The Frobenius norm is not an operator norm

Question

Definition: Let matrix $A$ be $m$-by-$n$, let $\| \cdot\|_{\tilde{m}}$ be a vector norm on $\mathbb R^{m}$ and $\| \cdot\|_{\tilde{n}}$ be a vector norm on $\mathbb R^{n}$. Then $$\| A\|_{\tilde{m}\tilde{n}} := \max\limits_{x \in \mathbb R^{n} \setminus \{0\}} \frac{\| Ax\|_{\tilde{m}}}{\| x\|_{\tilde{n}}}$$ is called an operator norm.

Question: prove that Frobenius norm is not an operator norm.

Remark: you cannot suppose that the norm on $\mathbb R^{m}$ and the norm on the $\mathbb R^{n}$ are the same norm but in different space.For example: $\| \cdot\|_2 $in $\mathbb R^{n}$ and $\| \cdot\|_2 $ in $\mathbb R^{m}$ .

Answer 1

Frobenius matrix norm $\Vert A \Vert_F$ is defined for an $m \times n$ real matrix $A = [a_{ij}]$ as follows:

$$ \Vert A \Vert_F = \sqrt{ \sum\limits_{i = 1}^m \ \sum\limits_{j = 1}^n \ \ | a_{ij} |^2 } $$

Operator norm $\Vert A \Vert $ is defined for an $m \times n$ real matrix $A = [a_{ij}]$ as follows:

$$ \Vert A \Vert = \max\limits_{\mathbf{x} \in \mathbf{R}^n \backslash \{\mathbf{0}\}} \ \left[ {\Vert A \mathbf{x} \Vert_m \over \Vert \mathbf{x} \Vert_n} \right] $$

An important property of the operator norm $\Vert \cdot \Vert$ is that $$ \Vert I \Vert = 1 $$ where $I$ is the $n \times n$ identity matrix.

For the Frobenius norm, we see that $$ \Vert I \Vert_F = \sqrt{1 + 1 + \cdots + 1} = \sqrt{n} $$

This calculation establishes that the Frobenius norm is not an operator norm for all $n \times n$ matrices for which $n \geq 2$.