The spectral norm of $A ∈ \Bbb R^{n\times n}$ is defined as $\|A\|_{*} = \max \{\|Ax\|_2 : \|x\|_2 = 1 \}$.
To show that $$\|A\|_{\ast} =\max \{y^{\top} Ax : \|x\|_2 = 1, \|y\|_2 = 1 \}$$
I have already proved that
$$\|A\|_{\ast} = \sigma_{1}$$
where $\sigma_{1}$ is the largest singular value of $A$. But I'm having difficulty proving above statement, can anyone give a hint to solve this?
For $x,y \in \Bbb{R}^n$ with $\|x\|_2=\|y\|_2 = 1$ by Cauchy-Schwarz we have $$y^T Ax \le \|Ax\|_2\|y\|_2 \le \|A\|_*$$ so $\max\{y^T Ax : \|x\|_2=\|y\|_2 = 1\} \le \|A\|_*$. Conversely, for $x \in \Bbb{R}^n$ with $\|x\|_2=1$ we get $$\|Ax\|_2 = \frac{(Ax)^TAx}{\|Ax\|_2} = y^TAx$$ where $y = \frac{Ax}{\|Ax\|_2}$ so $\|A\|_* \le \max\{y^T Ax : \|x\|_2=\|y\|_2 = 1\}$.