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\begin{document}
\centerline{Math 201 lecture for Wednesday, Week 11}
\bigskip

\centerline{\textbf{Inner product spaces}}  
\bigskip
 
We now add structure to a vector space allowing us to define length and angles.
\medskip

\textbf{Definition.}  Let~$V$ be a vector space over a field~$F$ where~$F$ is
either~$\R$ or~$\C$.  An \emph{inner product} on~$V$ is a function 
\begin{align*}
  \langle\ ,\ \rangle\colon V\times V&\to F\\
  (x,y)&\mapsto \langle x,y\rangle
\end{align*}
satisfying for all~$x,y,z\in V$ and $c\in F$:
\begin{enumerate}
  \item linearity: $\ips{x+y}{z}=\ips{x}{z}+\ips{y}{z}$ and
    $\ips{cx}{y}=c\ips{x}{y}$.
  \item conjugate symmetry: $\overline{\ips{x}{y}}=\ips{y}{x}$.
  \item positive-definiteness: $\ips{x}{x}\in\R_{\geq0}$, and $\ips{x}{x}=0$ iff~$x=0$.
\end{enumerate}
\medskip

\textbf{Note.} If~$F=\R$, then an inner product is known as a {\em non-degenerate symmetric form}.
If~$F=\C$, an inner product is known as a {\em non-degenerate Hermitian form}.
\medskip

\textbf{Examples.} 
\begin{itemize}
  \item The ordinary dot product on~$\R^n$:  Here, $V=\R^n$ and
    \[
      \ips{(x_1,\dots,x_n)}{(y_1,\dots,y_n)}=x\cdot y:=\sum_{i=1}^nx_iy_i=x_1y_1+\dots+x_ny_n.
    \]
    For example, in~$\R^3$, we would have
    \[
      \ips{(1,2,3)}{(2,3,4)}=2+6+12=20\quad\text{and}\quad
      \ips{(1,2,3)}{(-2,1,0)}=-2+2+0=0.
    \]
  \item The ordinary inner product on~$\C^n$: Here, $V=\C^n$ and
    \[
      \ips{(x_1,\dots,x_n)}{(y_1,\dots,y_n)}=x\cdot
      \bar{y}:=\sum_{i=1}^nx_i\bar{y_i}=x_1\bar{y_1}+\dots+x_n\bar{y_n}.
    \]
    For example, in~$\C^2$, we would have
    \begin{align*}
      \ips{(1+i,1-i)}{(1+2i,4)}
      &=(1+i)\overline{(1+2i)}+(1-i)\,\overline{4}\\
      &=(1+i)(1-2i)+(1-i)\,4\\
      &=(3-i)+(4-4i)=7-5i.
    \end{align*}
  \item Let~$V=\mathcal{C}_{\R}([0,1])=\left\{ f\colon[0,1]\to\R:
    \text{$f$ is continuous} \right\}$, the vector space of~$\R$-valued
    continuous functions on the interval~$[0,1]$, and
    \[
      \ips{f}{g}=\int_0^1f(t)g(t)\,dt.
    \]
    To check positive-definiteness, note that if~$f\neq0$, then~$f^2(t)>0$
    for~$t$ in some open interval in~$[0,1]$.  Hence,
    \[
      \ips{f}{f}=\int_0^1f^2(t)\,dt>0.
    \]
  \item $V=\R^2$, and
    \[
      \ips{(x_1,x_2)}{(y_1,y_2)}=3x_1y_1+2x_1y_2+2x_2y_1+4x_2y_2.
    \]
    For positive-definiteness, we have
    \[
      \ips{(x_1,x_2)}{(x_1,x_2)}=3x_1^2+4x_1x_2+4x_2^2.
    \]
    Complete the square:
    \begin{align*}
      3x_1^2+4x_1x_2+4x_2^2
      &= 3\left( x_1^2+\frac{4}{3}x_1x_2+\frac{4}{3}x_2^2 \right)\\
      &= 3\left( \left( x_1+\frac{2}{3}x_2 \right)^2-\frac{4}{9}x_2^2+\frac{4}{3}x_2^2 \right)\\
      &= 3\left( \left( x_1+\frac{2}{3}x_2 \right)^2+\frac{8}{9}x_2^2\right)\\
      &\geq0,
    \end{align*}
    with equality if and only if~$x_1=x_2=0$.
  \item Let~$F=\R$ or $\C$, and let $V=M_{m\times n}(F)$.  For~$A\in M_{m\times
    n}(F)$, define the \emph{conjugate transpose} of~$A$ by
    \[
      A^*=\overline{A^t},
    \]
    where the overline means taking the conjugate of each entry of~$A$.
    If~$A$ has only real entries, the~$A^*=A^t$.  Next, define the inner
    product,
    \[
      \ips{A}{B}=\tr(B^{*}A)=\sum_{i=1}^n(B^{*}A)_{ii}.
    \]
    (Note: The special case~$m=1$ gives the usual inner product on~$\R^n$
      or~$\C^n$.)
    Proof of positive-definiteness is left as an exercise.
\end{itemize}
\medskip

\textbf{Proposition.} Let~$(V,\ips{\,}{\,})$ be an inner product space
over~$F=\R$ or~$\C$.  Then for all~$x,y,z\in V$ and~$c\in F$,
\begin{enumerate}
  \item\label{11wedit1} $\ips{x}{y+z}=\ips{x}{y}+\ips{x}{z}$.
  \item\label{11wedit2} $\ips{x}{cy}=\overline{c}\ips{x}{y}$.
  \item\label{11wedit3} $\ips{x}{0}=\ips{0}{y}=0$.
  \item\label{11wedit4} If $\ips{x}{y}=\ips{x}{z}$ for all~$x\in V$, then~$y=z$.
\end{enumerate}

\begin{proof}
  For part~\ref{11wedit1}, notice that the definition of an inner product only
  guarantees sums on the left distribute.  However, using properties of
  conjugation,
  \begin{align*}
    \ips{x}{y+z}
    &=\overline{\ips{y+z}{x}}\\
    &=\overline{\ips{y}{x}}+\overline{\ips{z}{x}}\\
    &=\ips{x}{y}+\ips{x}{z}.
  \end{align*}
  Parts~\ref{11wedit2} and~\ref{11wedit3} are left as exercises.  For
  part~\ref{11wedit4}, $\ips{x}{y}=\ips{x}{z}$ for
  all~$x$ implies
  \begin{align*}
    0=\ips{x}{y}-\ips{x}{z}&= \ips{x}{y}+(-1)\ips{x}{z}\\
    &=\ips{x}{y}+\overline{(-1)}\ips{x}{z}\\
    &=\ips{x}{y}+\ips{x}{(-1)z}\\
  &=\ips{x}{y}+\ips{x}{-z}\\
  &=\ips{x}{y-z}
  \end{align*}
  for all~$x$.  In particular, let~$x=y-z$ to get
  $\ips{y-z}{y-z}=0$.  By positive-definiteness,~$y-z=0$.
\end{proof}
\end{document}  
