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I have two correlation matrices A and B. They are:

  • Real symmetric (with ones on the diagonal)
  • Positive semi-definite (eigenvalues are $\ge 0$)

I want to try to prove that the average of these two matrices $C={1\over2}A + {1\over2}B$ still has the same properties.

It was a few years since my linear algebra courses and would like some help along the way here. I am assuming it's possible as I have tested it numerically and for my 5000 randomly generated correlation matrices the property holds (at least within machine precision).

Grateful for any kind of help!

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Assume $A$ and $B$ are positive semidefinite ($z^TAz\geq0$ for all $z\in\mathbb{R}^n$). So then $$ z^T\frac12(A+B)z=\frac12(z^TAz+z^TBz)\geq0, $$ and therefore$\frac12(A+B)$ is positive semidefinite. Symmetry is also preserved under addition $\frac12(A+B)^T=\frac12(A^T+B^T)=\frac12(A+B)$.

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  • $\begingroup$ Ah of course, I had forgotten that $z^TAz \ge 0$ was a property and was only thinking of the eigenvalues. Thank you! $\endgroup$ – Fredrik Oct 4 '12 at 13:05

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