$\def\abs#1{\lvert#1\rvert}\def\Abs#1{\lVert#1\rVert}$ Let $\sigma_1$ be the largest singular value of $A\in M_{m,n}.$ Show that $$ \sigma_1(A)=\max\{\abs{x^*Ay}:\text{$x\in {\mathbb{C}^m}$, $y \in {\mathbb{C}^n}$ are unit vectors}\}. $$
I know that $$ \abs{x^*Ay}\leq\Abs x_2\Abs{A y}_2\leq \Abs x_2\Abs y_2 \sigma_1(A), $$ but I don't know that how to find vectors $x$ and $y$ for which this inequality is an equality.
By definition, a singular value $\lambda$ of $A$ is an eigenvalue of $A^*A$, say with eigenvector $v$. Now let $x$ and $y$ be scalings of $A v$ and $v$, respectively.