Proving that the spectral norm of a matrix is orthogonally invariant

Question

Does anyone know how to prove that the spectral norm:

$$\left \| A \right \|_{2}=\max\left \{ \sqrt \lambda : \lambda \in \sigma (A^*A)\right \},\; A\in\mathbb{R}^{n\times n}$$ (meaning that $\lambda$ is an eigenvalue of $A^*A$)

is orthogonally invariant?

Answer 1

I showed this in another answer, but I will copy it over.

Your definition is true, but we also have the equality (often the definition) $$\|A\|_2=\max_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2}.$$ Now, we just compute that if $U$ and $W$ are orthogonal, then

\begin{align*} \left\lVert UAW\right\lVert_2 &=\max_{x\neq 0}\frac{\left\lVert UAWx \right\lVert_2}{\left\lVert x\right\lVert_2}=\max_{x\neq 0}\frac{\sqrt{x^TW^TA^TU^TUAWx}}{\sqrt{x^Tx}}\\ &=\max_{z\neq 0}\frac{\sqrt{zA^TAz}}{\sqrt{z^Tz}}=\max_{z\neq 0}\frac{\left\lVert Az\right\lVert_2}{\left\lVert z\right\lVert_2}=\left\lVert A\right\lVert_2, \end{align*} where we used the substitution $z=Wx.$

Answer 2

Suppose $A^* A x = \lambda v$, and $U,V$ are orthogonal. Then $(UAV)^* (UAV)V^* x = V^* A^* A x = \lambda V^* x$. Hence $\sigma(A^*A) = \sigma ((UAV)^* (UAV))$ and so $\|A\|_2 = \|UAV\|_2$.