chap7.tex

%!TEX root = ./main.tex
\section{Operators on Inner Product Spaces}
\subsection{Self-Adjoint and Normal Operators}
\begin{definition}
	Suppose $T \in L(V,W)$, we define $T*$ by this formula 
	\[ \la T\vec v, \vec w \ra_W = \la \vec v, T^* \vec w \ra_V\]
\end{definition}
We can think og $\vec w$ as fixed, we note that $\la T \ast, \vec w \ra$ is a linear functiona;. hence it has a representer by Riesz, so we are entitled to call it $T^*\vec w$ such that 
\[ \la T\vec v, \vec w \ra = \la \vec v, T^* \vec w \ra \]
Hence we have $T^* \in \c L(W,V)$. We want to verify this property. Consider $T*(\vec w_1 + \lambda \vec w_2)$. We compute 
\begin{align*}
	\la \vec v, T*(\vec w_1 + \lambda \vec w_2) \ra &= \la T\vec v, \vec w_1 + \lambda \vec w_2\ra \\
	&= \la T\vec v, \vec w_1 \ra + \bar \lambda \la T\vec v + \vec w_2 \ra \\
	&= \la \vec v, T^*\vec w_1 \ra +  \bar \lambda \la \vec v, T^* \vec w_2 \ra \\
	&= \la \vec v, T^*\vec w_1 \ra +   \la \vec v, \lambda T^* \vec w_2 \ra \\
	&= \la \vec v, T^*\vec w_1 +  \lambda T^*w_2 \ \ra \forall \vec v \in V, \vec w_1, \vec w_2 \in W, \lambda \in \b F 
\end{align*}
So $T^*(\vec w_1 + \lambda \vec w_2) = T^* \vec w_1 + \lambda T^* \vec w_2$.

\begin{theorem}[Properties of the adjoint]
Let $T \in \c L(V,W)$, we have
	\begin{enumerate}
		\item $(S + T)^* = S^* + T^*$
		\item $(\lambda T)^* = \bar \lambda T^*$
		\item $(S \cdot T)^* = T^*S^*$
		\item $(\lambda T)^* = \bar \lambda T^*$
		\item $\b I^* = \b I$
	\end{enumerate}
\end{theorem}
\begin{proof} Refer to book page 206
\end{proof}
\begin{theorem}[Null space and range of adjoint]
Let $T \in \c L(V,W)$, then
	\begin{enumerate}
		\item $\nul T^* = (\range T)^\perp$
		\item $\range T^* = (\nul T)^\perp$
		\item $\nul T = (\range T*)^\perp$
		\item $\range T = (\nul T^*)^\perp$
	\end{enumerate}
\end{theorem}
\begin{proof}
Let $\vec w \in W$. Then 
\begin{align*}
	\vec w \in \nul T^* &\iff T^* \vec w = 0 \\
	&\iff \la \vec v, T^* \vec w \ra = 0 \ \forall \vec v \in V \\
	&\iff \la T\vec v, \vec w \ra \ \forall \vec v \in V \\
	&\iff \vec w \in (\range T)^\perp
\end{align*}
(b),(c),(d) follows by a similar logic and is left as an exercise.
\end{proof}

\begin{definition}
Let $T \in \c L(V)$. $T$ is called self-adjoint if $T^* = T$.
\end{definition}
\begin{definition}
	$T$ is called normal if $TT^* = T^*T$.
\end{definition}
%\begin{example}
%	Let $V$ be $\b R^3$ with the standard inner product, let $T \in \c L(V)$ such that $T: (x_1,x_2,x_3) \mapsto (x_1,2x_2,3x_3)$.
