# Spectral radius is not matrix norm.

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 $\|A\|=\epsilon$?

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

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.

• The spectral radius is only equal to the largest eigenvalue when the matrix is symmetric – information_interchange Mar 28 '20 at 1:50
• @information_interchange No. When $A$ is real symmetric, the spectral radius is always equal to the magnitude of the largest-sized eigenvalue, but not necessarily equal to the largest eigenvalue. E.g. when $A=\operatorname{diag}(0,-1)$, we have $\rho(A)=1\ne0=\lambda_\max(A)$. – user1551 Mar 28 '20 at 5:13

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.

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

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

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 $\|U_L^{-1} A U_L \| = r$.
• Choose $L$ large enough so that $\|A\|_L < \epsilon$. – copper.hat Dec 28 '17 at 20:20
• 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. – Ben Grossmann Dec 28 '17 at 20:24
• It doesn't disprove anything. It exhibits a norm such that $0=\rho(A) \le \|A\|_L \le \rho(A)+ \epsilon = \epsilon$. – copper.hat Dec 28 '17 at 20:28