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$\newcommand{\lmt}{\left[\begin{matrix}}$ $\newcommand{\rmt}{\end{matrix}\right]}$

Hi,

I was reading through a proof of the number of domino tilings of a $(2n)\times(2n)$ chessboard, and somewhere in the proof was the following unjustified claim:

Let $A=\lmt 0&1&0&\cdots&0\\ -1&0&1&\ddots&\vdots\\ 0&-1&0&\ddots&0\\ \vdots&\ddots&\ddots&\ddots&1\\ 0&\cdots&0&-1&0 \rmt$ and $B=\lmt 0&1&0&\cdots&0\\ 1&0&1&\ddots&\vdots\\ 0&1&0&\ddots&0\\ \vdots&\ddots&\ddots&\ddots&1\\ 0&\cdots&0&1&0 \rmt$. In case my notation isn't clear, they have $\pm1$ on the superdiagonal and subdiagonal and $0$ everywhere else.

Let $C$ be the matrix with blocks $\lmt -A&I&0&\cdots&0\\ I&-A&I&\ddots&\vdots\\ 0&I&-A&\ddots&0\\ \vdots&\ddots&\ddots&\ddots&I\\ 0&\cdots&0&I&-A \rmt$.

Let $p_B$ be the characteristic polynomial of $B$. Then, the claim is that

$$ \det C = \det p_B(A). $$

This was stated without proof, so I'm wondering if this follows from a well-known theorem, or if there's a slick proof of it.

Also, I'm curious to what extent this claim generalizes. Thanks!

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Yes, this works fairly generally. (I don't know if this is "well known," maybe to others, I didn't know it before.)

Since $B$ is self-adjoint, we can diagonalize it by a unitary transformation: $U^*BU=\textrm{diag}(b_1, \ldots , b_{2n})$. Moreover, this also works for the block versions of these matrices, in the following sense: if we let $$ \mathcal U = \begin{pmatrix} u_{11}I & u_{12}I & \ldots & u_{1,2n}I\\ && \ldots & \\ u_{2n,1}I & u_{2n,2}I & \ldots & u_{2n,2n}I \end{pmatrix} , $$ with the matrix elements $u_{jk}$ of $U$, then this $\mathcal U$ is still unitary and $\mathcal U^*\mathcal{B}\mathcal U=\textrm{diag}(b_1I, \ldots , b_{2n}I)$. Here, I write $\mathcal B$ for the block version of $B$, where we replace the $1$'s with $I$. Similarly, $\mathcal A$ will denote the matrix with $A$ on the "diagonal" and $0$'s everywhere else.

Then $$ \det C = \det (\mathcal{B-A})=\det(\mathcal{U^*(B-A)U}) =\det\textrm{diag}(b_1I-A, b_2I-A, \ldots , b_{2n}I-A) . $$ The last step uses that $\mathcal{U^*AU}=\mathcal A$, and this follows because $\mathcal U$ is unitary with blocks that are multiples of $I$ and $\mathcal A$ is block diagonal (just writing it out will give it quickly).

Thus $\det C=\prod \det (b_jI-A) = \det \prod (b_j I-A) = \det p_B(A)$, as claimed.

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  • $\begingroup$ Thanks! Clear and concise proof. Looks like the only requirement is that $B$ is Hermitian. $\endgroup$ – lilreddragon May 31 '16 at 1:27

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