Is linear independence preserved with positive semidefinite product Let $\mathbf{A}$ and $\mathbf{B}$ be $m\times n$ real matrices with $m\geq n$ and $\mathbf{\Sigma}$ be a $n\times n$ symmetric positive definite matrix. Is the following statement true or false:
If there does not exist $\lambda\in\mathbb{R}$ such that $$\mathbf{A}\mathbf{A}'=\lambda \mathbf{B}\mathbf{B}',$$ then there does not exist $\delta\in\mathbb{R}$ such that $$\mathbf{A}\mathbf{\Sigma}\mathbf{A}'=\delta\mathbf{B}\mathbf{\Sigma}\mathbf{B}'$$
The statement is trivially true when $n=1$. I could not find an example in which it fails, nor could I show it holds universally. I suspect it could be shown to be true/false by decomposing $\mathbf{\Sigma}$ using a Cholesky or Eigenvalue decomposition, but was not able to reach a conclusion.
 A: For simplicity let's restrict our attention to the case that $m = n$ and that $B$ is invertible. I assume that $(-)'$ means the transpose, which I'll write $(-)^T$. Then
$$A A^T = \lambda B B^T \Leftrightarrow B^{-1} A A^T (B^T)^{-1} = \lambda$$
where by $\lambda$ I mean $\lambda$ times the identity matrix, and similarly
$$A \Sigma A^T = \delta B \Sigma B^T \Leftrightarrow B^{-1} A \Sigma A^T (B^T)^{-1} = \delta \Sigma.$$
Now we can remove a variable by writing $C = B^{-1} A$ so that $C^T = A^T (B^T)^{-1}$; the question becomes, if there does not exist $\lambda$ such that
$$C C^T = \lambda$$
then is it also true that there does not exist $\delta$ such that
$$C \Sigma C^T = \delta \Sigma.$$
This is false. The first condition says that $C$ is not an orthogonal matrix up to scale, or equivalently that the singular values of $C$ are not all equal. And the second says a similar thing, but with respect to a different quadratic form, namely the one given by $\Sigma$. And it is certainly possible for a matrix to be orthogonal with respect to one quadratic form but not another. Explicitly, take $n = 2$ and
$$\Sigma = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array} \right], C = \left[ \begin{array}{cc} 2 \cos t & 2 \sin t \\ -\sin t & \cos t \end{array} \right].$$
Then we can take $\delta = 4$. Hopefully it's very clear now.
