Show that $\rho(A) \geq \max_{1\leq i \leq n}|a_{ii}|.$

Question

Let $A=(a_{ij})$ be normal. Show that $\rho(A) \geq \max_{1\leq i \leq n}|a_{ii}|,$ where $\rho(.)$ is the spectral radius of $A$.

I know that $\rho(A) \leq \sqrt{\lambda_{\text{max}}(A^{^*}A)}:=\sigma_{\text{max}}(A)=||A||.$ Is there any relation between diagonal element and the eigenvalue? I know the relation with the trace but can not connect with each diagonal element.

Answer 1

It $A$ is a bounded operator on a Hilbert space (including the case of $n\times n$ matrices), one defines:

• the norm: $\|A\|=\sup_{\|x\|\leq1}\|Ax\|$,
• the spectral radius: $\rho(A)=\sup_{\lambda\in\sigma(A)}|\lambda|$, and
• the numerical radius: $w(A)=\sup_{\|x\|\leq1}|\langle Ax,x\rangle|$.

For every $A$ $$ \rho(A)\leq w(A)\leq \|A\| \leq 2w(A). $$ but when $A$ is normal one has that $$ \rho(A) = w(A) =\|A\|. $$

Since a diagonal entry of an $n\times n$ matrix is given by $A_{ii}=\langle Ae_i,e_i\rangle$, we then have $$ |a_{ii}| \leq w(A) = \rho(A), $$ provided $A$ is normal.

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EDIT

Lemma. If $A$ is a normal matrix, then the numerical radius of $A$ coincides with its spectral radius.

Proof. Let $\{e_i\}_{i=1}^n$ be an orthonormal basis consisting of eigenvectors, with corresponding eigenvalues $\{\lambda _i\}_{i=1}^n$. For $\|x\|\leq 1$, we then have $$ \langle Ax, x\rangle = \Big\langle A\Big(\sum_{i=1}^n x_ie_i\Big),\sum_{i=1}^n x_ie_i\big\rangle = \Big\langle \sum_{i=1}^n x_i\lambda _ie_i,\sum_{i=1}^n x_ie_i\big\rangle = \sum_{i=1}^n |x_i|^2\lambda _i. $$ Therefore $$ |\langle Ax, x\rangle | \leq \sum_{i=1}^n |x_i|^2|\lambda _i| \leq \big (\sup_{1\leq i\leq n} |\lambda _i|\big ) \sum_{i=1}^n |x_i|^2 = \rho (A)\|x\|^2 \leq \rho (A). $$ Taking the supremum over $x$ we get $$ w(A) = \sup_{\|x\|\leq 1}|\langle Ax, x\rangle | \leq \rho (A). $$ On the other hand, for every $i≤n$ we have that $$ w(A) \geq |\langle Ae_i, e_i\rangle | = |\langle \lambda _ie_i, e_i\rangle | = |\lambda _i|. $$ Taking the supremum over $i$ we get $$ w(A) \geq \sup_{1≤i≤n} |\lambda _i| = \rho(A). $$ QED

Answer 2

Correct me if I'm wrong. Don't we have the fact that :
$ \rho(A) = \max_{ \Vert v \Vert \le 1} \Vert \langle v, Av \rangle\Vert $ for any normal matrix $A$ ?
If this is true, isn't your problem solved?
Disclaimer: I don't know, just asking because to me it is feasible due to the fact that $A$ is unitarily diagonalizable.