How to find eigenvalues of following block matrix $M$ in terms of eigenvalues of matrix $A$?
$M=\begin{bmatrix} A & B \\ B & O \\ \end{bmatrix}$
Where matrices $A$,$B$ are symmetric matrices.Also $B=A+I_n$ and $O$ is zero matrix of order $n$
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asked Apr 15, 2016 at 10:35
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1$\begingroup$ You will find information on this by looking up "saddle point problem". In general, $B\neq B^T$, so you may be able to simplify the results you find. Sherman-Morrisson formula may be helpful too if$A$ is invertible. $\endgroup$– DarylCommented Apr 15, 2016 at 10:41
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$\begingroup$ The following holds: $\det(M-\lambda I_{2n}) = \det( (A-\lambda I_n)(-\lambda I_n) - (A+I_n)^2)=0$. I don't think it can be factorised, but maybe it helps anyway. $\endgroup$– CycloneCommented Apr 19, 2016 at 13:12
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$\begingroup$ OK,thanks for your comment $\endgroup$– kalpeshmpopatCommented Apr 19, 2016 at 16:29
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1 Answer
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First we diagonalize $A = SDS^{-1}$ with $S^TS=I$ and $D$ diagonal. Then $$ \pmatrix{S^{-1}&0\\0&S^{-1}} M \pmatrix{S&0\\0&S} =\pmatrix{D & D+I \\D+I & 0}. $$ Applying a suitable permutation $P$, we obtain the block diagonal structure $$ P\pmatrix{D & D+I \\D+I & 0}P^{-1} = \pmatrix{ \lambda_1 & \lambda_1+1& \\ \lambda_1& 0 \\ &&&\lambda_2& \lambda_2+1\\ &&&\lambda_2+1 & 0 \\ &&&&&\ddots } $$ with $\lambda_i$ denoting the eigenvalues of $A$. These small matrices $$\pmatrix{ \lambda_i & \lambda_i+1& \\ \lambda_i+1& 0 } $$ have characteristic polynomials $$ t^2 -\lambda_i t -(\lambda_i+1)^2=0, $$whose roots are then the eigenvalues of $M$.
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$\begingroup$ What is permutation matrix $P$ is? $\endgroup$ Commented Apr 22, 2016 at 1:26