Find all the eigenvalues of the following matrix:

$$ M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} $$

where $A = diag\{a_1, \cdots, a_k\}$, $B = diag\{b_1, \cdots, b_k\}$, $C = diag\{c_1, \cdots, c_k\}$, $D = diag\{d_1, \cdots, d_k\}$, where $a_i, b_i, c_i, d_i \in \mathbb{C} (i=1,\cdots,k)$.


Since we have $AB = BA$, $CD = DC$, we have

$$ \mathrm{det} (M) = \mathrm{det} (AD - BC) $$

and the characteristic polynomial is

$$ p_M(\lambda) = (a_1d_1 - \lambda)\cdots(a_kd_k - \lambda) - (b_1c_1 + \cdots b_kc_k) $$

however, I cannot proceed from here, since I am stuck at solving $p_M(\lambda) = 0$.

I have tried applying the fact : the eigenvalues of Kronecker product of matrices are the product of the eigenvalues of each matrix. But I think $M$ does not seem to be factored into a Kronecker product.

Any help will be appreciated.

  • 1
    $\begingroup$ Rearrange the rows and columns to bring $M$ into a more convenient form. $\endgroup$ – kimchi lover Apr 17 '19 at 23:48

Because the blocks commute, we have $$ \det(M - \lambda I) = \det[(A - \lambda I)(D - \lambda I) - BC] \\ = \det \operatorname{diag}[(a_1-\lambda)(d_1 - \lambda) - b_1c_1, \dots, (a_n-\lambda)(d_n - \lambda) - b_nc_n]\\ = [(a_1-\lambda)(d_1 - \lambda) - b_1c_1] \cdots [(a_n-\lambda)(d_n - \lambda) - b_nc_n]\\ = \det \left[\pmatrix{a_1&b_1\\c_1&d_1} - \lambda I\right] \cdots \det \left[\pmatrix{a_n&b_n\\c_n&d_n} - \lambda I\right] $$ So, the eigenvalues of $M$ are the eigenvalues of the matrices of the form $\pmatrix{a_k&b_k\\c_k&d_k}$ for $k = 1, \dots, n$.

To reach the same conclusion, it also suffices to argue that there is a permutation matrix $P$ such that $$ PMP^T = \pmatrix{a_1&b_1\\c_1&d_1\\ &&\ddots\\&&&a_n&b_n\\ &&&c_n &d_n} $$


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