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I need help with the following proof:

Given a symmetric positive-definite matrix, show that any diagonal submatrix of full rank must also be symmetric positive-definite.

Thanks

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  • $\begingroup$ Hint: if $A$ is an $n\times n$ matrix, then the upper left $k\times k$ submatrix of $A$ is $M^T A M$ for some $n\times k$ matrix $M$ (what is the matrix $M$?). $\endgroup$ – Minus One-Twelfth May 24 at 21:22
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An $n \times n$ matrix $Q$ is SPD if $x^T Q x> 0$ for every vector $x$. This is also true the vector $x$ contains zeros.

Let now $P= Q_{i:j,i:j}$ some submatrix around the matrix diagonal with $i<j<n $. Then we have $y^T P y = x^T Q x > 0$ where $x$ contains the the vector $y$ and $x_k = 0$ for $k<i$ and $k>j$.

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