# Which matrix/operator in a cone has the largest negative spectral part?

• Background:

Let $\mathcal{K}$ be set (convex cone, if you like) of symmetric matrices of order $n$. Each matrix $A \in \mathcal{K}$ can be decomposed in a unique way as $A=A_{+}-A_{-}$, where $A_{+}$ and $A_{-}$ are mutually orthogonal positive definite matrices.

• The question

Let us define $m(A)=\frac{||A_{-}||}{||A||}$, using the Frobenius norm. I am interested in finding a matrix $A \in \mathcal{K}$ that maximizes $A$.

• Notes:

My original motivation comes from the case when $\mathcal{K}$ is the set of nonnegative matrices. However, the general version at the very least seems to make sense.

Consider a (0,1) adjacency matrix of a bipartite graph. The famous Coulson-Rushbrook result says its spectrum is symmetric so we get $m(A)=1/\sqrt{2}$. This is a strong contender (optimal in the order two case, by a result of Hirriart-Urruty and Seeger), but it possible to do better in some cases...

A possible generalization could be to decompose $A$ with respect to a more general cone, instead of the cone of positive definite matrices, using Moreau decomposition.

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you mean "maximizes m(A)" right? isn't this maximized by a matrix for which $A_+=0$, then $m(A)=1$. – Suvrit Mar 6 '14 at 2:55
@Suvrit But if $A_{+}=0$ then $A$ could land outside $\mathcal{K}$. – Felix Goldberg Mar 6 '14 at 7:07
You wrote about: $K$ is the set of symmetric matrices, in which case this is fine. If $K$ is a closed convex cone such as nonengative orthant or psd matrices, then $A_- = 0$ for all matrices, so I guess I'm confused about your terminology a bit... – Suvrit Mar 6 '14 at 14:53