what will be eigenvalues of this block matrix $ A=\begin{pmatrix}
O & C\\C^T & D
\end{pmatrix}$, where $O$ is null matrix of order $2$
$C=\begin{pmatrix}
1&0&0&0&0\\0&1&0&0&0
\end{pmatrix}$, and $D$ is a $0,1$ matrix whose eigenvalues are known.
What can be said about eigenvalues of $A$ in terms of eigenvalues of $D$? 
I am aware that eigenvalues of $D$ are interlaced between those of $A$. but here C is also of some special form and $D$ is also $0,1$. So can we say something more?
 A: Consider the eigenvalue equation
\begin{equation*}
A\mathbf{f}=\lambda \mathbf{f}
\end{equation*}
Write
\begin{equation*}
\mathbf{f}=\left(
\begin{array}{c}
\mathbf{f}_{1} \\
\mathbf{f}_{2}
\end{array}
\right)
\end{equation*}
and work things out noting that $C^{T}C$ is the $2\times 2$ identity matrix.
Then you find
\begin{equation*}
D\mathbf{f}_{2}=(\lambda ^{2}+1)\mathbf{f}_{2}
\end{equation*}
so $\mathbf{f}_{2}$ is an eigenvector of $D$, say at the eigenvalue $\mu $.
Then
\begin{equation*}
\mu =\lambda ^{2}+1
\end{equation*}
There are two $\mu $ 's so also two $\lambda $'s.
Edit: Here is the full story. 
We split
\begin{equation*}
\mathbf{f}=\left(
\begin{array}{c}
\mathbf{f}_{1} \\
\mathbf{f}_{2}
\end{array}
\right)
\end{equation*}
where $\mathbf{f}_{1}$ has two components and $\mathbf{f}_{2}$ three.
\begin{eqnarray*}
A\mathbf{f} &=&\lambda \mathbf{f} \\
0\mathbf{f}_{1}+C\mathbf{f}_{2} &=&\lambda \mathbf{f}_{1} \\
C^{T}\mathbf{f}_{1}+D\mathbf{f}_{2} &=&\lambda \mathbf{f}_{2} \\
\mathbf{f}_{1} &=&\frac{1}{\lambda }C\mathbf{f}_{2} \\
\frac{1}{\lambda }C^{T}C\mathbf{f}_{2}+D\mathbf{f}_{2} &=&\lambda \mathbf{f}
_{2}
\end{eqnarray*}
\begin{equation*}
\left(
\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0
\end{array}%
\right) \left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}
\right) =\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)
\end{equation*}
\begin{eqnarray*}
\frac{1}{\lambda }\mathbf{f}_{2}+D\mathbf{f}_{2} &=&\lambda \mathbf{f}_{2} \\
D\mathbf{f}_{2} &=&(\lambda ^{2}+1)\mathbf{f}_{2}
\end{eqnarray*}
I do not know what you mean by $D$ is a $0,1$ matrix. It seems it is $
3\times 3$ in which case it has three eigenvalues $\mu _{j}$ with $j=1,2,3$.
Then
\begin{equation*}
\mu _{j}=\lambda _{j}^{2}+1
\end{equation*}
from which $\lambda _{j}^{2}$ follows. In my earlier answer I said that $D$
had two eigenvalues. This must be three.