%\end{example}
\begin{example}
	Suppose $T \in \c L(V)$, we can define $T: \vec v \mapsto \la \vec , \vec x \ra \vec y$ for some fixed $\vec x, \vec y$ in $V$. Compute $T^*$. \\
	We compute 
	\begin{align*}
		\langle \vec v, T^* \vec w \rangle &= \langle T \vec v, \vec w \rangle  \\
		&= \langle \langle \vec v, \vec x \rangle \vec y, \vec w \rangle  \\
		&= \langle \vec v, \vec x \rangle \langle \vec y, \vec w \rangle \\
		 &= \langle \vec v, \overline{\langle \vec y, \vec w \rangle} \vec x \rangle \\
		 &= \langle \vec v, \langle \vec w, \vec y \rangle \vec x \rangle
	\end{align*}
	hence we can concldue $T^* \vec w = \la \vec w, \vec y \ra \vec x$ for all $\vec w \in W$.
\end{example}
\newpage
\subsubsection{Matrix representation}
Suppose $T \in \c L(V,W)$, where $V,W$ are finite-dimensional vector spaces. Let $\li{\vec e}n$ be an orthonormal basis for $V$ and $\li{\vec f}m$ be a orthonormal basis for $W$. We can see that $\c M(T)$ is obtained through
\[ T \vec e_j = \la T\vec e_j, \vec f_1 \ra \vec f_1 + \la T\vec e_j, \vec f_2 \ra \vec f_2 + \cdots + \la T\vec e_j, \vec f_m \ra \vec f_m\]
\[ T^* \vec f_k = \la T^* \vec f_k, \vec e_1 \ra \vec e_1 + \la T^* \vec f_k, \vec e_2 \ra \vec e_2 + \cdots + \la T^* \vec f_k, \vec e_n \ra \vec e_n\]
Then 
\[ \c M(T^*)(l,k) = \la T^* \vec f_k, \vec e_l \ra \implies \c M(T^*)(j,i) = \la T^* \vec f_i, \vec e_j \ra \]
Therefore we have $\c M(T^*) = \bar{\c M(T)}^T$
\begin{remark}
	The above statement only holds if the basis for $V$ and $W$ are orthonormal.
\end{remark}
\begin{remark}
	If an operator $T$ is self-adjoint, then $T$ is normal, but not the converse.
\end{remark}

\begin{proposition}
	The eigenvalue of any self-adjoint operator is real.
\end{proposition}
\begin{proof}
Suppose self-adjoint $T \in \c L(V)$. and $\lambda$ is an eigenvalue of $T$ and let $\vec v$ be the eignevector corespond to $\lambda$. We compute 
\[
	\lambda \la \vec v, \vec v \ra = \la \lambda \vec v, \vec v \ra 
	= \la T\vec v, \vec v \ra 
	= \la \vec v, T\vec v \ra 
	= \la \vec v, \lambda \vec v \ra 
	= \bar \lambda \la \vec v, \vec v \ra
\] We can see that $\lambda = \bar \lambda \implies \lambda \in \b R$.
\end{proof}
\begin{question}
	Suppose $\la T\vec v,\vec v \ra = 0 \forall \vec v \in V$. Does the statement implies $T$ is the zero map?
\end{question}
\begin{answer}[Surprisingly]
	Yes over $\b C$ and no over $\b R$.
\end{answer}
\begin{proof}
	Supoose $\b F = \b C$, the following holds.
	\[ \la T(\vec v + \vec w), \vec v + \vec w \ra = \la T\vec v, \vec v \ra + \la T\vec v, \vec w \ra + \la T\vec w,\vec v \ra + \la T\vec w, \vec w \ra \]
	\[ \la T(\vec v - \vec w), \vec v + \vec w \ra = \la T\vec v, \vec v \ra - \la T\vec v, \vec w \ra - \la T\vec w,\vec v \ra + \la T\vec w, \vec w \ra \]
	Subtratc the first equation by the second we have 
	\[ \la T(\vec v + \vec w), (\vec v + \vec w) \ra - \la T(\vec v - \vec w), (\vec v + \vec w) \ra =\boxed{ 2\la T\vec v, \vec w\ra + 2\la T\vec w, \vec v \ra} \]
	We also compute 
	\[ \la T(\vec v + i\vec w) , (\vec v + i\vec w)\ra - \la T(\vec v - i\vec w) , (\vec v - i\vec w)\ra = \boxed{2i\left(\la T\vec w, \vec v\ra - \la T\vec v, \vec w \ra\right)}\]
	Take the two boxed equation and divide the second one by $i$ then subtract from first gives us 
	\[ 4\la T \vec v, \vec w \ra = 0\]
	Suppose $\b F = \b R$. Consider $\b R^2$. Take $T\vec v$ and rotate $\pi / 2$ gives us $T(x_1, x_2) := (-x_2 , x_1)$. We can see that $\la T\vec v, \vec v \ra = 0 \forall \vec v$ but $T \neq 0$. However, if $T$ is self-adjoint then $T$ is $0$.
