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
2026-03-26 09:17:42.1774516662
Given an SPD matrix, any diagonal submatrix of full rank must be SPD.
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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 $k>j$.