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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 $\|A\|=\epsilon$?

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    $\begingroup$ You can conjugate $A$ to $\pmatrix{0&\epsilon\\0&0}$. $\endgroup$ – Lord Shark the Unknown Dec 28 '17 at 19:26
  • $\begingroup$ what is $\rho$? $\endgroup$ – mathreadler Dec 28 '17 at 19:27
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    $\begingroup$ @mathreadler its spectral radius. $\endgroup$ – Tien Dec 28 '17 at 19:31
  • $\begingroup$ @Lord Shark the Unknown why I conjugate A to that matrix? $\endgroup$ – Tien Dec 28 '17 at 19:33
  • $\begingroup$ @Tien his point is that $\rho(\epsilon A) = \rho(A)$ $\endgroup$ – Omnomnomnom Dec 28 '17 at 19:42
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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.

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From this Wikipedia page, the spectral norm of a matrix $A\in\mathbb{C}^{n\times n}$ is defined as

$$\rho\left(A\right)=\max_{1\leq i\leq n}\left\{\left|\lambda_{i}\right|\right\}$$

where the $\lambda_{i}$'s are the eigenvalues of the matrix. In your case

$$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$

is triangular, so the diagonal entries are the eigenvalues. Thus you have $\lambda_{1}=\lambda_{2}=0$, and it follows immediately that

$$\rho\left(A\right)=0$$

This is not a norm since $A\neq 0$ and a norm $\left\|\cdot\right\|$ must satisfy

$$\left\|A\right\|=0\:\Longleftrightarrow\: A=0$$

by definition.

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Define the norm $\| A \|_L = \|V_L^{-1} A V_L\|$ where $V_L = \begin{bmatrix} 1 & 0 \\ 0 & {1 \over L}\end{bmatrix}$.

Note that $\lim_{L \to \infty} V_L^{-1} A V_L = 0$

Addendum:

Note that the above norm is an induced norm. If we let $\|x\|_L = \|V_L^{-1} x\|$ then $\|A\|_L = \sup_{\|x\|_L \le 1} \|Ax\|_L$.

Another addendum (the question changed):

With $U_L=\begin{bmatrix} L & 0 \\ 0 & 1\end{bmatrix}$ show that $\lim_{L \downarrow 0} \|U_L^{-1} A U_L\| = 0$ and $\lim_{L \to \infty} \|U_L^{-1} A U_L\| = \infty$.

Now, for any $r>0$ use the intermediate value theorem to choose some $L$ such that $\|U_L^{-1} A U_L \| = r$.

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  • $\begingroup$ I don't understand how this answers the question $\endgroup$ – Omnomnomnom Dec 28 '17 at 20:19
  • $\begingroup$ Choose $L$ large enough so that $\|A\|_L < \epsilon$. $\endgroup$ – copper.hat Dec 28 '17 at 20:20
  • $\begingroup$ @Omnomnomnom: Am I missing something? $\endgroup$ – copper.hat Dec 28 '17 at 20:22
  • $\begingroup$ that doesn't disprove of a norm such that $\rho(A) \leq |||A|||\leq \rho(A)+\epsilon$ for some fixed $\epsilon$. If you took $L \to )^+$, on the other hand, that would accomplish what we want. $\endgroup$ – Omnomnomnom Dec 28 '17 at 20:24
  • $\begingroup$ It doesn't disprove anything. It exhibits a norm such that $0=\rho(A) \le \|A\|_L \le \rho(A)+ \epsilon = \epsilon$. $\endgroup$ – copper.hat Dec 28 '17 at 20:28

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