Is Frobenius norm induced up to a scalar factor?

Question

I know that the Frobenius norm is not induced since $||I||_F=\sqrt n\neq 1$.

But what if we consider the norm $\frac 1 {\sqrt n} ||\cdot ||_F$?

Thank you!

Answer 1

Any induced norm satisfies $\|A\|\ge \rho(A)$, where $\rho(A)$ is the spectral radius of $A$, i.e. the largest absolute value of its eigenvalues. This fact is stated without proof in Wikipedia; here is a proof.

Any induced norm is of the form $$\|A\|=\max_x\frac{\|Ax\|}{\|x\|}.$$ Let $x^*$ be an eigenvector corresponding to the eigenvalue $\lambda^*$ with the largest absolute value. Then $$\|A\|\ge\frac{\|Ax^*\|}{\|x^*\|}=\frac{\|\lambda^*x^*\|}{\|x^*\|}=|\lambda^*|\frac{\|x^*\|}{\|x^*\|}=|\lambda^*|.$$

Let $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$. Then your norm is $\|A\|=\dfrac1{\sqrt2}\|A\|_F=\dfrac1{\sqrt2}$ while $\rho(A)=1$, contradicting the above inequality.