Can someone help with this exercise?
Given a symmetric n x n positive definite matrix A, and an arbitrary non-
singular n x n matrix P, show that P'AP is positive definite.
We know that $\displaystyle (Ax,x)\geq0, \forall x\in V $, $\displaystyle A'=A$, and $\displaystyle (Ax,x)=0 \implies x=0$.
So we want to show P'AP>0.
Let $\displaystyle x\in V,
(P'APx,x)=(A(Px),(Px))\geq0$, since $\displaystyle Px \in V$
Similarly, $\displaystyle (A(Px),(Px))=0 \implies Px=0$ and since P is invertible this implies $\displaystyle x=0$.
Now self-adjoint should be given probably from a theorem, but here's a proof.
If A=A', then (P'AP)'=P'A'P''=P'AP.