I have seen some disagreement online and was wondering if anyone could clarify for me:
If $X$ is an arbitrary $n \times n$ matrix and $A$ is an arbitrary orthogonal $n \times n$ matrix, is it true that $$\| AX \|_p = \|X\|_p$$
For all $p \in \mathbb{Z}_+\cup{\infty}$, where $\|\cdot\|_p$ is the matrix $p$-norm defined as:
$$\| A \|_p = \sup_{x \neq 0} \frac{|Ax|_p}{|x|_p}$$
Where $|\cdot|_p$ is the $p$-norm (of vector).
Best Answer
Let's start with the Frobenius norm: Using the trace characterization,
\begin{align*} \left\lVert UAW\right\lVert_F&=\text{tr} (UAWW^TA^TQ^T)=\text{tr}(UAA^TU^T)\\ &=\text{tr}(A^TU^TU^TA)=\text{tr}(AA^T)\\ &=\left\lVert A\right\lVert_F\end{align*}.
Next, for the operator norm induced by the $2$-norm:
\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 $z=Wx.$
For an example to show that the infinity and $1$-norm do not work, take $U$ to be rotation matrix which rotates counterclockwise by $60$ degrees, $A=\begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix},$ and $W=I$. Then, $$UAW=\begin{bmatrix} \frac{1}{2}-\frac{3\sqrt{3}}{2} & 1-2\sqrt{3} \\ \frac{3}{2}+\frac{\sqrt{3}}{2} & 2+\sqrt{3}\end{bmatrix}.$$