Eigen value of a block matrix I have following block matrices:
$$M_1 = \left(\begin{array}{cc}A & B\\B' & D\end{array}\right)$$
and 
$$M_2 = \left(\begin{array}{cc}A & -B\\-B' & D\end{array}\right)$$
I want to show that $\mathrm{eig}(M_1) = \mathrm{eig}(M_2)$.
How can I prove that?
 A: If you conjugate $M_1$ with
$$
\pmatrix{I&0\cr0&-I\cr},
$$
where the block sizes of this matrix match those of $M_1$ and $M_2$, what do you get?
What does conjugation do to the eigenvalues?
A: Let me explicate Jyrki's answer, somewhat (feel free to upvote my answer, but I think you should give his answer precedence as far as accepting an answer). First of all, recall that the eigenvalues of a matrix $M$ are the zeroes of the characteristic polynomial $\mathrm{char}(P):=\det(M-tI)$, where $I$ is the identity matrix of the same size as $M$. Secondly, recall that $\det(AB)=\det(A)\det(B)$, whenever $A,B$ are same-sized square matrices. Thirdly, if $P$ is an invertible matrix (that is, $\det(P)\neq 0$), then $\det(P^{-1})=\frac{1}{\det(P)}$ (this actually follows from the second fact, since an identity matrix has determinant of $1$.
Now, as a consequence of all that, we see that if $P$ is any invertible matrix of the same size as a matrix $M$, then we have
$\begin{eqnarray*}
\mathrm{char}(PMP^{-1}) & = & \det(PMP^{-1}-tI)\\
& = & \det(PMP^{-1}-tPP^{-1})\\
& = & \det\bigl(P(M-tI)P^{-1}\bigr)\\
& = & \det(P)\det(M-tI)\det(P^{-1})\\
& = & \det(M-tI)\\
& = & \mathrm{char}(M),
\end{eqnarray*}$
so "conjugating" a matrix will not change its characteristic polynomial, and so will not change its eigenvalues.
In particular, let's suppose that $A$ is a $k\times k$ matrix and $D$ is a $m\times m$ matrix, and let $I_k$ and $I_m$ denote the $k\times k$ and $m\times m$ identity matrices, respectively. Jyrki is suggesting, then, that you set $$P=\left(\begin{array}{cc} I_k & 0_{k\times m}\\0_{m\times k} & -I_m\end{array}\right),$$ then observe that $M_2=PM_1P^{-1}$. From this, the desired conclusion follows by the work above.

In a comment below, p.s. pointed out yet another way to show that conjugation doesn't change eigenvalues, and it's so nice that I'm going to go ahead and append it to my answer. Suppose $(\lambda,x)$ is an eigenpair of $M$, and that $P$ is an invertible matrix of the same size as $M$. Putting $y=Px$, we see that $$(PMP^{-1})y=PM(P^{-1}P)x=PMx=P(\lambda x)=\lambda(Px)=\lambda y,$$ so every eigenvalue of $M$ is an eigenvalue of any given conjugate of $M$. Since $M=P^{-1}(PMP^{-1})(P^{-1})^{-1}$, then by similar reasoning, every eigenvalue of a conjugate of $M$ is also an eigenvalue of $M$.
The only drawback to this approach (that I can see) is that it doesn't make immediately obvious the fact that the respective algebraic multiplicities of the eigenvalues are conjugation-invariant, as well as the eigenvalues, themselves. On the other hand, this approach shows much more readily that the geometric multiplicities of the eigenvalues is conjugation-invariant. Both of these are, of course, more information than you were actually seeking, but I wanted to point out that these stronger results hold.
