
\documentclass[12pt]{amsart}

\usepackage{241}

\title{21-241 Matrices and Linear Transformations \\
       Summer 1 2012 \\
       Course notes \\
       June 12}
\author{Paul McKenney}


\begin{document}

\maketitle

\begin{defn*}
  If $x,y\in\CC^n$, their \emph{inner product} is defined to be
  \[
    \brak{x}{y} = \sum_{k=1}^n x_k \overline{y_k}
  \]
  We say that $x$ and $y$ are \emph{orthogonal}, and write $x\perp y$, if $\brak{x}{y} = 0$.
\end{defn*}

Note that, viewing $y^H$ and $x$ as $1\times n$ and $n\times 1$ size matrices respectively, we have
\[
  \brak{x}{y} = y^H x
\]
In the case where $x,y\in\RR^n$, all the complex conjugates drop away, and we can write $y^\top$ instead of $y^H$.

\begin{example*}
  Let $p$ and $q$ be points in the plane $\RR^2$.  Show that if $p$ and $q$ lie on the unit circle, then $\brak{p}{q}$ is exactly $\cos(\theta)$, where $\theta$ is the angle between $p$ and $q$ on the unit circle (equivalently, the length of the arc between $p$ and $q$, centered at $0$.)  Show that $p\perp q$ if and only if the lines $L$ and $R$, going through $0$ and $p$, and $0$ and $q$ respectively, are perpendicular.
\end{example*}
\begin{proof}
  Suppose $p$ and $q$ lie on the unit circle; then we can write their coordinates down as $p = (\cos\vp,\sin\vp)$ and $q = (\cos\psi,\sin\psi)$ for some $\vp$ and $\psi$, and the angle between them is exactly $\theta = |\vp - \psi|$.  Now,
  \[
    \brak{p}{q} = \cos\vp\cos\psi + \sin\vp\sin\psi = \cos(\vp - \psi)
  \]
  Since $\cos$ is an even function, $\cos(\vp - \psi) = \cos\theta$.

  The other part of the problem follows from this part; $p\perp q$ if and only if $\theta$ is a zero of $\cos$, if and only if $\theta$ is one of $\pi/2$ or $3\pi/2$, if and only if the lines $L$ and $R$ described are orthogonal.
\end{proof}

Before we continue we'll need to recall some facts about complex conjugation, ie the map $z\mapsto \bar{z}$ given by $a + bi\mapsto a - bi$.  In short, it's as nice as you would want.
\begin{fact}
  For all $z,w\in\CC$,
  \begin{enumerate}
    \item  $\overline{z + w} = \bar{z} + \bar{w}$,
    \item  $\overline{zw} = \bar{z}\bar{w}$,
    \item  $\overline{\bar{z}} = z$.
  \end{enumerate}
\end{fact}

\begin{example*}
  Which of the above algebraic properties hold for the hermitian operator?  Ie,  which of
  \begin{enumerate}
    \item  $(A + B)^H = A^H + B^H$,
    \item  $(AB)^H = A^H B^H$,
    \item  $(A^H)^H = A$
  \end{enumerate}
  is true for all $A,B\in M_n(\CC)$?  If one of them is wrong, what's the ``right'' version?
\end{example*}

\begin{fact}
  \begin{enumerate}
    \item  For all $x,y,z\in\CC^n$, $\brak{x+y}{z} = \brak{x}{z} + \brak{y}{z}$.
    \item  For all $x,y\in\CC^n$ and $\lambda\in\CC$, $\brak{\lambda x}{y} = \lambda\brak{x}{y}$.
    \item  For all $x,y\in\CC^n$, $\brak{x}{y} = \overline{\brak{y}{x}}$.
  \end{enumerate}
\end{fact}
This fact is often summarized in the following way.  (We often say that the inner product is linear in its first argument, and \emph{conjugate}-linear in its second argument.)
\begin{cor}
  For all $x_1,x_2,y_1,y_2\in\CC^n$ and $\lambda_1,\lambda_2,\mu_1,\mu_2\in\CC$,
  \[
    \brak{\lambda_1 x_1 + \lambda_2 x_2}{\mu_1 y_1 + \mu_2 y_2} = \lambda_1 \overline{\mu_1} \brak{x_1}{y_1} + \lambda_1 \overline{\mu_2}\brak{x_1}{y_2} + \lambda_2\overline{\mu_1}\brak{x_2}{y_1} + \lambda_2\overline{\mu_2}\brak{x_2}{y_2}
  \]
\end{cor}
It's useful to keep in mind that when $x,y\in\RR^n$, all of the complex conjugates above disappear.

