I am very interested in finding the eigenvalues of the matrix $\bf{A}$ below, which consists of $4$ block matrices of the same size $n \times n$ ,
$
\textbf{A}=\begin{pmatrix} {\bf{A}}_1 & {\bf{A}}_2 \\ {\bf{A}}_3 & {\bf{A}}_4 \end{pmatrix}
$, with ${\bf{A}}_2=({\bf{A}}_3)^T$, where no entry is equal to $\bf{0}$ (matrix with only zeros as its entries). The matrices are allowed to have complex entries.
Furthermore following relations hold true:
$$ \begin{aligned} {\bf A}_1 &= {\bf B}_1^T {\bf B}_1 + {\bf B}_2^T {\bf B}_2\\ {\bf A}_2 &= {\bf B}_2^T {\bf B}_3\\ {\bf A}_3 &= {\bf B}_3^T {\bf B}_2\\ {\bf A}_4 &= {\bf B}_3^T {\bf B}_3 \end{aligned} $$
I know that the determinant of this matrix $\bf{A}$ is obtained as
$$ \det({\bf{A}} ) = \det({\bf{A}}_1) \det \left({\bf{A}}_4-{\bf{A}}_3{\bf{A}}_1^{-1}{\bf{A}}_2 \right),$$
however I do not know how I would compute the eigenvalues of this matrix ${\bf{A}}.$ I am thinking of using:
$$ \det(\textbf{A}_1-\lambda \textbf{I}) \det(\textbf{A}_4-\lambda \textbf{I} - \textbf{A}_3 (\textbf{A}_1-\lambda \textbf{I})^{-1} \textbf{A}_2) = 0 $$
However, could I analytically solve for $\lambda$? Can anyone please provide me with some advice? I would highly appreciate any kind of help.
I come up with a partial solution.
First, let us simplify notation settings $${\bf{B}}={\bf{B}}_1, {\bf{C}}={\bf{B}}_2,{\bf{D}}={\bf{B}}_3.$$ The matrix is $${\bf{A}}=\begin{pmatrix}{\bf{B}}^T{\bf{B}}+{\bf{C}}^T{\bf{C}} & {\bf{C}}^T{\bf{D}} \\ {\bf{D}}^T{\bf{C}} & {\bf{D}}^T{\bf{D}} \end{pmatrix},$$ which is nothing but a product $${\bf{A}}=\begin{pmatrix}{\bf{C}}^T & {\bf{B}}^T \\ {\bf{D}}^T & \bf{0} \end{pmatrix}\begin{pmatrix} {\bf{C}} & {\bf{D}} \\ {\bf{B}} & \bf{0}\end{pmatrix}:={\bf{M}}^T{\bf{M}}$$ The two matrices are transpose to each other, hence they have equal determinants.
The determinant is $$\text{det}({\bf{A}})=\text{det}({\bf{M}})^2=\Big(\text{det}({\bf{C}})\;\text{det}\left(-{\bf{B}}{\bf{C}}^{-1}{\bf{D}} \right)\Big)^2=\Big(\text{det}({\bf{B}})\;\text{det}({\bf{D}}) \Big)^2$$ This result fits with the fact that the matrix $\begin{pmatrix} {\bf{C}} & {\bf{D}} \\ {\bf{B}} & \bf{0}\end{pmatrix}$ is anti-triangular.
NOTE
If you can provide more details about the matrices, we could find even more.
EDIT
The matrix can also be written with the use of transpose triangular block matrices: $${\bf{A}}=\begin{pmatrix}{\bf{B}}^T & {\bf{C}}^T \\ \bf{0} & {\bf{D}}^T \end{pmatrix}\begin{pmatrix} {\bf{B}} & {\bf{0}} \\ {\bf{C}} & \bf{D}\end{pmatrix}:={\bf{\Omega}}^T{\bf{\Omega}}$$ Eigenvalues of $\bf{A}$ are singular values of ${\bf{\Omega}}^T$ or ${\bf{\Omega}}.$ In this question is their matrix $M$ equal to my ${\bf{\Omega}}^T$ except the names of blocks. This is a next small step towards a solution you are looking for.