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Based on one previous question: Why is the function $\|\mathbf{J}\|_{\infty}$ $1$-Lipschitz w.r.t to the Euclidean norm?.

For a real symmetric matrix $J$, let $\|J\|_{\infty}$ be the spectral radius of $J$. Why do we have $$ \|J\|_{\infty}=\sup_{\|v\|=1}\langle v, Jv\rangle ? $$ (Here I think that the sup is taken over $\|v\|_2=1$?)

I check the definition of the spectral radius of a matrix, which is $\rho(J):=\max_{1\le i\le n}|\lambda_i|=\|J\|_2$ with eigenvalues $\lambda_i$.

And the right hand side on the first display looks like the operator norm of the matrix $J$?

Here I get $$ \|J\|_2=\sup_{\|x\|_2=1}\|Jx\|_2=\sup_{\|x\|_2=1}\langle Jx, Jx\rangle=\sup_{\|x\|_2=1}\langle x, J^TJx\rangle $$ but this is not equal to $\sup_{\|x\|_2=1}\langle x, Jx\rangle$?

Hermi
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  • Since $J$ is symmetric and spectral norm and operator norm are invariant under multiplication by orthogonal matrices, you can assume $J$ is diagonal. In this case the norms are obvious, and it seems that they are indeed both equal in this case. – Mason May 30 '22 at 23:36
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    @Hermi This post contains quite some inconsistencies. First of all, the spectral radius is not a norm for general matrices. Also, as defined $||J||_\infty$ is not necessarily positive. Second of all, the spectral radius of a matrix is not equal to $||J||_2$ in general. Finally, $||J||_2$$ is equal the square root of any of the expressions on the right-hand side. – KBS May 31 '22 at 10:24

1 Answers1

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The spectral radius of a matrix $J$ is defined as

$$\rho(J):=\max\{|\lambda|:\lambda\in\mathrm{spec}(J)\},$$

where $\mathrm{spec}(J)$ is the spectrum of the matrix $J$. In the case of a positive semidefinite matrix $J$, then the spectral radius simply coincides with the maximum eigenvalue $\lambda_{\mathrm{max}}(J)$ which is also equal to

$$\rho(J)=\lambda_{\mathrm{max}}(J)=\sup_{\|v\|_2=1}\langle v, Jv\rangle.$$

Note that this is not the case of other symmetric matrices, such as, $J=-1$ for which the spectral radius is 1 but the maximum eigenvalue is -1. This could be fixed by considering

$$\rho(J)=\max\{\lambda_{\mathrm{max}}(J),-\lambda_{\mathrm{min}}(J)\}=\max\{\sup_{\|v\|_2=1}\langle v, Jv\rangle,-\inf_{\|v\|_2=1}\langle v, Jv\rangle\}.$$

On the other hand, $||J||_2$ is the maximum singular value of the matrix $J$ which is given by

$$||J||_2:=\left(\sup_{\|v\|_2=1}\langle Jv, Jv\rangle\right)^{1/2}.$$

It is, in general, different from the spectral radius of $J$. However, in the symmetric case, we do have that

$$\rho(J)=\|J\|_2,$$

which follows from the definitions of the spectral radius and the spectral norm, and the fact that the eigenvalues of $J^2$ are the squares of those of $J$.

KBS
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  • Thanks! For a symmetric matrix, why do we have $\rho(J)=|J|_2$? – Hermi May 31 '22 at 19:21
  • @Hermi Left as an exercise. Just consider the definitions. – KBS May 31 '22 at 19:44
  • As in https://en.wikipedia.org/wiki/Matrix_norm, it says that $|J|_2$ is the largest singular value of $J$. But $rho(J)$ is the largest absolute value of eigenvalue. Are they the same? – Hermi May 31 '22 at 20:04
  • @Hermi No they are not. I have already mentioned that in my answer. But they are when the matrix is symmetric. – KBS May 31 '22 at 22:33
  • Oh, I see. Thanks! – Hermi May 31 '22 at 23:33
  • I read a paper. Why does the author say $|J|{\infty}=\sup{|x|=1}<x.Jx>$ Is this a typo? If $|J|_{\infty}$ means the sup-norm of $J$, this is not true, right? – Hermi Jun 01 '22 at 19:02
  • What is $J$ and how is that norm defined? Please be precise in your statements. Also what paper are you talking about? If you mention a resource, provide a link to it. – KBS Jun 01 '22 at 20:03