\end{proof}
\begin{remark}
	Suppose $\b F = \b R$ and $T = T^*$. We have 
	\[ 4\la T\vec v, \vec w \ra = \la T(\vec v + \vec w), (\vec v + \vec w) \ra - \la T(\vec v - \vec w), (\vec v - \vec w )\ra\]
	Hence $T = 0$. 
\end{remark}

\begin{corollary}
	$\la T \vec v, \vec v \ra \in \b R$ in a complex space is equivalent to $T$ being self adjoint.
\end{corollary}
\begin{proof}
	\[ \la T \vec v, \vec v \ra \in \b R \iff \la T\vec v, \vec v\ra = \la T^*\vec v, \vec v \ra \implies \la (T - T^*)\vec v, \vec v\ra = 0 \implies T - T^* = 0\]
	We can see that $T = T^*$. Hence $T$ is self-adjoint.
\end{proof}

\begin{theorem}
	$T$ is normal if anf only if $||T\vec v|| = ||T^* \vec v|| \ \forall \vec v \in V$.
\end{theorem}
\begin{proof}
	\[ ||T\vec v|| = ||T^* \vec v|| \implies \la T\vec v, T\vec v\ra = \la T^* \vec v, T^* \vec v \ra \implies \la T^* T \vec v, \vec v \ra = \la TT^* \vec v, \vec v \ra\]
	Hence $T$ is normal sicne $TT^* = T^*T$.
\end{proof}
\begin{theorem}
	Say $\lambda, \vec v$ is an eigenpair of a normal oeprator $T$, then 
	\[ ||(T - \lambda \b I) \vec v ||=||(T^* - \bar \lambda \b I)\vec v||\]
\end{theorem}
\subsection{Spectral Theorem}
\subsubsection*{Over Complex Vector Space}
\begin{theorem}[Spectral Theorem over Complex Vector Space]
	Suppose $T \in \c L(V)$ where $V$ is finite dimensioanl vector space and $\b F = \b C$ and $T$ is normal. Then $V$ is a orthonormal basis of eigenvactors of $T$, and vice versa,  if $T$ has a diagonal representation with respect to some orthonormal basis, then $T$ is normal.
\end{theorem}
\begin{proof}
	Suppose $T$ has a diagonal matrix representation with some repsect to some orthonormal basis. i.e. 
	\[ \c M(T) = \begin{bmatrix}
		\lambda_1 & & & 0 \\
		& \lambda_2 \\
		& & \ddots \\
		0& & & \lambda_n 
	\end{bmatrix}\qquad \c M(T^*) = \begin{bmatrix}
		\bar \lambda_1 & & & 0 \\
		& \bar \lambda_2 \\
		& & \ddots \\
		0& & & \bar \lambda_n 
	\end{bmatrix}\]
	Since any two diagonal matrices commute, we can see that 
	\[ \c M(T) \c M(T^*) = \c M(T^*) \c M(T) =  \begin{bmatrix}
		|\lambda_1|^2 & & & 0 \\
		& |\lambda_2|^2 \\
		& & \ddots \\
		0& & & |\lambda_n|^2 
	\end{bmatrix}\]
	We have $TT^* = T^*T$, hence $T$ is normal.

