How to show $\sup_{x,y \neq 0} \frac {x^TAy} {\|x\|_2\|y\|_2}=\sup_{y \neq 0} \frac{\|Ay\|_2} {\|y\|_2}$

Question

I can figure out that the maximum of singular value: $\sigma_{max}=\sup_{y \neq 0} \frac{\|Ay\|_2} {\|y\|2}$ for a real matrix A of $m \times n$. But I cannot achieve that: $$\sup_{x,y \neq 0} \frac {x^TAy} {\|x\|_2\|y\|_2}=\sup_{y \neq 0} \frac{\|Ay\|_2} {\|y\|_2}$$ Can you help?

Answer 1

By Cauchy-Schwartz inequality, $$\sup_{\|\mathbf{u}\|_2=1}\mathbf{u}^T \mathbf{v}=\|\mathbf{v}\|_2.$$

By letting $\mathbf{v} = \mathbf{x}/\|\mathbf{x}\|_2$ and $\mathbf{v}=\mathbf{Ay}/\|\mathbf{y}\|_2$, we immediately arrive at $$\sup_{\mathbf{x,y} \neq 0} \frac {\mathbf{x}^T\mathbf{Ay}} {\|\mathbf{x}\|_2\|\mathbf{y}\|_2}=\sup_{\mathbf{y} \neq 0} \frac{\|\mathbf{Ay}\|_2} {\|\mathbf{y}\|_2}.$$