TITLE: Why does the spectral norm equal the largest singular value?
QUESTION [46 upvotes]: This may be a trivial question yet I was unable to find an answer:
$$\left \| A \right \| _2=\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A)$$
where the spectral norm $\left \| A \right \| _2$ of a complex matrix $A$ is defined as $$\text{max} \left\{ \|Ax\|_2 : \|x\| = 1 \right\}$$
How does one prove the first and the second equality?

REPLY [34 votes]: First of all,
$$\begin{align*}\sup_{\|x\|_2 =1}\|Ax\|_2 & = \sup_{\|x\|_2 =1}\|U\Sigma V^Tx\|_2 = \sup_{\|x\|_2 =1}\|\Sigma V^Tx\|_2\end{align*}$$
since $U$ is unitary, that is, $\|Ux_0\|_2^2 = x_0^TU^TUx_0 = x_0^Tx_0 = \|x_0\|_2^2$, for some vector $x_0$.
Then let $y = V^Tx$. By the same argument above, $\|y\|_2 = \|V^Tx\|_2 = \|x\|_2 = 1$ since $V$ is unitary.
$$\sup_{\|x\|_2 =1}\|\Sigma V^Tx\|_2 = \sup_{\|y\|_2 =1}\|\Sigma y\|_2$$
Since $\Sigma = \mbox{diag}(\sigma_1, \cdots, \sigma_n)$, where $\sigma_1$ is the largest singular value. The max for the above, $\sigma_1$, is attained when $y = (1,\cdots,0)^T$. You can find the max by, for example, solving the above using a Lagrange Multiplier.