# When is the Frobenius norm equal to the spectral radius?

I know that the spectral radius is $\rho(A) = max |\lambda_l| = \S_{max}^2$ and that the Frobenius norm is $||A|| = \sqrt{tr(A^*A)} = (\sum_{k}S_k^2)^{1/2}$, which means I want to find the matrix A for which the following is true

$$||A||_F = \sqrt{tr(A^*A)} = (\sum_{k}S_k^2)^{1/2} = S_{max}^2$$

So is the spectral radius equal to the frobenius norm if A is a square matrix with its largest eigenvalue equal to |1|?

(S are the singular values)

## 1 Answer

We assume that $A\not= 0$.

Let $spectrum(A)=(\lambda_i)$ where $|\lambda_1|\geq |\lambda_2|\geq\cdots$ and $\Sigma(A)=(\sigma_i)$ where $\sigma_1\geq \sigma_2\geq\cdots$. Then $\rho(A)=[\lambda_1|\leq \sigma_1=||A||_F=\sqrt{\sum_i \sigma_i^2}$. Thus $|\lambda_1|=\sigma_1$ and $\sigma_2=\cdots=\sigma_n=0$.

Finally $rank(A)=1$ and $A=uv^*$ where $u,v$ are non-zero vectors. One has $AA^*=||v||^2uu^*$, $\lambda_1=trace(uv^*)=v^*u$ and $\sigma_1^2=||v||^2trace(uu^*)=||v||^2||u||^2$. Moreover $|\lambda_1|=|v^*u|\leq ||u||||v||=\sigma_1$, one deduces that the Schwartz inequality is an equality, that implies that $u,v$ are parallel.

Conclusion: $A=0$ or $A=\alpha uu^*$ where $\alpha\in \mathbb{C}^*$ and $u$ is a non-zero vector. (The converse is easy)

EDIT. @ Sandra , I see your edit. I think that you did not understand one word of my answer.