I have question about the proof about positive semidefinite (p.s.d) of a matrix. Let's say $M$ is a d by d p.s.d matrix, $H$ is any d by n matrix with n larger(or much larger) than d. Then how about the p.s.d property of $H^{T}MH$? I am interested a formal proof rather than a quick answer or a counter example.
I gave some attempts by following the property of p.s.d matrix, i.e for any (d by 1)vector $x$, $x^{T}Mx$ is non-negative.
Therefore, for any (n by 1) vector $y$, $y^{T}H^{T}MHy$ can be simplified as $x^{T}Mx$ by replacing $Hy$ with x. Since $x^{T}Mx$ is non-negative, $y^{T}H^{T}MHy$ is non-negative, which implies the p.s.d property of $H^{T}MH$
Is there any flaw in my proof?
many thanks
henry
That is correct. If $y$ is an arbitrary vector, let $x=Hy$, so that $$y^{*}H^{*}MHy=\left(Hy\right)^{*}M\left(Hy\right)=x^{*}Mx\geq0$$ by the PSD property of $M$.