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I run into the following fact:

Let $X$ be $n \times n$ positive semidefinite matrix, and let $a$ be an $n$ dimensional vector consisting of diagonal entries of $X$ (i.e. $a_i = X_{ii}$). Suppose additionally that sum of all entries of $X$ is 1. Under these assumptions, it seems to be true that if we define: $$ Y = \begin{bmatrix} 1 & a^\top \\ a & X \end{bmatrix} $$ then $Y$ is positive semidefinite. I'm strongly convinced that it is true, but can't seem to find a proof.

Using a criterion of positivity, and assuming some additional generic conditions, one can prove that positivity of $Y$ is equivalent to positivity of the matrix $X - aa^\top$ or of the number $1 - a^\top X^{-1} a$ (assuming $X$ - invertible), but either of these is a bit elusive.

I'd be grateful for any further ideas or references. Also (with relation to the issue, but not exclusively for that reason): does the inner prodct defined on matrices by $\left<A,B \right> = \mathrm{Tr}(AB)$ have any nice properties in terms of matrix multiplication?

Edit: I was a bit vague about the statement: in a number of places, when it should have been "semidefinite" there was "definite". The ideas for an attempt at solution are just ideas, so I think it's OK to leave them a little imprecise. Thank you for pointing this out, as well as for the counterexamples showing that $Y$ can be only semidefinite rather than definite.

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Seems to fail if $n=1$; then $X=(1)$, $a=(1)$, $Y=\pmatrix{1&1\cr1&1\cr}$ is not positive definite. – Gerry Myerson Mar 6 '12 at 3:55

Seems to fail if $X=(1/n)I$ where $I$ is the $n\times n$ identity matrix. The matrix $Y$ then has zero as an eigenvalue.

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