Given a matrix space $\mathbb{R}^{m\times n}$, a vector norm $\|\cdot\|_p$ in $\mathbb{R}^{n}$ and a vector norm $\|\cdot\|_q$ in $\mathbb{R}^{m}$ with $p,q \ge 1$. The matrix norm induced by the two vector norms is defined as $$\| A \|_{p,q} = \sup{\frac{\|Ax\|_q}{\|x\|_p}}$$ The Frobenius norm is defined as $$\| A \|_F = \sqrt{\sum_{i,j} a_{ij}^2}$$ I want to prove that the Frobenius norm cannot be induced by any two vector norms. It is an old question though. I tried to follow the idea in Algebraic Pavel's solution, but found out that it is not quite trivial to show that the two vector norms are unitarily invariant. Here is my attempt:
Suppose the Frobenius norm is induced by $\|\cdot\|_p,\|\cdot\|_p$. Since the Frobenius norm has unitary invariance property, so for any orthogonal matrix $P,Q$, $$ \sup_{x\neq 0}\frac{\|Ax\|_q}{\|x\|_p}=\|A\|_F=\|PAQ\|_F=\sup_{x\neq 0}\frac{\|PAQx\|_q}{\| x\|_p}=\sup_{x\neq 0}\frac{\|PAx\|_q}{\|Q^T x\|_p}$$ Then I want to show that $(p,q)$ that satisifies these equalities for any orthogonal $P,Q$ can only be $(2,2)$. Since we can fix one of $P,Q$ to be an identity matrix, it suffices to show the following statement:
If for any orthogonal matrix $Q$, any matrix $A$ and any $q\ge 1$, it holds that $$ \sup_{x\neq 0}\frac{\|Ax\|_q}{\|x\|_p}=\sup_{x\neq 0}\frac{\|Ax\|_q}{\|Q x\|_p}$$ then $p$ must be $2$.
The difficulty I met when proving this statement is that I cannot ignore the numerator and say $\sup_{x\neq 0}\|x\|_p=\sup_{x\neq 0}\|Q x\|_p$ immediately. Could somebody help? Thank you very much.