Eigenvalues of block matrix related What are the eigenvalues of following block matrix?
$$\begin{bmatrix}
A & B \\ 
B^T & O
\end{bmatrix}$$
Here, $A$ and $B$ are any square matrices of order $n$, $O$ is zero matrix of order $n$.
 A: Generally, determinant of block matrix can be obtained following formula.
$$
\det
\begin{bmatrix}
A_{0} & B_{0} \\
C_{0} & D_{0}
\end{bmatrix}
=
\det(D_{0})
\det(A_{0}-B_{0}D_{0}^{-1}C_{0})
$$
Then, this problem's case: $(A_{0},B_{0},C_{0},D_{0})=(A-\lambda E,B,B^{\text{T}},-\lambda E)$, where $\lambda$ is eigenvalue. Therefore,
$$
\begin{align}
\det
\begin{bmatrix}
A-\lambda E & B \\
B^{\text{T}} & -\lambda E
\end{bmatrix}
=&
\det(-\lambda E)
\det((A-\lambda E)-B(-\lambda E)^{-1}B^{\text{T}}) \\
=&
\det(-\lambda(A-\lambda E)-B B^{\text{T}}) \\
=&
\det(\lambda^2 E -\lambda A -B B^{\text{T}})
\end{align} 
$$
This determinant only cannot identify the eigenvalues as far as $A$ and $B$ are not supplied concrete elements. However, we can get ingredients of matrices how this quadratic matrix polynomial is made to decompose. To sum up, we can see what eigenvalues are constructed by $n$ orders matrices. Hence, we attempt to factorize above determinant.
$$
\begin{align}
\det(\lambda^2 E -\lambda A -B B^{\text{T}})=&\det(X-\lambda E)(Y - \lambda E) \\
=&\det(X-\lambda E)\det(Y - \lambda E) \\
\end{align}
$$
Above determinants are changeable so that following conditions can be applied to solve these matrices as $X,Y$. Especially, secondly condition is allowed commutative.  
$$
\begin{cases}
X+Y=A \\
XY=YX=-BB^{\text{T}}
\end{cases}
$$
Because,
$$
\begin{cases}
\det(X-\lambda E)(Y - \lambda E)=\det(\lambda^2 E-\lambda(X+Y)+XY) \\
\\
\det(Y-\lambda E)(X - \lambda E)=\det(\lambda^2 E-\lambda(X+Y)+YX)
\end{cases}
$$
Thus, $X,Y$ are estimated below.
$$
\det\biggl(\cfrac{A+\sqrt{A^2+4BB^{\text{T}}}}{2}-\lambda E\biggr)
\det\biggl(\cfrac{A-\sqrt{A^2+4BB^{\text{T}}}}{2}-\lambda E\biggr)
=0
$$
In conclusion, eigenvalues $\lambda$ show these matrices.
Its means that the block matrix includes $X$'s eigenvalues $\lambda_{X}$ and $Y$'s eigenvalues $\lambda_{Y}$.
Incidentally, $A=\pm 2i\sqrt{BB^{\text{T}}}$ are special cases owing to multiple root as $X=Y$. Then, eigenvalues $\lambda$ correspond to just half $A$'s eigenvalues $\lambda_{0.5A}$.
