Let $X \in \mathbb{R}^{n \times n}$, $C \in \mathbb{R}^{m \times n}$ and $C \neq 0$ where $X > 0$ means $X$ is positive definite. I have worked out a proof that $X > 0 \Rightarrow CXC' > 0$ but I am not sure whether it is correct. Here it goes:
Let $v \in \mathbb{R}^m$ be an arbitrary vector and $u=C'v$. Then $v'(CXC')v = (C'v)'X(C'v) = u'Xu$. Since $X>0$, $u'Xu > 0$ which implies $v'(CXC')v > 0$ $\forall v$ then $CXC' > 0$.
Is there any problem with this proof? The reason why I am asking this is that I was also asked to determine the sufficient condition for $CXC' > 0$ given $X>0$, but isn't $X>0$ already the sufficient condition?
Thank you very much.
If $v \ne 0$, you can not garantee that $u=C'v \ne 0$.
Example: $n=2$ , $X=I_2$ and $ C=\left[\begin{matrix}1 & 0\\0 & 0\\\end{matrix}\right]$.
Then $ CXC'=C$ and $C$ is not positive definite.