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Let $K$ be a field of characteristic 0, and consider the following block matrix $$M=\left(\begin{array}{cc} A & B\\ -B&D\end{array}\right),$$ where each block is an $n\times n$ matrix with coefficients in $K$. I am looking for a relation between the eigenvalues of $M$ and those of $A$ and $D$.

Context: Here, I'm assuming that $M$ is invertible and semisimple. I was wondering if there is a way to show that both $A$ and $D$ are invertible and semisimple as well. Moreover, we can also assume that $B$ is $2\times2$ and has one of the following forms: $$\left\{\left(\begin{array}{cc} 1 & 0\\ 0&1\end{array}\right),\left(\begin{array}{cc} 1 & 0\\ 0&0\end{array}\right),\left(\begin{array}{cc} 0 & 0\\ 0&0\end{array}\right)\right\}.$$

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I am not sure whether this is much related, but I can say that in general, there is no relation. A recent paper by R. Turkmena, V. E. Paksoyb and F. Zhang, Some inequalities of majorization type, Linear Algebra and its Applications, in press. says that if $M$ is positive definite and $B$ is symmetric, then the eigenvalues of $M$ are majorized by those of $A+D$. –  Sunni Jun 5 '12 at 20:14
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It's certainly not necessary for $A$ and $D$ to be invertible, e.g. with $B = \pmatrix{1 & 0\cr 0 & 1\cr}$ you could have $A = D = \pmatrix{0 & 0\cr 0 & 0\cr}$, or with $B = \pmatrix{1 & 0\cr 0 & 0\cr}$ you could have $A = \pmatrix{0 & 0\cr 0 & 1\cr}$ and $D = \pmatrix{1 & 1\cr 1 & 1\cr}$.

Of course with $B = \pmatrix{0 & 0\cr 0 & 0\cr}$ the eigenvalues of $M$ are the union of the eigenvalues of $A$ and of $D$.

In all cases $\text{Tr}(M) = \text{Tr}(A) + \text{Tr}(D)$, so the sum of the eigenvalues of $M$ is the sum for $A$ plus the sum for $D$.

In the case $B = \pmatrix{1 & 0\cr 0 & 1\cr}$, the coefficient of $\lambda^2$ in the characteristic polynomial of $M$ (which is $\sum_{i < j} \lambda_i \lambda_j$ where $\lambda_i$ are the eigenvalues of $M$) is $a_{{1}}a_{{2}}+a_{{1}}d_{{1}}+a_{{1}}d_{{2}}+a_{{2}}d_{{1}}+a_{{2}}d_{ {2}}+d_{{1}}d_{{2}}+2$, where $a_i$ and $d_i$ are the eigenvalues of $A$ and $D$ respectively.

In the case $B = \pmatrix{1 & 0\cr 0 & 0\cr}$, that coefficient would be $a_{{1}}a_{{2}}+a_{{1}}d_{{1}}+a_{{1}}d_{{2}}+a_{{2}}d_{{1}}+a_{{2}}d_{ {2}}+d_{{1}}d_{{2}}+1$.

In the case $B = \pmatrix{1 & 0\cr 0 & 0\cr}$, I think these are the only equations linking the eigenvalues of $M$ with those of $A$ and $D$: you can choose $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ arbitrarily satisfying the two constraints on $\sum_i \lambda_i$ and $\sum_{i<j} \lambda_i \lambda_j$ and find a suitable $A$ and $D$ with eigenvalues $a_i$ and $d_j$ that works.

In the case $B = \pmatrix{1 & 0\cr 0 & 1\cr}$, it looks to me like there is an additional constraint: you can only choose one eigenvalue (say $\lambda_1$) arbitrarily, and then the other three will satisfy

$$\eqalign{{\lambda}^{3}&+ \left( -a_{{1}}-a_{{2}}-d_{{1}}-d_{{2}}+\lambda_{{1}} \right) {\lambda}^{2}\cr + \left( a_{{1}}a_{{2}} \right. & \left. +a_{{1}}d_{{1}}+a_{{1}}d_ {{2}}-a_{{1}}\lambda_{{1}}+a_{{2}}d_{{1}}+a_{{2}}d_{{2}}-a_{{2}} \lambda_{{1}}+d_{{1}}d_{{2}} -d_{{1}}\lambda_{{1}} -d_{{2}}\lambda_{{1}} +{\lambda_{{1}}}^{2}+2 \right) \lambda \cr +a_{{2}}d_{{2}}\lambda_{{1}} &-a_{ {1}}a_{{2}}d_{{2}}-a_{{2}}d_{{1}}d_{{2}}-a_{{1}}a_{{2}}d_{{1}}-a_{{1}} d_{{1}}d_{{2}}-a_{{1}}-a_{{2}}+{\lambda_{{1}}}^{3}+2\,\lambda_{{1}}-d_ {{1}}-d_{{2}}-d_{{1}}{\lambda_{{1}}}^{2}\cr -d_{{2}}{\lambda_{{1}}}^{2} & +a_ {{1}}a_{{2}}\lambda_{{1}}+a_{{1}}d_{{2}}\lambda_{{1}}+a_{{2}}d_{{1}} \lambda_{{1}}+a_{{1}}d_{{1}}\lambda_{{1}}+d_{{1}}d_{{2}}\lambda_{{1}}- a_{{2}}{\lambda_{{1}}}^{2}-a_{{1}}{\lambda_{{1}}}^{2}=0\cr} $$

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Ok, I see. If I assume that all eigenvalues are distinct, can we say something more? (In your first example, there are only 2 distinct eigenvalues, but I believe that in the second example, all eigenvalues are distinct. Hence, I would be perfectly happy with the case $B=I$.) –  M Turgeon Jun 5 '12 at 21:06
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With $B = I$ try e.g. $A = D = \pmatrix{0 & 0\cr 0 & t\cr}$. The characteristic polynomial of $M$ is $P(\lambda) = {\lambda}^{4}-2\,t{\lambda}^{3}+ \left( 2+{t}^{2} \right) {\lambda}^{2 }-2\,t\lambda+1+{t}^{2} $. Thus $B$ is invertible if $t^2 + 1\ne 0$, and has distinct (complex) eigenvalues if $t (t^2 + 1) \ne 0$. –  Robert Israel Jun 5 '12 at 22:49
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