I have seen an example of matrix
$$A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}$$
whose spectral radius is zero therefore the spectral radius is not matrix norm. Why the spectral radius is not matrix norm in this case Is it possible that $\|A\|=\epsilon$?
Best Answer
You seem to have confused spectral radius with spectral norm. The word spectral in spectral norm actually refers not to the spectrum of $A$, but to the spectrum of $A^\ast A$. In my opinion, we should all abandon this misleading name, but adopt the more appropriate name, the operator norm, instead.
The operator norm $\|A\|_2$ is defined by $$ \max_{\|x\|_2\ne0}\frac{\|Ax\|_2}{\|x\|_2}=\max_{\|u\|_2=1}\|Au\|_2.\tag{1} $$ It is identical to the largest singular value, i.e. the square root of the largest eigenvalue of $A^\ast A$.
Unlike spectral radius, the operator norm is defined for all square or non-square complex matrices. The norm can be more accurately called the spectral norm when $A$ is square and self-adjoint. In that case, the operator norm coincides with the spectral radius, and the spectral radius is a (submultiplicative) norm on the real linear space of all self-adjoint matrices.
The operator norm is always bounded below by the spectral radius, but as your example shows, the two quantities can be unequal. Your $A$ is nonzero in the first place, so its norm cannot possibly be zero. Using the equivalent definition on the RHS of $(1)$, or the singular value of $A$, it is not hard to show that $\|A\|_2=1$.
Square matrices whose operator norms equal their spectral radii are called radial. The name originates from the fact that the equality of $\|A\|_2$ and $\rho(A)$ is equivalent to the equality of $\|A\|_2$ and $r(A)=\max_{\|u\|=1}|\langle Au,u\rangle|$, the numerical radius of $A$. Radial matrices includes normal matrices as subclass, while the latter includes self-adjoint matrices as an even smaller subclass.