$\operatorname{Rank}\left(C^{T} AC\right)=\operatorname{rank}(C)$ if $A$ is a $n \times n$ symmetric and positive definite matrix.

Question

Let $A \in R^{n \times n}$ be symmetric and positive definite and let $C \in R^{n \times m}$. Show that $\operatorname{Rank}\left(C^{T} A C\right)=\operatorname{rank}(C)$

We know that $Cx=0 \Rightarrow C^TACx=0$. This gives $nullity(C)\leq nullity (C^{T} A C)$. Now how to proceed?

Answer 1

Let $U=C^{-1/2}$ be symmetric square root of $A$.

Then $C^TAC = (UC)^T UC$.

So $\text{rank}(C^TAC) = \text{UC} $ (since $\text{rank}(B^TB) = \text{rank}(B)$ for all $B$)

and $\text{rank}(UC) = \text{rank}(C)$ as $U$ is invertible.

Answer 2

Note that $Cx =0$ iff $(Cx)^T A (Cx) = 0$ iff $C^T A Cx = 0$.

Hence $\ker C = \ker C^TAC$.