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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?

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Similar matrices have the same eigenvalues. Find a non-singular matrix $P$ such that $P M_1 = M_2 P,$ or $M_1 = P^{-1} M_2 P.$ This operation is called conjugation. Jyrki's answer provides such a $P.$ – user2468 Jul 10 '12 at 21:03

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?

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I didn't get what you ask me do? M1 and M2 are real block matrices, with appropriate dimensions of A, B and D. – Mohsin Jul 10 '12 at 20:25
@Mohsin: This may help. – user26872 Jul 10 '12 at 20:51
Sorry, I initially wanted to give an extended hint only. Then I forgot about this site for an hour :-(. Meanwhile Cameron has explained in detail what I wanted you to do. Thanks also to J.D. for digging out a couple of relevant links. – Jyrki Lahtonen Jul 10 '12 at 21:29
@Jyrki Forgetting about the site is good! – Dylan Moreland Jul 10 '12 at 21:34
Thanks for the answer, i have got the point :) – Mohsin Jul 11 '12 at 12:28

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.

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+1 No need not to expect upvotes! It may well happen that your answer turns out to be more helpful to somebody than my terse but perhaps a bit obscure hint. – Jyrki Lahtonen Jul 10 '12 at 21:32
That was my thought, as well, Jyrki (which is why I decided to post an answer, myself, ultimately). Still, it doesn't really feel like the answer is mine, so if it were my choice, I'd accept your answer and just upvote mine. /shrug/ – Cameron Buie Jul 10 '12 at 21:37
Rather than resort to the determinant, in this case I think it's much easier just to note $Ax = \lambda x \Leftrightarrow (PAP^{-1}) (Px) = \lambda (Px)$. – p.s. Jul 10 '12 at 21:37
Excellent point, @p.s.! – Cameron Buie Jul 10 '12 at 21:38

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