Let $n \geq m$ and let $C$ be a $n \times m$ full rank matrix, that is $rank(C) =m$. Considering that $D$ is a diagonal positive semidefinite matrix, under which conditions is the $ m \times m$ matrix $X=C^\mathsf{T}DC$ invertible? Is it enough to have $rank(D) \geq m$?
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$\begingroup$Since $X=C^TDC$ is generally positive semi-definite, $X$ is nonsingular if and only if it is positive definite, that is, $v^TXv=v^TC^TDCv>0$ for all nonzero $v$. Another way, how to say this, is that $D$ is positive definite on the column-span (range) of $C$. Also, $X$ is nonsingular if and only if the intersection of the range of $C$ and the nullspace of $D$ is trivial ($\mathrm{Im}(C)\cap\mathrm{Ker}(D)=\{0\}$). Note that this is true for any positive semi-definite $D$, not necessarily diagonal.
Certainly, $\mathrm{rank}(D)\geq m$ is not sufficient for $X$ to be nonsingular. Consider $$ C=\begin{bmatrix}1 \\ 0\end{bmatrix}, \quad D=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}. $$
$\endgroup$ 0 $\begingroup$If $D$ is positive definit, and $C$ has full rank ($C\in \mathbb R^{n,m}$, $n\ge m$, $rank(C)=m$), then $C^TDC$ is positive definite, hence invertible: Take $x\ne 0$ then $$ x^TC^TDCx = (Cx)^TD(Cx) >0 $$ as $D$ is positive definite and $Cx\ne 0$.
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