q^d = D p + h$ represents the same equations as {eq}`two_eq_demand`.) +（请检查 $q^d = D p + h$ 是否表示与 {eq}`two_eq_demand` 相同的方程。） -We rewrite {eq}`two_eq_supply` as +我们将 {eq}`two_eq_supply` 重写为 ```{math} :label: two_eq_supply_mat + q^s = C p \quad \text{where} \quad q^s = @@ -741,76 +716,71 @@ We rewrite {eq}`two_eq_supply` as \end{bmatrix}. ``` -Now equality of supply and demand can be expressed as $q^s = q^d$, or +现在供给和需求的相等可以表示为 $q^s = q^d$，或 $$ C p = D p + h. $$ -We can rearrange the terms to get +我们可以重新排列这些项得到 $$ (C - D) p = h. $$ -If all of the terms were numbers, we could solve for prices as $p = h / -(C-D)$. - -Matrix algebra allows us to do something similar: we can solve for equilibrium -prices using the inverse of $C - D$: +如果所有项都是数字，我们可以求解价格为 $p = h / (C-D)$。 +矩阵代数允许我们做类似的事情：我们可以使用 $C - D$ 的逆矩阵来求解均衡价格： ```{math} :label: two_matrix - p = (C - D)^{-1} h. -``` - -Before we implement the solution let us consider a more general setting. +p = (C - D)^{-1} h. +``` +在我们实施解决方案之前，让我们考虑一个更一般的设置。 -### More goods +### 更多商品 -It is natural to think about demand systems with more goods. +考虑有更多商品的需求系统是很自然的。 -For example, even within energy commodities there are many different goods, -including crude oil, gasoline, coal, natural gas, ethanol, and uranium. +例如，即使在能源商品中也有许多不同的商品， +包括原油、汽油、煤炭、天然气、乙醇和铀。 -The prices of these goods are related, so it makes sense to study them -together. +这些商品的价格是相关的，所以一起研究它们是有意义的。 -Pencil and paper methods become very time consuming with large systems. +对于大型系统，纸笔方法会变得非常耗时。 -But fortunately the matrix methods described above are essentially unchanged. +但幸运的是，上面描述的矩阵方法基本上保持不变。 -In general, we can write the demand equation as $q^d = Dp + h$, where +一般来说，我们可以将需求方程写为 $q^d = Dp + h$，其中 -* $q^d$ is an $n \times 1$ vector of demand quantities for $n$ different goods. -* $D$ is an $n \times n$ "coefficient" matrix. -* $h$ is an $n \times 1$ vector of constant values. +* $q^d$ 是一个 $n \times 1$ 的向量，表示 $n$ 种不同商品的需求量。 +* $D$ 是一个 $n \times n$ 的"系数"矩阵。 +* $h$ 是一个 $n \times 1$ 的常数值向量。 -Similarly, we can write the supply equation as $q^s = Cp + e$, where +类似地，我们可以将供给方程写为 $q^s = Cp + e$，其中 -* $q^s$ is an $n \times 1$ vector of supply quantities for the same goods. -* $C$ is an $n \times n$ "coefficient" matrix. -* $e$ is an $n \times 1$ vector of constant values. +* $q^s$ 是一个 $n \times 1$ 的向量，表示相同商品的供给量。 +* $C$ 是一个 $n \times n$ 的"系数"矩阵。 +* $e$ 是一个 $n \times 1$ 的常数值向量。 -To find an equilibrium, we solve $Dp + h = Cp + e$, or +为了找到均衡，我们求解 $Dp + h = Cp + e$，或 ```{math} :label: n_eq_sys_la + (D- C)p = e - h. ``` -Then the price vector of the n different goods is +那么，n 种不同商品的价格向量是 $$ p = (D- C)^{-1}(e - h). $$ +### 一般线性系统 -### General linear systems - -A more general version of the problem described above looks as follows. +上述问题的一个更一般版本看起来如下。 ```{math} :label: la_se @@ -822,22 +792,21 @@ A more general version of the problem described above looks as follows. \end{matrix} ``` -The objective here is to solve for the "unknowns" $x_1, \ldots, x_n$. +这里的目标是解出"未知数" $x_1, \ldots, x_n$。 -We take as given the coefficients $a_{11}, \ldots, a_{nn}$ and constants $b_1, \ldots, b_n$. +我们给定系数 $a_{11}, \ldots, a_{nn}$ 和常数 $b_1, \ldots, b_n$。 -Notice that we are treating a setting where the number of unknowns equals the -number of equations. +注意，我们处理的是未知数数量等于方程数量的情况。 -This is the case where we are most likely to find a well-defined solution. +这是我们最有可能找到明确定义解的情况。 -(The other cases are referred to as [overdetermined](https://en.wikipedia.org/wiki/Overdetermined_system) and [underdetermined](https://en.wikipedia.org/wiki/Underdetermined_system) systems -of equations --- we defer discussion of these cases until later lectures.) +（其他情况被称为[超定](https://en.wikipedia.org/wiki/Overdetermined_system)和[欠定](https://en.wikipedia.org/wiki/Underdetermined_system)方程组 --- 我们将在后续讲座中讨论这些情况。） -In matrix form, the system {eq}`la_se` becomes +用矩阵形式表示，方程组 {eq}`la_se` 变为 ```{math} :label: la_gf + A x = b \quad \text{where} \quad A = @@ -854,50 +823,43 @@ In matrix form, the system {eq}`la_se` becomes b_n \end{bmatrix}. ``` +例如，{eq}`n_eq_sys_la` 具有这种形式，其中 -```{prf:example} -:label: le_ex_gls -For example, {eq}`n_eq_sys_la` has this form with - -$$ - A = D - C, +$$ + A = D - C, \quad b = e - h - \quad \text{and} \quad + \quad \text{和} \quad x = p. $$ -``` - -When considering problems such as {eq}`la_gf`, we need to ask at least some of -the following questions -* Does a solution actually exist? -* If a solution exists, how should we compute it? +当考虑诸如 {eq}`la_gf` 这样的问题时，我们至少需要问以下一些问题： +* 解是否真的存在？ +* 如果解存在，我们应该如何计算它？ - -## Solving systems of equations +## 解方程组 ```{index} single: Matrix; Solving Systems of Equations ``` -Recall again the system of equations {eq}`la_se`, which we write here again as +再次回顾方程组 {eq}`la_se`，我们在此重新写为 ```{math} :label: la_se2 - A x = b. + + A x = b ``` -The problem we face is to find a vector $x \in \mathbb R^n$ that solves -{eq}`la_se2`, taking $b$ and $A$ as given. +我们面临的问题是找到一个向量 $x \in \mathbb R^n$，使其解决 {eq}`la_se2` ，其中 $b$ 和 $A$ 是给定的。 -We may not always find a unique vector $x$ that solves {eq}`la_se2`. +我们可能并不总能找到一个唯一的向量 $x$ 来解决 {eq}`la_se2` 。 -We illustrate two such cases below. +我们在下面举例说明两种这样的情况。 -### No solution +### 无解 -Consider the system of equations given by, +考虑由以下给出的方程组： $$ \begin{aligned} @@ -906,9 +868,9 @@ $$ \end{aligned} $$ -It can be verified manually that this system has no possible solution. +可以手动验证这个系统没有可能的解。 -To illustrate why this situation arises let's plot the two lines. +为了说明为什么会出现这种情况，让我们绘制这两条直线。 ```{code-cell} ipython3 fig, ax = plt.subplots() @@ -921,15 +883,16 @@ plt.show() +++ {"tags": []} -Clearly, these are parallel lines and hence we will never find a point $x \in \mathbb{R}^2$ -such that these lines intersect. -Thus, this system has no possible solution. +显然，这些是平行线，因此我们永远无法找到一个点 $x \in \mathbb{R}^2$ 使得这些线相交。 + +因此，这个系统没有可能的解。 -We can rewrite this system in matrix form as +我们可以将这个系统用矩阵形式重写为 ```{math} :label: no_soln + A x = b \quad \text{where} \quad A = @@ -945,25 +908,21 @@ We can rewrite this system in matrix form as \end{bmatrix}. ``` -It can be noted that the $2^{nd}$ row of matrix $A = (2, 6)$ is just a scalar multiple of the $1^{st}$ row of matrix $A = (1, 3)$. - -The rows of matrix $A$ in this case are called **linearly dependent.