Eigenvalues of positive semidefinite matrix

Question

Today I found a problem concerning positive semidefinite matrix:

Suppose $A$ is a positive semidefinite matrix, with eigenvalues: $\lambda_1\geq\dots\geq\lambda_n$, and diagonal elements: $a_{11}\geq\dots\geq a_{nn}$. Now prove that, for any $k\leq n$, $\sum_{i=1}^k\lambda_i\geq\sum_{i=1}^ka_{ii}$.

Now for $k=1,n-1,n$, the problem is trivial. But what about the other $k$?

This proposition is very likely to be true as I've run some simulation on MATLAB, without finding any exception.

However this seems hard to prove. I think maybe some transformation needs to be done. Anyone has any clue? Thanks~

Besides there are some background in PCA. I shall provide some supplement if necessary later.

Consider the fact that $\lambda_1I-A$ is positive semidefinite. — DiaryofNewton, Jul 26 '19 at 00:19
yeah of course, that's the requirement of "positive semidefinite". — DiaryofNewton, Jul 26 '19 at 04:30
The definition I know says that a matrix $A$ is positive semidefinite iff $x^T A x \ge 0$, without requiring $A$ to be symmetric — Sudix, Jul 26 '19 at 05:36

score 0 · Answer 1 · answered Jul 25 '19 at 23:17

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By Sylvester's criterion, a symmetric matrix is positive semi-definite if and only if all its principal minors are non-negative. This in particular means that all the principal submatrices are also positive semi-definite themselves. You checked the result for $k=1$, so it is true in general by induction.

answered Jul 25 '19 at 23:17

Ivo Terek

77,665

I was about to post a similar answer, but I struggled to clarify the relationship between the eigenvalues of the principle minors and the eigenvalues of the full matrix. Could you elaborate on this point? – Theo Bendit Jul 25 '19 at 23:21
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You're right, that's actually a non-trivial step and the eigenvalues of the principal submatrices do not need to be the same as the eigenvalues of the original matrix, I was a bit hasty. It seems that we can bound the eigenvalues of the submatrix between the eigenvalues of the original matrix, according to this answer. – Ivo Terek Jul 25 '19 at 23:36
yeah it's true, only considering the minors seems not strong enough – DiaryofNewton Jul 26 '19 at 00:22
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Ok I think Ivo Terek got the key. Using those bounds I can indeed solve the problem. Thanks A Lot !!!!!!!!!!!!!!! This problem bothers me for a whole day lol. – DiaryofNewton Jul 26 '19 at 02:56

Eigenvalues of positive semidefinite matrix

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