	\noindent Suppose $T$ is normal. By Schur's Theorem, there exists an orthonormal basis such that
	\[ \c M(T) = \begin{bmatrix}
	a_{11} & a_{12} & \cdots & a_{1n} \\
	0 & a_{22} & \cdots & a_{2n} \\
	\vdots & \vdots & \ddots & \vdots\\
	0 & 0 & \cdots & a_{nn} 
	\end{bmatrix} \implies \c M(T^*) = \begin{bmatrix}
	\bar a_{11} & \bar a_{12} & \cdots & \bar a_{1n} \\
	0 & \bar a_{22} & \cdots & \bar a_{2n} \\
	\vdots & \vdots & \ddots & \vdots\\
	0 & 0 & \cdots & \bar a_{nn} 
	\end{bmatrix} \]
	Rall that $||T\vec v|| = ||T^* \vec v|| \qquad \forall \vec v \in V$. Call this this orthornormal basis $\li{\vec v}n$. We have $T\vec e_1 = a_{11}\vec e_1$, so $||T\vec e_1|| = |a_{11}|$, we then compute 
	\begin{align*}
		T^* \vec e_1 &= \bar a_{11} \vec e_1 + \bar a_{12} \vec e_2 + \cdots + \bar a_{1n} \vec e_n \\
		||T^*\vec e_1 || &= \sqrt{|a_{11}|^2 + |a_{12}|^2 + \cdots + |a_{1n}|^2}
	\end{align*}
	Since $||T\vec e_1|| = ||T^* \vec e_1||$, we get $|a_{12}| = |a_{13}| = \cdots = |a_{1n}| = 0$. 

	\noindent
	Using a similar logic, we have $||T\vec e_j|| = ||T^*\vec e_j||$ implies $|a_{jj+1}| = |a_{jj+2}| = \cdots = |a_{jn}| = 0$. Hence $T$ is diagonal
\end{proof}
\begin{remark}
	So actaully the schur form of a normal operator is neccisarily diagonal.
\end{remark}
\newpage
\subsubsection*{Over Real Vector Space}
\begin{lemma}
	Suppose $T \in \c L(V)$ is self-adjoint and $\beta, \gamma \in \b R$ such that $b\beta^2 - 4 \gamma$ then
	\[ T^2 + \beta T + \gamma I\] is invertible.
\end{lemma}
\begin{proof}
Consider nonzero $\vec v \in V$. We can factor 
	\begin{align*}
		\la (T^2 + \beta T - \gamma I) \ra &= \la T^2 \vec v, \vec v \ra + \la \beta T \vec v, \vec v \ra + \gamma \la \vec v, \vec v\ra \\
		&= \la T\vec v, T\vec v \ra + \beta \la T\vec v, \vec v\ra + \gamma||\vec v||^2 \\
		&\geq ||T\vec v||^2 - |\beta| \cdot ||T\vec v|| \cdot ||\vec v|| + \gamma||\vec v||^2  \\
		&= \left( ||T\vec v|| - \frac{|\beta| \cdot ||\vec v||}{2} \right)^2 + \left( \gamma - \frac{\beta^2}{4} \right)||\vec v||^2 \\
		&> 0
	\end{align*}
	hence we can see that $\nul (T^2 + \beta T - \gamma I) = \lb 0 \rb$. Hence it's injective. Since $T^2 + \beta T + \gamma I \in \c L(V)$, we know $(T^2 + \beta T + \gamma I)$ is invertible. 
\end{proof}
\begin{theorem}
	$T$ has a eigenvalue if $T$ is self-adjoint in any vector space.
