Modification of Poincaré Separation Theorem We know from  Poincaré's separation theorem that for semi-orthogonal $\mathbf{B} \in \mathbb{R}^{n\times k}$ and real, symmetric $\mathbf{A} \in \mathbb{R}^{n \times n}$ with eigenvalues $\lambda_1 > \lambda_2 > ... > \lambda_n$, the product $\mathbf{B}^T \mathbf{AB}$ has eigenvalues $\mu_i$ such that,
$$ \lambda_i \geq \mu_i \geq \lambda_{n-k+i}, \quad i = 1,2,...,k$$
From searching online there seems to be a proof for this theorem in Magnus and Neudecker, "Matrix differential calculus with applications in statistics and econometrics," but I do not access to the text. I'm sure this proof would shed light on my following question.
My question: is there any way to find a similar result for the product $\mathbf{ABB}^T$, i.e. can we bound the eigenvalues of the product relative to the eigenvalues of the matrix $\mathbf{A}$?
 A: Yes. In fact, even though $\mathbf{ABB}^T$ need not be symmetric, we can show that its non-zero eigenvalues will be real and identical to the non-zero eigenvalues of $\mathbf{B}^T\mathbf{AB}$. 
First, observe that for any non-zero eigenvalue, $\lambda$, of $\mathbf{ABB}^T$, there exists a corresponding left eigenvector, $\mathbf{v}$, such that
$$\mathbf{v}^T\mathbf{ABB}^T=\lambda\mathbf{v}^T \iff \mathbf{v}^T = \frac{\mathbf{v}^T\mathbf{AB}}{\lambda}\mathbf{B}^T$$
which implies that $\mathbf{v}$ will necessarilly belong in the columnspace of $\mathbf{B}$. However, since $\mathbf{B}$ is semi-orthogonal, $\mathbf{BB}^T$ will be an orthogonal projector on its columnspace. Therefore, we have that  
$$\mathbf{v}^T\mathbf{BB}^T\mathbf{ABB}^T=\lambda\mathbf{v}^T $$
which in turn implies that $\mathbf{BB}^T\mathbf{ABB}^T$ and $\mathbf{ABB}^T$ have identical non-zero eigenvalues.
Now consider the eigenvalue decomposition, $\mathbf{U\Lambda U}^T$, of $\mathbf{B}^T\mathbf{AB}$. Then, $\mathbf{BB}^T\mathbf{ABB}^T = \mathbf{BU\Lambda U}^T\mathbf{B}^T$, which is also an eigendecomposition, since $\mathbf{U}^T\mathbf{B}^T\mathbf{BU}= \mathbf{I}$
Hence, $\mathbf{BB}^T\mathbf{ABB}^T$ and $\mathbf{B}^T\mathbf{AB}$ will also have identical non-zero eigenvalues, which concludes our proof.
