• 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.

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.

• 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.