\end{theorem}
\begin{proof}
	Assume $\dim V = n$. Consider any $\vec v \in V$. Then $\vec v, T\vec v, T^2 \vec v, \ldots, T^n \vec n$ are linearly dependent ( I.e, there exist $\li an \in \b R$ such that 
	\[a_0 \vec v + a_1 T \vec v + \cdots + a_n T^\vec v = 0\]
	Consider $f(x) = a_0x + a_1x + \cdots + a_nx^n$. We know from chapter 4 we can factor $f(x)$ as 
	\[ f(x) = c(x^2 + \beta_1 x + \gamma_1) \cdots (x^2 + \beta_m x + \gamma_m) (x - \lambda_1) \cdots (x - \lambda_n)\]
	where all coefficients are real and $\beta_i^2 - 4 \gamma < 0$. By lemma we know that the qaudratice term is invertible, then we can simply factor them out. Therefore we have 
	\[ 0 = (T - \lambda_1 I) \cdots (T - \lambda_m I)\vec v\]
	Hence one of the $(T - \lambda_j I)$ is not injective. Hence $T$ has an eigenvalue.
\end{proof}
\begin{theorem}
	Suppose $T \in \c L(V)$, where $V$ is a fininite dimensional vector space and $\b F = \b R$ and $T$ is self-adjoint. Then $T$ has a digona matrix representation with some orthonorma basis for $V$. And conversely, if $T$ has a diagonal matrix representation eith repsetc to some orthonormal basis, then $T = T^*$.
\end{theorem}
\begin{proof}
	Suppose $T$ has a diagonal matrix representation with some repsect to some orthonormal basis. i.e. 
	\[ \c M(T) = \begin{bmatrix}
		\lambda_1 & & & 0 \\
		& \lambda_2 \\
		& & \ddots \\
		0& & & \lambda_n 
	\end{bmatrix}\qquad \c M(T^*) = \begin{bmatrix}
		\bar \lambda_1 & & & 0 \\
		& \bar \lambda_2 \\
		& & \ddots \\
		0& & & \bar \lambda_n 
	\end{bmatrix}\]
	We know $\c M(T) = \c M(T^*)$ since $\lambda = \bar \lambda$ in reals. Hence $T$ is self-adjoint.
	\noindent
	Conversely, suppose $T = T^*$. We just found out $T$ has at least one eigenvalue eigenvector pair. Say $T\vec u = \lambda \vec u$. Without the loss of generality $||\vec u|| = 1$. If $\vec w \perp \vec u$, then $\la T\vec u, \vec w \ra = 0 = \la \vec u, T\vec w \ra$. So $T\vec w \perp \vec u$. Notice that $T\vert_{\spa(\vec u)^\perp}$ is still self-adjoint. 
	\[ \la T_{\spa(\vec u)^\perp} \vec w_1, \vec w_2 \ra = \la T \vec w_1, \vec w_2 \ra = \la \vec w_1, T\vec w_2\ra = \la \vec w_1, T\vert_{\spa(\vec u)^\perp} w_2 \ra \qquad \forall \vec w_1, \vec w_2 \in \spa(\vec u)^\perp\]
	Hence $\c M(T) = \begin{bmatrix}
		\lambda & 0 & \cdots & 0 \\
		0 & \ast & \cdots & \ast \\\
		0 & \vdots & \ddots & \vdots \\
		0 & \ast & \cdots & \ast 
	\end{bmatrix}$ Now the problem is reduce to that for $T\vert_{\spa(\vec u)^\perp}$, which has a dimensional of $\dim V - 1$. By induction we can build a orthonomal basis of $V$ which consists of eigenvectors.
\end{proof}
\subsection{Positive Operators and Isometries}
\begin{definition}
	Suppose $T \in \c L(V)$ is self-adjoint and satisfies 
	\[ \la T\vec v, \vec v \ra \geq 0 \qquad \forall \vec v \in V\]
	Then $T$ is called \textbf{nonnegative}.

	\noindent If instead $\la T\vec v, \vec v\ra > 0 \qquad \forall \vec v \in V$ then $T$ is called \textbf{positive}.