** - +可以注意到，矩阵 $A$ 的第 $2$ 行 $(2, 6)$ 只是第 $1$ 行 $(1, 3)$ 的标量倍数。 +在这种情况下，矩阵 $A$ 的行被称为**线性相关的。** ```{note} -Advanced readers can find a detailed explanation of linear dependence and -independence [here](https://python.quantecon.org/linear_algebra.html#linear-independence). +高级读者可以在[这里](https://python.quantecon.org/linear_algebra.html#linear-independence)找到关于线性相关和线性无关的详细解释。 -But these details are not needed in what follows. +但在接下来的内容中不需要这些细节。 ``` +### 多解 -### Many solutions - -Now consider, +现在考虑， $$ \begin{aligned} @@ -972,14 +931,15 @@ $$ \end{aligned} $$ -Any vector $v = (x,y)$ such that $x = 2y - 4$ will solve the above system. +任何满足 $x = 2y - 4$ 的向量 $v = (x,y)$ 都将解决上述系统。 -Since we can find infinite such vectors this system has infinitely many solutions. +由于我们可以找到无限多个这样的向量，这个系统有无穷多个解。 -This is because the rows of the corresponding matrix +这是因为对应矩阵的行 ```{math} :label: many_solns + A = \begin{bmatrix} 1 & -2 \\ @@ -987,17 +947,15 @@ This is because the rows of the corresponding matrix \end{bmatrix}. ``` -are linearly dependent --- can you see why? +是线性相关的 --- 你能看出为什么吗？ -We now impose conditions on $A$ in {eq}`la_se2` that rule out these problems. +我们现在对 {eq}`la_se2` 中的 $A$ 施加条件，以排除这些问题。 +### 非奇异矩阵 -### Nonsingular matrices +对于每个方阵，我们都可以指定一个唯一的数，称为[行列式](https://en.wikipedia.org/wiki/Determinant)。 -To every square matrix we can assign a unique number called the -[determinant](https://en.wikipedia.org/wiki/Determinant). - -For $2 \times 2$ matrices, the determinant is given by, +对于 $2 \times 2$ 矩阵，行列式由以下公式给出： $$ \begin{bmatrix} @@ -1008,56 +966,49 @@ $$ {\color{red}{ad}} - {\color{blue}{bc}}. $$ -If the determinant of $A$ is not zero, then we say that $A$ is *nonsingular*. +如果 $A$ 的行列式不为零，我们就说 $A$ 是*非奇异的*。 -A square matrix $A$ is nonsingular if and only if the rows and columns of $A$ -are linearly independent. +当且仅当 $A$ 的行和列是线性无关的，方阵 $A$ 才是非奇异的。 -A more detailed explanation of matrix inverse can be found [here](https://www.mathsisfun.com/algebra/matrix-inverse.html). +关于矩阵逆的更详细解释可以在[这里](https://www.mathsisfun.com/algebra/matrix-inverse.html)找到。 -You can check yourself that the in {eq}`no_soln` and {eq}`many_solns` with -linearly dependent rows are singular matrices. +你可以自己检查 {eq}`no_soln` 和 {eq}`many_solns` 中具有线性相关行的矩阵是奇异矩阵。 -This gives us a useful one-number summary of whether or not a square matrix -can be inverted. +这为我们提供了一个有用的单数值概括，用来判断一个方阵是否可逆。 -In particular, a square matrix $A$ has a nonzero determinant, if and only if -it possesses an *inverse matrix* $A^{-1}$, with the property that $A A^{-1} = -A^{-1} A = I$. +特别地，方阵 $A$ 具有非零行列式，当且仅当它具有*逆矩阵* $A^{-1}$，满足 $A A^{-1} = A^{-1} A = I$。 -As a consequence, if we pre-multiply both sides of $Ax = b$ by $A^{-1}$, we -get +因此，如果我们用 $A^{-1}$ 左乘 $Ax = b$ 的两边，我们得到 ```{math} :label: la_se_inv + x = A^{-1} b. ``` -This is the solution to $Ax = b$ --- the solution we are looking for. - +这是对 $Ax = b$ 的解答 --- 这就是我们要寻找的解。 - -### Linear equations with NumPy +### 使用NumPy求解线性方程 ```{index} single: Linear Algebra; SciPy ``` -In the two good example we obtained the matrix equation, +在两个好的例子中，我们得到了矩阵方程： $$ p = (C-D)^{-1} h. $$ -where $C$, $D$ and $h$ are given by {eq}`two_eq_demand_mat` and {eq}`two_eq_supply_mat`. +其中 $C$、$D$ 和 $h$ 由 {eq}`two_eq_demand_mat` 和 {eq}`two_eq_supply_mat` 给出。 -This equation is analogous to {eq}`la_se_inv` with $A = (C-D)^{-1}$, $b = h$, and $x = p$. +这个方程类似于 {eq}`la_se_inv`，其中 $A = (C-D)^{-1}$，$b = h$，且 $x = p$。 -We can now solve for equilibrium prices with NumPy's `linalg` submodule. +我们现在可以使用NumPy的`linalg`子模块求解均衡价格。 -All of these routines are Python front ends to time-tested and highly optimized FORTRAN code. +所有这些程序都是经过时间检验和高度优化的FORTRAN代码的Python前端。 ```{code-cell} ipython3 -C = ((10, 5), # Matrix C +C = ((10, 5), # 矩阵 C (5, 10)) ``` @@ -1068,68 +1019,64 @@ C = np.array(C) ``` ```{code-cell} ipython3 -D = ((-10, -5), # Matrix D +D = ((-10, -5), # 矩阵 D (-1, -10)) D = np.array(D) ``` ```{code-cell} ipython3 -h = np.array((100, 50)) # Vector h -h.shape = 2,1 # Transforming h to a column vector +h = np.array((100, 50)) # 向量 h +h.shape = 2,1 # 将h转换为列向量 ``` ```{code-cell} ipython3 from numpy.linalg import det, inv A = C - D -# Check that A is nonsingular (non-zero determinant), and hence invertible +#检查A是否为奇异矩阵（行列式是否为零），是否可逆 det(A) ``` ```{code-cell} ipython3 -A_inv = inv(A) # compute the inverse +A_inv = inv(A) #计算逆矩阵 A_inv ``` ```{code-cell} ipython3 -p = A_inv @ h # equilibrium prices +p = A_inv @ h #均衡价格 p ``` ```{code-cell} ipython3 -q = C @ p # equilibrium quantities +q = C @ p # 均衡数量 q ``` -Notice that we get the same solutions as the pencil and paper case. +注意，我们得到的解与纸笔计算的情况相同。 -We can also solve for $p$ using `solve(A, h)` as follows. +我们还可以使用 `solve(A, h)` 来求解 $p$，如下所示。 ```{code-cell} ipython3 from numpy.linalg import solve -p = solve(A, h) # equilibrium prices +p = solve(A, h) # 均衡价格 p ``` ```{code-cell} ipython3 -q = C @ p # equilibrium quantities +q = C @ p # 均衡数量 q ``` +观察我们如何通过 `inv(A) @ y` 或使用 `solve(A, y)` 来求解 $x = A^{-1} y$。 -Observe how we can solve for $x = A^{-1} y$ by either via `inv(A) @ y`, or using `solve(A, y)`. - -The latter method uses a different algorithm that is numerically more stable and hence should be the default option. - - +后一种方法使用了一种不同的算法，在数值上更加稳定，因此应该是默认选项。 -## Exercises +## 练习 ```{exercise-start} :label: lin_eqs_ex1 ``` +让我们考虑一个有3种商品的市场 - 商品0、商品1和商品2。 -Let's consider a market with 3 commodities - good 0, good 1 and good 2. - -The demand for each good depends on the price of the other two goods and is given by: +每种商品的需求取决于其他两种商品的价格，由以下公式给出： $$ \begin{aligned} @@ -1139,9 +1086,9 @@ $$ \end{aligned} $$ -(Here demand decreases when own price increases but increases when prices of other goods increase.) +（这里，当自身价格上涨时需求下降，但当其他商品价格上涨时需求增加。） -The supply of each good is given by: +每种商品的供给由以下公式给出： $$ \begin{aligned} @@ -1151,19 +1098,17 @@ $$ \end{aligned} $$ -Equilibrium holds when supply equals demand, i.e, $q_0^d = q_0^s$, $q_1^d = q_1^s$ and $q_2^d = q_2^s$. +当供给等于需求时，市场达到均衡，即 $q_0^d = q_0^s$，$q_1^d = q_1^s$ 和 $q_2^d = q_2^s$。 -1. Set up the market as a system of linear equations. -2. Use matrix algebra to solve for equilibrium prices. Do this using both the `numpy.linalg.solve` - and `inv(A)` methods. Compare the solutions. +1. 将市场设置为线性方程组。 +2. 使用矩阵代数求解均衡价格。