Proof $C = A \cdot B \cdot A^{\top}$ nonsingular if $A$ has full rank.

Question

Given are the matrices $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times n}$. $B$ is known to be symmetric positive definite.

I made the following observation: The matrix $C = A \cdot B \cdot A^{\top} \in \mathbb{R}^{m \times m}$ is singular for $\text{rank}\,{(A)} < m$. Example:

$$ C = \begin{pmatrix} 1 & 2 \\ 4 & 8 \end{pmatrix} \cdot \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 4 & 8 \end{pmatrix}^{\top} = \begin{pmatrix} 25 & 100 \\ 100 & 400 \end{pmatrix} $$

with $\det C = 25 \cdot 400 - 100^2 = 0$.

Is there a proof for the claim: If $A$ has maximal rank, then C is nonsingular?

Any hints or help is highly appreciated.

Answer 1

Because $B$ is symmetric and positive definite, there exists $D$ with $B=DD^T$ with $D$ symmetric and positive definite. Then $$ C=ADD^TA^T=AD(AD)^T. $$ From $\operatorname{null}(X)=\operatorname{null}(XX^T)$, and the rank-nullity theorem, the rank of $C$ is equal to the rank of $AD$; as $D$ is invertible, this is equal to the rank of $A$.

So $C$ will be non-singular precisely when $A$ has rank $m$. For the rank of $A$ to be equal to $m$ and maximal, you need $m\leq n$.