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Let $A\in\mathbb C^{n\times n}$ be a Hermitian matrix such that all its principal minors are non-negative (i.e. for $B=\left(a_{l_il_j}\right)_{1≤i,j≤k}$ with $1≤l_1<...<l_k≤n$ we have $\det(B)≥0$). Then how to show that $A$ is positive semi-definite?

I thought maybe we could use induction since the condition is also satisfied for every submatrix, but I couldn't find an easy way. Can prove it elegantly?

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  • $\begingroup$ Yes. Use the Schur complement. $\endgroup$ – Chris Godsil Jun 19 '16 at 13:50
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Here is one way to prove the statement by mathematical induction. The base case $n=1$ is trivial. For the inductive step, let $B$ be any principal submatrix of $A$ and define $f(t)=\det(B+tI)$. By Jacobi's formula, $f'(t)=\operatorname{tr}\left(\operatorname{adj}(B+tI)\right)$. Since the diagonal entries of $\operatorname{adj}(B+tI)$ are some proper principal minors of $A+tI$, and by induction assumption, all proper principal submatrices of $A$ are positive semidefinite, $f'(t)$ must be positive whenever $t>0$. It follows that $f$ is strictly increasing on $[0,\infty)$. Consequently, $f(t)>f(0)=\det(B)\ge0$ for all $t>0$.

Hence all principal minors of $A+tI$ are positive when $t>0$. By the original Sylvester's criterion for positive definite matrices, $A+tI$ is positive definite. Taking $t\to0^+$, we conclude that $A$ is positive semidefinite.

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