\end{definition}
\begin{theorem}[Characterzation Theorem]
	The following are equivalent
	\begin{enumerate}
		\item $T$ is nonnegative
		\item $T = T^*$ and all its eigenvalyes are nonnegative.
		\item $T$ has a nonnegative square root, i.e. there exists $T = R^* \in \c L(V)$ such that $R^2 = T$. 
		\item $T$ has a self-adjoint square root. i.e. $\exists S = S^*$ such that $S^2 = T$.
		\item There exists $Q \in \c L(V)$ such that $Q^*Q = T$.
	\end{enumerate}
\end{theorem}
\newpage
\begin{proof}
	(e) $\implies$ (a). Suppose $T = Q^*Q$, so i
	\[ \la T\vec v, \vec v \ra = \la Q^*Q\vec v, \vec v \ra = \la Q\vec v, Q\vec v \ra = ||Q\vec v||^2 \geq 0\]
	(a) $\implies$ (b) Suppose $T$ is nonnegative. We know that nonnegative already satisfies self-adjoint. TODO
\end{proof}
\begin{definition}
	Suppose $S \in \c L(V)$. $S$ is called an \textbf{isometry} if 
	\[ ||S\vec v|| = ||\vec v|| \qquad \forall \vec v \in V\]
\end{definition}
\begin{remark}
	Observe that isometry necessarily preserves all inner products.
	\[ \la S\vec u, S \vec v \ra = \la \vec u, \vec v \ra \qquad \forall \vec u, \vec v \in V\] 
	This following from polar polarization from 7.11. for $\b F = \b R$ we have 
	\[ 4\la T\vec u, \vec v \ra = \la T(\vec u + \vec v), (\vec u + \vec v) \ra - \la T(\vec u - \vec v), (\vec u - \vec v )\ra\]
\end{remark}
\begin{corollary}
	An isometry maps an orthonormal to another orthonormal basis.
\end{corollary}
\begin{proof}
	If $\la \vec e_i, \vec e_j \ra = \delta_{ij}$. Then $\la S\vec e_i, S\vec e_j \ra = \la \vec e_i, \vec e_j \ra = \delta_{ij}$ where $\delta_{ij} = \left\{ \begin{array}{cc} 
	1 & \text{ if } i = j \\ 0 & \text{ if } i \neq j \end{array} \right.$. \\
	So if $(\li{\vec e}n)$ is an northnormal basis then $\li{S\vec e}n$ is an orthonormal basis.  
\end{proof}
\newpage
\subsection{Polar Decomposition and Singular Value Decomposition}
\begin{theorem}[Polar Decomposition]
	Take $T \in \c L(V)$. There exist an isometry $S \in \c L(V)$ such that
	\[ T = S \sqrt{T^*T}\]
	where $\sqrt{T^*T}$ is the nonnegative square root of $T^*T$.
\end{theorem}
\begin{proof}
	Observe that $||T\vec v || = \left|\left|\sqrt{T^*T} \vec v\right|\right| \qquad \forall \vec v \in V$. Indeed 
	\begin{align*}
		||T\vec v|| &= \la T\vec v, T \vec v \ra \\
		&= \la T^*T \vec v, \vec v \ra \\
		&= \la \sqrt{T^*T} \cdot \sqrt{T^*T} \vec v, \vec v \ra \\
		&= \la \sqrt{T^*T} \vec v, \sqrt{T^*T} \vec v \ra \\
		&= \left|\left| \sqrt{T^*T} \vec v \right|\right|
	\end{align*}
	It's clearly that there exists an isometry between $T$ and $\sqrt{T^*T}$.
