TITLE: Largest singular value of non square matrix
QUESTION [7 upvotes]: Let $B$ be an $m\times n$ matrix with complex number as its element.
Let $\sigma$ denotes the largest singular value of $B$
Prove that 
\begin{equation}
\sigma = \max\limits_{\|u\|_2=1,\|v\|_2=1} |u^*Bv|.
\end{equation}
My partial solution:
Let $x$ be the eigenvector of $B^*B$ that corespond to eigen value $\sigma^2$.
Define $z=\frac{x}{\|x\|_2}$ and $w=\frac{1}{\sigma} Bz$.
Note that $\|z\|_2=1$ and $\|w\|_2=\sqrt{w^*w}=\sqrt{\frac{1}{\sigma^2} z^*B^*Bz}=\frac{1}{\sigma}\sqrt{z^*\sigma^2 z}=1$.
Thus
\begin{align*}
\sigma=\frac{1}{\sigma} z^*\sigma^2z=\frac{1}{\sigma} z^* B^*Bz=w^*Bz\leq \max\limits_{\|u\|_2=1,\|v\|_2=1} |u^*Bv|.
\end{align*}
My question is that how to prove the reverse inequality.
If $m=n$, then I can solve the problem by using the Rayleigh-Ritz theorem for $B^*B$. But, I have no idea for $m\neq n$.
Thank you.

REPLY [3 votes]: The reverse inequality follows from the fact that $\|B\|_2 = \sigma$.
Let $u$ and $v$ be unit vectors. Then
\begin{align*}
| u^* B v | &\leq \|u \|_2 \|B v\|_2 \qquad \text{(by Cauchy-Schwarz inequality)}\\
&\leq \|u\|_2 \|B\|_2 \|v\|_2 \\
&= \sigma.
\end{align*}
It follows that 
\begin{equation}
\max_{\|u\|_2=1,\|v\|_2=1} | u^* B v| \leq \sigma.
\end{equation}