How to prove the equivalent condition of normal matrix? In Wikipedia, https://en.wikipedia.org/wiki/Normal_matrix, it says that a complex square matrix $A$ is normal if and only if the following condition holds:
$$||A||_2=\max_{||x||_2=1}|<Ax,x>|=\max\{|\lambda||\lambda\in \sigma(A)\}$$ where $\sigma(A)$ means the spectrum of $A$ of course consisting of all eigenvalues of $A$. I can prove that $A$ is normal implies the condition. But I can not see how to prove the condition implies $A$ is normal. Could any one please help or hint on this?
 Let $$A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&0&1\\
0&0&0
\end{array}} \right]$$. Then for any $x = {[{x_1},{x_2},{x_3}]^T}$ ,$||x|{|_2} = 1$and ${\sup _{||x|{|_2} = 1}}| < Ax,x > | = ||{x_1}{|^2} + {{\bar x}_2}{x_3}| \le |{x_1}{|^2} + |{x_2}||{x_3}| \le |{x_1}{|^2} + \frac{{|{x_2}{|^2} + |{x_3}{|^2}}}{2} \le 1$. And also if $x_1=1$, then it attains the supremum. So we can conclude $\sup_{||x||_2=1}|<Ax,x>|=1$. ${A^H}A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&0&0\\
0&0&1
\end{array}} \right]$ which shows that $||A||_2=1$ since $||A||_2$ is the largest singular value of $A$. Then it is also obviously true that the spectral radius is just $1$. So is the equivalent condition in Wikipedia not correct?
 A: That alleged equivalent condition is indeed wrong. Denote by $r(A)=\max_{\|x\|_2=1}|\langle Ax,x\rangle|$ the numerical radius of a complex square matrix $A$. In general, we have
$$
\rho(A)\le r(A)\le\|A\|_2.
$$
Since $r(A)$ is sandwiched between $\rho(A)$ and $\|A\|_2$, the condition mentioned in your question is simply saying that $\rho(A)=\|A\|_2$.
A matrix $A$ with $\rho(A)=\|A\|_2$ is called radial. It is known that $A$ is radial if and only if it is unitarily similar to a matrix of the form
$$
\pmatrix{D&0\\ 0&L}
$$
for some diagonal matrix $D$ and some lower triangular matrix $L$ such that $\|L\|_2\le\rho(A)$. See M. Goldberg and G. Zwas (1974), On matrices having equal spectral radius and spectral norm, Linear Algebra and Its Application, 8: 427-434.
Obviously, $A$ is normal if and only if $L$ is normal. However, $A$ becomes radial when $D\ne0$ and $\|L\|_2$ is sufficiently small, regardless of whether $L$ is normal or not. The two notions are therefore non-equivalent. The matrix transpose of your example $A$ is one such case that $D\ne0$ and $L$ is small but non-normal.
By the way, a matrix $A$ with $\rho(A)=r(A)$  is called spectral. It is known that $A$ is spectral if and only if it is unitarily similar to the above $D\oplus L$ form with $r(L)\le\rho(A)$. See Goldberg et al. (1975), The numerical radius and spectral matrices, Linear and Multilinear Algebra, vol. 2, pp.317-326.
A: A useful observation: by Schur triangularization, proving that your statement only holds for normal matrices $A$ is equivalent to showing that if $A$ is upper-triangular, then the statement implies that $A$ is diagonal.