\end{proof}
\begin{remark}[Construction of $S$]
	For any $\vec v \in V$, define
	\[ S_1\left( \sqrt{T^*T} \vec v\right) := T \vec v\]
	We first need to check this is well-defined. That is  if 
	\[ \sqrt{T^*T} \vec v_1 = \sqrt{T^*T} \vec v_2 \implies \vec v_1 = \vec v_1\]
	This is true because 
	\[ \sqrt{T^*T}(\vec v_1 - \vec v_2) = \vec 0 \implies 0 = \left|\left| \sqrt{T^*T} (\vec v_1 - \vec v_2)\right|\right| = ||T(\vec v_1 - \vec v_2)||\]
	hence $T\vec v_1 = T\vec v_2$. \\
	So $S_1$ is now defined as an element of $\c L\left(\range \sqrt{T^*T}, \range T\right)$ and $S_1$ is actually invertiable and an isometry. \\
	So $\dim \range \sqrt{T^*T} = \dim \range T$. Now we need to extend $S_1$ to an operator on $V$. \\ 
	Take $\left(\range \sqrt{T^*T}\right)^\perp$ and $(\range T)^\perp$. Send any orthonormal basis of $\left(\range \sqrt{T^*T}\right)^\perp$ to any orthonormal basis of $(\range T)^\perp$. This defines another isometry $S_2$.
	Finally define  
	\[ S\vec v = S_1 \vec u + S_2 \vec w\] where $\vec v = \vec u + \vec w , \vec u \in \range (\sqrt{(T^*T)}), \vec w \in \range \left(\sqrt{(T^*T)}\right)^\perp$. This creates $S$ which is now an isometry on entire $V$. Hence $T = S \sqrt{T^*T}$.
\end{remark}
\begin{theorem}
	Let $T \in \c L(V)$, for finite dimensional $V$. Then there exists orthonormal basis $\li {\vec e}n$ and $\li {\vec f}n$ and values $\li sn$, all nonnegative such that 
	\[ T\vec v = s_1 \la \vec v, \vec e_1 \ra \vec f_1 + s_2 \la \vec v, \vec e_2 \ra \vec f_2 + \cdots + s_n \la \vec v, \vec e_n \ra \vec f_n\]
	The $s_j$ are called singular values of $T$.
\end{theorem}

\begin{proof}[Proof. (derivation for polar decomposition)]
	Say $T = S\sqrt{T^*T}$. By the charaterization theorem w eknow that $V$ has an orthonormal eigenbasis $\li{\vec e}n$ consisiting of eigenvectors of $\sqrt{T^*T}$ corresponding to (nonnegative) eigenvalues $\li sn$ 
	\begin{align*}
	\vec v &= \la \vec v, \vec e_1 \ra + \la \vec v, \vec e_2 \ra + \cdots + \la \vec v, \vec e_n \ra \\ 
	\sqrt{T^*T} \vec v &= s_1\la \vec v, \vec e_1 \ra + s_2\la \vec v, \vec e_2 \ra + \cdots + s_n\la \vec v, \vec e_n \ra \\
	S\sqrt{T^*T}\vec v &= s_1\la \vec v, \vec e_1 \ra\vec f_1 + s_2\la \vec v, \vec e_2 \ra\vec f_2 + \cdots + s_n\la \vec v, \vec e_n \ra\vec f_n \\
	T\vec v &= s_1\la \vec v, \vec e_1 \ra\vec f_1 + s_2\la \vec v, \vec e_2 \ra\vec f_2 + \cdots + s_n\la \vec v, \vec e_n \ra\vec f_n
	\end{align*}
	and $\li{\vec f}n$ is also orthonormal.
\end{proof}
\begin{example}
	Take $T(x_1, x_2) = (2x_1 + x_2, -x_1 + 2x_2)$.
	Find its polar decomposition.
\end{example}
\begin{answer}
	\[T = \bml 2 & -1 \\ 1 & 2 \bmr \qquad T^* = \bml 2 & 1 \\ -1 & 2 \bmr \implies T^*T = \bml 5 & 0 \\ 0 & 5 \bmr\]
	Therefore we have $s_1 = s_2 = \sqrt 5$ and $f_1 = \left(2 / \sqrt 5, -1 / \sqrt 5\right),f_2 = \left(1 / \sqrt 5, 2 / \sqrt 5\right)$.
\end{answer}