Determinant of a large block matrix $\newcommand{\lmt}{\left[\begin{matrix}}$
$\newcommand{\rmt}{\end{matrix}\right]}$
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!
 A: 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.