分别使用 `numpy.linalg.solve` 和 `inv(A)` 方法来做这个。比较这两种解法。 ```{exercise-end} ``` ```{solution-start} lin_eqs_ex1 :class: dropdown ``` - -The generated system would be: +生成的系统将是： $$ \begin{aligned} @@ -1173,11 +1118,11 @@ $$ \end{aligned} $$ -In matrix form we will write this as: +用矩阵形式，我们将其表示为： $$ Ap = b -\quad \text{where} \quad +\quad \text{其中} \quad A = \begin{bmatrix} 35 & -5 & -5 \\ @@ -1190,7 +1135,7 @@ A = p_1 \\ p_2 \end{bmatrix} -\quad \text{and} \quad +\quad \text{且} \quad b = \begin{bmatrix} 100 \\ @@ -1203,18 +1148,18 @@ $$ import numpy as np from numpy.linalg import det -A = np.array([[35, -5, -5], # matrix A +A = np.array([[35, -5, -5], # 矩阵 A [-5, 25, -10], [-5, -5, 15]]) -b = np.array((100, 75, 55)) # column vector b +b = np.array((100, 75, 55)) # 列向量 b b.shape = (3, 1) -det(A) # check if A is nonsingular +det(A) # 检查A是否为奇异矩阵 ``` ```{code-cell} -# Using inverse +# 使用inverse from numpy.linalg import det A_inv = inv(A) @@ -1224,13 +1169,14 @@ p ``` ```{code-cell} -# Using numpy.linalg.solve +# 使用 numpy.linalg.solve from numpy.linalg import solve p = solve(A, b) p ``` -The solution is given by: +答案为： + $$ p_0 = 4.6925, \; p_1 = 7.0625 \;\; \text{and} \;\; p_2 = 7.675 $$ @@ -1241,64 +1187,61 @@ $$ ```{exercise-start} :label: lin_eqs_ex2 ``` -Earlier in the lecture we discussed cases where the system of equations given by $Ax = b$ has no solution. +在讲座的早些时候，我们讨论了$Ax = b$这个方程组没有解的情况。 -In this case $Ax = b$ is called an _inconsistent_ system of equations. +在这种情况下，$Ax = b$被称为*不相容*方程组。 -When faced with an inconsistent system we try to find the best "approximate" solution. +面对不相容系统时，我们尝试找到最佳的"近似"解。 -There are various methods to do this, one such method is the **method of least squares.** +有多种方法可以做到这一点，其中一种是**最小二乘法**。 -Suppose we have an inconsistent system +假设我们有一个不相容系统 ```{math} :label: inconsistent + Ax = b ``` -where $A$ is an $m \times n$ matrix and $b$ is an $m \times 1$ column vector. -A **least squares solution** to {eq}`inconsistent` is an $n \times 1$ column vector $\hat{x}$ such that, for all other vectors $x \in \mathbb{R}^n$, the distance from $A\hat{x}$ to $b$ -is less than the distance from $Ax$ to $b$. +其中$A$是一个$m \times n$矩阵，$b$是一个$m \times 1$列向量。 + +对于{eq}`inconsistent`，**最小二乘解**是一个$n \times 1$列向量$\hat{x}$，使得对于所有其他向量$x \in \mathbb{R}^n$，$A\hat{x}$到$b$的距离小于$Ax$到$b$的距离。 -That is, +即， $$ \|A\hat{x} - b\| \leq \|Ax - b\| $$ -It can be shown that, for the system of equations $Ax = b$, the least squares -solution $\hat{x}$ is +可以证明，对于方程组$Ax = b$，最小二乘解$\hat{x}$是 ```{math} :label: least_squares + \hat{x} = (A^T A)^{-1} A^T b ``` -Now consider the general equation of a linear demand curve of a good given by: +现在考虑一种商品的线性需求曲线的一般方程： $$ p = m - nq $$ -where $p$ is the price of the good and $q$ is the quantity demanded. +其中$p$是商品的价格，$q$是需求量。 -Suppose we are trying to *estimate* the values of $m$ and $n$. +假设我们正试图*估计*$m$和$n$的值。 -We do this by repeatedly observing the price and quantity (for example, each -month) and then choosing $m$ and $n$ to fit the relationship between $p$ and -$q$. +我们通过重复观察价格和数量（例如，每个月）来做到这一点，然后选择$m$和$n$来拟合$p$和$q$之间的关系。 -We have the following observations: +我们有以下观察结果： -| Price | Quantity Demanded | +| 价格 | 需求量 | |:-----:|:-----------------:| | 1 | 9 | | 3 | 7 | | 8 | 3 | - -Requiring the demand curve $p = m - nq$ to pass through all these points leads to the -following three equations: +要求需求曲线$p = m - nq$通过所有这些点，得到以下三个方程： $$ \begin{aligned} @@ -1308,17 +1251,18 @@ $$ \end{aligned} $$ -Thus we obtain a system of equations $Ax = b$ where $A = \begin{bmatrix} 1 & -9 \\ 1 & -7 \\ 1 & -3 \end{bmatrix}$, -$x = \begin{bmatrix} m \\ n \end{bmatrix}$ and $b = \begin{bmatrix} 1 \\ 3 \\ 8 \end{bmatrix}$. +因此，我们得到一个方程组$Ax = b$，其中$A = \begin{bmatrix} 1 & -9 \\ 1 & -7 \\ 1 & -3 \end{bmatrix}$， +$x = \begin{bmatrix} m \\ n \end{bmatrix}$，$b = \begin{bmatrix} 1 \\ 3 \\ 8 \end{bmatrix}$。 +可以验证这个系统没有解。 + +（问题在于我们有三个方程但只有两个未知数。） -It can be verified that this system has no solutions. +因此，我们将尝试找到$x$的最佳近似解。 -(The problem is that we have three equations and only two unknowns.) +1. 使用{eq}`least_squares`和矩阵代数找到最小二乘解$\hat{x}$。 -We will thus try to find the best approximate solution for $x$. +2. 使用`numpy.linalg.lstsq`找到最小二乘解，并比较结果。 -1. Use {eq}`least_squares` and matrix algebra to find the least squares solution $\hat{x}$. -2. Find the least squares solution using `numpy.linalg.lstsq` and compare the results. ```{exercise-end} ``` @@ -1333,14 +1277,14 @@ from numpy.linalg import inv ``` ```{code-cell} ipython3 -# Using matrix algebra -A = np.array([[1, -9], # matrix A +# 运用线性代数 +A = np.array([[1, -9], # 矩阵 A [1, -7], [1, -3]]) -A_T = np.transpose(A) # transpose of matrix A +A_T = np.transpose(A) # 矩阵A的转置 -b = np.array((1, 3, 8)) # column vector b +b = np.array((1, 3, 8)) # 列向量 b b.shape = (3, 1) x = inv(A_T @ A) @ A_T @ b @@ -1348,7 +1292,7 @@ x ``` ```{code-cell} ipython3 -# Using numpy.linalg.lstsq +# 使用 numpy.linalg.lstsq x, res, _, _ = np.linalg.lstsq(A, b, rcond=None) ``` @@ -1358,10 +1302,10 @@ x, res, _, _ = np.linalg.lstsq(A, b, rcond=None) print(f"x\u0302 = {x}") print(f"\u2016Ax\u0302 - b\u2016\u00B2 = {res[0]}") ``` +这是一个可视化图，展示了最小二乘法如何近似一组点之间连线的方程。 -Here is a visualization of how the least squares method approximates the equation of a line connecting a set of points. +我们也可以将此描述为在一组点之间"拟合"一条直线。 -We can also describe this as "fitting" a line between a set of points. ```{code-cell} ipython3 fig, ax = plt.subplots() @@ -1370,10 +1314,10 @@ q = np.array((9, 7, 3)) a, b = x -ax.plot(q, p, 'o', label='observations', markersize=5) -ax.plot(q, a - b*q, 'r', label='Fitted line') -plt.xlabel('quantity demanded') -plt.ylabel('price') +ax.plot(q, p, 'o', label='观测点', markersize=5) +ax.plot(q, a - b*q, 'r', label='拟合线') +plt.xlabel('需求数量') +plt.ylabel('价格') plt.legend() plt.show() ``` @@ -1381,9 +1325,8 @@ plt.show() ```{solution-end} ``` +### 延伸阅读 -### Further reading - -The documentation of the `numpy.linalg` submodule can be found [here](https://numpy.org/devdocs/reference/routines.linalg.html). +`numpy.linalg` 子模块的文档可以在[这里](https://numpy.org/devdocs/reference/routines.linalg.html)找到。 -More advanced topics in linear algebra can be found [here](https://python.quantecon.org/linear_algebra.html#id5). +线性代数的更高级主题可以在[这里](https://python.quantecon.org/linear_algebra.html#id5)找到。