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I am wondering if the spectral radius of a matrix is may be some kind of a norm ($l_{\infty}$-norm?) of it and if that is convex. Any pointers to related ideas would be helpful too.

Rodrigo de Azevedo
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InterlacingStudent
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    For the matrix $\left[\matrix{0 & 1\0 & 0}\right]$ the spectral radius is zero, but the matrix is not. Too bad for a norm. – A.Γ. Aug 30 '15 at 20:44

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The spectral radius is the biggest eigenvalue of a matrix.

There is a linked norm called the spectral norm, which is infact the square root of the biggest eigenvalue of the matrix $A^*A$. (So not linked to its own spectral radius, but to the spectral radius of $A^*A$)

Norms are always convex. Due to triangle inequality and to the homogeneity.

Edit:

Biggest eigenvalue is of course not completly correct. You have to consider the absolute values of the eigenvalues...

HendryXT
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  • Thanks. Is there any literature you can point me to which talks of trying to minimize the spectral norm of a matrix among a set of possible perturbations by trying to minimize the spectral norm of each perturbation to be added in every step. Does this ring a bell? coordinate descent and rounding with spectral norm? – InterlacingStudent Aug 30 '15 at 20:56
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Spectral radius is not itself a norm, but it is a lower bound of any matrix norm. That is, $\rho(A)\leq \| A\|$ for any matrix norm $\| \cdot \|.$ (See here)

Also, for symmetric matrices $A \in \mathbb{S}^n$, $\rho(A)=\| A\|_2$. (See here)

For some more bounds on $\rho(A)$, see this paper.

William Zheng
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