If $A$ is positive definite then any principal submatrix of $A$ is positive definite.


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In the proof i dont understand about $j_i$'s.. Can some one interpret the proof in a simpler way(may be the original one is simpler)

if someboy is rewriting the proof it would give me a better understanding.(if possible)

  • 1
    $\begingroup$ Looks like the author defined principal submatrix as one with the same indices of columns and rows removed. Could that be a problem? $\endgroup$
    – Element118
    Nov 29 '15 at 10:36

The $j_i$'s represent the coordinates removed from $A$ to obtain the principal submatrix $B$. So we want to show $y^{\intercal}By > 0$ for all non-zero $y \in \mathbb{R}^{n-s}$. The argument shows that any such $y$ can be exported to some $x \in \mathbb{R}^n$ (by choosing the missing coordinates as zero) such that $y^{\intercal}By = x^{\intercal}Ax$. Then the result follows.

  • $\begingroup$ can you please elaborate on it more? $\endgroup$
    – David
    Nov 29 '15 at 13:42

I think the proof is saying that since x can be anything, we can let the first part of x be 0s and the rest be any number (z). The final result depends now only on the submatrix and since z can be anything, that means x is an arbitrary vector.


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