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content/notes/math/linear_algebra.mdx

@@ -5661,12 +5661,64 @@ $$
 Let $V$ be an $\mathbb{F}$-vector space. The map $\langle\cdot, \cdot\rangle: V\times V \to\mathbb{F}$ is called an inner product on $V$ if it satisfies for all $\boldsymbol{u},\boldsymbol{v}, \boldsymbol{w}\in V$ and $\alpha, \beta \in\mathbb{F}$
 
 1. **Positive definiteness:** $\langle \boldsymbol{v}, \boldsymbol{v} \rangle \geq 0$ and $\langle \boldsymbol{v}, \boldsymbol{v} \rangle = 0 \implies \boldsymbol{v} = 0$
-2. **Linearity in the second argument:** $\langle\boldsymbol{u}, \alpha\boldsymbol{v} + \beta\boldsymbol{w}\rangle = \alpha\langle\boldsymbol{u}, \boldsymbol{v}\rangle + \beta\langle\boldsymbol{u}, \boldsymbol{w}\rangle$
+2. **Linearity in the first argument:** $\langle \alpha\boldsymbol{u} + \beta\boldsymbol{v}, \boldsymbol{w} \rangle = \alpha\langle\boldsymbol{u}, \boldsymbol{w}\rangle + \beta\langle\boldsymbol{v}, \boldsymbol{w}\rangle$
 3. **Conjugate symmetry:** $\langle \boldsymbol{v}, \boldsymbol{w}\rangle = \overline{\langle\boldsymbol{w}, \boldsymbol{v}\rangle}$
+    - For $\mathbb{F} = \R$, the inner product is symmetric, i.e. $\langle \boldsymbol{v}, \boldsymbol{w}\rangle = \langle\boldsymbol{w}, \boldsymbol{v}\rangle$
+
+The pair $(V,\langle\cdot, \cdot\rangle)$ is called an *inner product space*. Note that the inner product can also be defined with linearity the in second argument instead.
+</MathBox>
+
+If $\mathbb{F} = \R$, the inner product is symmetric making it linear in both arguments. In this case, the inner product is *bilinear*. However, if $\mathbb{F} = \mathbb{C}$, linearity in the first argument and conjugate symmetry implies conjugate linearity in the second argument, i.e.
+
+$$
+\begin{align*}
+  \langle \mathbf{u}, \alpha\mathbf{v} + \beta\mathbf{w} \rangle =& \overline{\langle \alpha\mathbf{v} + \beta\mathbf{w}, \mathbf{u}, \rangle} \\
+  =& \bar{\alpha}\overline{\langle\mathbf{v},\mathbf{u}\rangle} + \bar{\beta}\overline{\langle\mathbf{w},\mathbf{u}\rangle} \\
+  =& \bar{\alpha}\langle\mathbf{u},\mathbf{v}\rangle + \bar{\beta}\langle\mathbf{u},\mathbf{w}\rangle
+\end{align*}
+$$
+
+Thus, a complex inner product is linear in the first argument and conjugate linear in the second argument. This property is referred to as *sesquilinearity*.
+
+<MathBox title='' boxType='lemma'>
+If $V$ is an inner product space and $\langle\mathbf{u},\mathbf{x}\rangle = \langle\mathbf{v},\mathbf{x}\rangle$ for all $\mathbf{x}\in V$, then $\mathbf{u} = \mathbf{v}$.
+</MathBox>
+
+<MathBox title='' boxType='proposition'>
+Let $V$ be an $\mathbb{F}$-inner product space and let $\mathrm{T}\in\mathcal{L}(V)$ be a linear operator on $V$.
+1. $\langle\mathrm{T}\mathbf{v},\mathbf{w}\rangle = 0 \implies \mathrm{T} = 0,\; \forall \mathbf{v},\mathbf{w}\in V$
+2. If $\mathbb{F} = \mathbb{C}$, then $\langle\mathrm{T}\mathbf{v},\mathbf{v}\rangle = 0 \implies \mathrm{T} = 0,\; \forall \mathbf{v}\in V$. This does not hold in general for real inner product spaces.
+
+<details>
+<summary>Proof</summary>
+
+**(1):** This follows directly from the previous lemma.
+
+**(2):** Let $\mathbf{v} = \alpha\mathbf{x} + \mathbf{y}$ for $\mathbf{x},\mathbf{y}\in V$ and $\alpha\in\mathbb{C}$, then
+
+$$
+\begin{align*}
+  0 =& \langle \mathrm{T}(\alpha\mathbf{x} + \mathbf{y}), \alpha\mathbf{x} + \mathbf{y} \rangle \\
+  =& |\alpha^2| \langle \mathrm{T}\mathbf{x}, \mathbf{x}\rangle + \langle\mathrm{T}\mathbf{y}, \mathbf{y}\rangle + \alpha\langle\mathrm{T}\mathbf{x},\mathbf{y}\rangle + \bar{\alpha}\langle\mathrm{T}\mathbf{y},\mathbf{x}\rangle \\
+  =& \alpha\langle\mathrm{T}\mathbf{x},\mathbf{y}\rangle + \bar{\alpha}\langle\mathrm{T}\mathbf{y},\mathbf{x}\rangle
+\end{align*}
+$$
+
+Setting $r = 1$ and $r = i$ gives, respectively
+
+$$
+\begin{align*}
+  \langle\mathrm{T}\mathbf{x},\mathbf{y}\rangle + \langle\mathrm{T}\mathbf{y},\mathbf{x}\rangle =& 0 \\
+  \langle\mathrm{T}\mathbf{x},\mathbf{y}\rangle - \langle\mathrm{T}\mathbf{y},\mathbf{x}\rangle =& 0 \\
+\end{align*}
+$$
+
+These two equations imply that $\langle\mathrm{T}\mathbf{x},\mathbf{y}\rangle = 0$ for all $\mathbf{x},\mathbf{y}\in V$ and it follows by **(1)** that $\mathrm{T} = 0$. For the last statement, rotation by $90\degree$ in the $\R^2$ plane has the property that $\langle\mathrm{T}\mathbf{v}, \mathbf{v}\rangle = 0$ for all $\mathbf{v}\in V$.
+</details>
 </MathBox>
 
 <MathBox title='Norm' boxType='definition'>
-Let $\langle\cdot, \cdot\rangle: V\times V \to\mathbb{F}$ be an inner product on the $\mathbb{F}$-vector space $V$. The *norm* on $V$ is a map $\lVert\cdot\rVert: V\to[0,\infty)$ defined by
+Let $V$ be an $\mathbb{F}$-inner product space. The *norm* on $V$ is a map $\lVert\cdot\rVert: V\to[0,\infty)$ defined by
 
 $$
   \lVert \boldsymbol{x} \rVert := \sqrt{\langle \boldsymbol{x}, \boldsymbol{x} \rangle}