# $S^tDS$ positive semidefinite if $D$ is so

Let $\mu_i \in \mathbb{R}_{\ge0}$, $i=1,...,n$, and $D=(\delta_{ij}\mu_i)\in\text{M}(n,\mathbb{R})$.

$D$ itself is surely positve semidefinite, this can be proved via prinipal minors.

But why is $S^tDS$ also positive semidefinite for a matrix $S$ with real entries? Is there a proof using principal minors?

• how about looking at $x^T S^T D S^T x$? – LinAlg Jun 27 '18 at 11:35
• Could you explain please? I guess you mean $x^tS^tDSx=(Sx)^tDSx$? – user337073 Jun 27 '18 at 11:37
• right, why the interest in principal minors? – LinAlg Jun 27 '18 at 11:38
• Just thought it would be easier using positive semidefinte <=> all principal minors $\ge 0$. – user337073 Jun 27 '18 at 11:40