Given an SPD matrix, any diagonal submatrix of full rank must be SPD.

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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• 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 $M$?). – Minus One-Twelfth May 24 at 21:22

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 $$i. Then we have $$y^T P y = x^T Q x > 0$$ where $$x$$ contains the the vector $$y$$ and $$x_k = 0$$ for $$k and $$k>j$$.