# Is positively weighted sum of eigenvalues of a matrix X, convex function of X?

Is positively weighted sum of eigenvalues of a matrix X, convex function of the matrix X?

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1. Is this homework? Smells like homework. Please tag it if it is. 2. What have you tried? –  Inquest Oct 16 '12 at 19:00
Given a matrix, there is no cannonical ordering of the eigenvalues unless you specify an ordering of $\mathbb C$. If you take a continuous function which depends only on the eigenvalues, it must be symmetric. Therefore, the only weighted sum of the eigenvalues you can take (subject to continuity in the entries of the matrix) is the trace. –  Aaron Oct 16 '12 at 19:01
It is not homework, it is a valid research question. I am tending to believe that only weight vectors whose values are in decreasing order make this function convex, but I am not sure. –  Lee Oct 16 '12 at 19:03
@Aaron: I don't quite understand why I cannot form such a function, it seems quite natural to me. For example, the maximum of a trace of a product of two matrices is such a function. It is a weighted sum of eigenvalues. –  Lee Oct 16 '12 at 19:08
The eigenvalues of two matrices is not determined by the eigenvalues of the two matrices, not even when the matrices commute. Even if it were, it would not be a linear function of the eigenvalues (it would be quadratic). I'm not sure what you mean by the maximum of a trace, though. The trace gives you one number, not a set of numbers. I think that I might not be fully understanding what you are asking. Can you rephrase your question in a more rigorous fashion? –  Aaron Oct 16 '12 at 19:37