\begin{fact}
  For all $x\in\CC^n$, $\brak{x}{x}$ is nonnegative.
\end{fact}

\begin{defn*}
  The \emph{norm} of a vector $x\in\CC^n$ is defined to be
  \[
    \norm{x} = \sqrt{\brak{x}{x}} = \left(\sum_{k=1}^n |x_k|^2\right)^{\frac{1}{2}}
  \]
  The \emph{distance between} two vectors $x,y\in\CC^n$ is $\norm{x - y}$.
\end{defn*}

\begin{fact}
  \begin{enumerate}
    \item  If $x\in\CC^n$ and $\lambda\in\CC$, then $\norm{\lambda x} = |\lambda| \norm{x}$,
    \item  If $x,y\in\CC^n$ and $x\perp y$, then $\norm{x + y}^2 = \norm{x}^2 + \norm{y}^2$.
  \end{enumerate}
\end{fact}

\begin{fact}(Cauchy-Schwartz Inequality)
  For any $x,y\in\CC^n$, we have
  \[
    |\brak{x}{y}|^2 \le \norm{x}\norm{y}
  \]
  Moreover, the equality above holds if and only if $x$ and $y$ are linearly independent.
\end{fact}
\begin{proof}
  If $y = 0$ then we're done.  (Why?)  So assume $y\neq 0$; then $\norm{y} \neq 0$.  Let
  \[
    z = x - \frac{\brak{x}{y}}{\brak{y}{y}} y = x - \frac{\brak{x}{y}}{\norm{y}^2} y
  \]
  Then,
  \[
    \brak{z}{y} = \brak{x - \frac{\brak{x}{y}}{\brak{y}{y}} y}{y} = \brak{x}{y} - \frac{\brak{x}{y}}{\brak{y}{y}}\brak{y}{y} = 0
  \]
  So $z\perp y$, and it follows that $\frac{\brak{x}{y}}{\brak{y}{y}}y$ and $z$ are orthogonal.  So by the above fact,
  \[
    \norm{\frac{\brak{x}{y}}{\brak{y}{y}}y + z}^2 = \frac{|\brak{x}{y}|^2}{|\brak{y}{y}|^2}\norm{y}^2 + \norm{z}^2 \ge \frac{|\brak{x}{y}|^2}{\norm{y}^2}
  \]
  Finally, note that
  \[
    x = \frac{\brak{x}{y}}{\brak{y}{y}}y + z
  \]
  So we have
  \[
    \norm{x}^2 \ge \frac{|\brak{x}{y}|^2}{\norm{y}^2}
  \]
  Moving stuff around and taking a root, we get
  \[
    \norm{x}\norm{y} \ge \brak{x}{y}
  \]
  as desired.
\end{proof}

\begin{thm}
  Suppose $x_1,\ldots,x_k\in\CC^n$ are pairwise orthogonal, ie, $x_i\perp x_j$ for $i\neq j$, and each $x_i$ is nonzero.  Then $\{x_1,\ldots,x_k\}$ is linearly independent.
\end{thm}
\begin{proof}
  Suppose
  \[
    \lambda_1 x_1 + \cdots + \lambda_k x_k = 0
  \]
  Now take the inner product with $x_i$;
  \[
    0 = \brak{0}{x_i} = \brak{\sum_{j=1}^k \lambda_j x_j}{x_i} = \sum_{j=1}^k \lambda_j \brak{x_j}{x_i} = \lambda_i \norm{x_i}^2
  \]
  Since $x_i$ is nonzero, $\norm{x_i}^2 \neq 0$.  Then $\lambda_i = 0$.
\end{proof}

\begin{thm}
  Suppose $A\in M_n(\CC)$, and $x,y\in\CC^n$.  Then $\brak{Ax}{y} = \brak{x}{A^H y}$.
\end{thm}
\begin{proof}
  The $i$th entry of $Ax$ is
  \[
    (Ax)_i = \sum_{j=1}^n A_{ij} x_j
  \]
  Similarly, the $j$th entry of $A^H y$ is
  \[
    (A^H y)_j = \sum_{i=1}^n (A^H)_{ji} y_i = \sum_{i=1}^n \overline{A_{ij}} y_i
  \]
  So,
  \begin{align*}
    \brak{Ax}{y} & = \sum_{i=1}^n \left(\sum_{j=1}^n A_{ij} x_j\right) \overline{y_i} \\
      & = \sum_{i=1}^n \sum_{j=1}^n A_{ij} x_j \overline{y_i} \\
      & = \sum_{i=1}^n \sum_{j=1}^n x_j\overline{\overline{A_{ij}} y_i} \\
      & = \sum_{j=1}^n x_j \sum_{i=1}^n \overline{\overline{A_{ij}} y_i} \\
      & = \sum_{j=1}^n x_j \overline{(A^H y)_j} \\
      & = \brak{x}{A^H y} \\
  \end{align*}
\end{proof}


\end{document}
