I have been asked to prove that for $A\in M_n(\mathbb{C})$, with $||A||:=\sup_{x\in\mathbb{C}^n,|x|=1}|Ax|$, $$||A||=\sqrt{\lambda}$$ where $\lambda$ is the eigen value of largest modulus of $A^*A$.
I know that this supremum is achieved on the set.
So far, I have that $A^*A$ is self adjoint, and so there is $P$ unitary and $D$ a diagonal matrix of eigen values of $A^*A$ such that $A^*A=P^*DP$.
So, for $x\in \mathbb C^n, |x|=1$ $$|Ax|^2 = \langle Ax,Ax \rangle = \langle x, A^*Ax \rangle = \langle x ,P^*DPx \rangle = \langle Px,DPx \rangle$$
I feel that I should be able to use that $\langle Px,Py\rangle=\langle x,y\rangle$ as $P$ is unitary to get somewhere.
Initial I had that $|Ax|^2=\langle Px,DPx \rangle\leq\langle y, \lambda y \rangle=\lambda\langle y,y\rangle=\lambda$ where $y$ is an eigen vector for $\lambda$ and $y=Px$ for some $x$, but I realised that $\lambda$ need not be non-negative (it must be real though), so the inequality doesn't make sense.
Any help would be appreciated.
Edit 1) Looking at this, I think I missed something obvious. As $||\cdot||$ is a norm, it must map to $[0,\infty)$, so $\lambda$ is non negative else, it would not. So I'll look for a way to show $A^*A$ is pos.def.
Edit 2) On further consideration, it needs only that $\lambda$, the eigen value of largest absolute value, is non-negative, so it may not be pos.def.
Note that $A^*A$ is necessarily positive semidefinite. To show this, note that for any $x \in \Bbb C^n$, $$ x^*(A^*A)x = (x^*A^*)Ax = \|Ax\|^2 \geq 0 $$ The result you showed regarding $(x,Mx)$ is called Rayleigh's theorem. See also the min-max theorem (AKA the Courant-Fischer principle).