Today I would like to bring up a simple result of which I was unaware until a few days ago: Sylvester's determinant theorem
– which is not Sylvester's determinant identity, the one that relates the determinant of a matrix with some special submatrices –. It is
more often referred to as the Weinstein-Aronszajn identity (see [1]
and [2]), and overall it is very simple to state:
Given a field $\mathbf F$ and $A \in \mathscr M_{m \times n} (\mathbf F), B \in \mathscr M_{n \times m} (\mathbf F)$, then
$$
\det (\mathbf 1 + AB) = \det (\mathbf 1 + BA).
$$
As a remark, notice that the first $\mathbf 1$ is the $m\times m$ identity, which differs from the second, $n\times n$ identity; I will be calling both
indistinguishably henceforth. This theorem is not particularly hard to prove either:
Recall first the determinant of a block-partitioned matrix; if $A, B, C, D$ are $m\times m, m\times n, n\times m$ and $n\times n$ matrices respectively with
$A$ and $D$ invertible, then:
$$
\det A \det (D - CA^{-1}B) = \det \begin{pmatrix}
A & B\\
C & D
\end{pmatrix} = \det D \det (A - B D^{-1} C).
$$
Applying this result in the two different fashions above yields the desired equality:
$$
\det(\mathbf 1 + AB) = \det \begin{pmatrix}
\mathbf 1 & -A\\
B & \mathbf 1
\end{pmatrix} = \det(\mathbf 1 + BA).
$$
Since the previous determinants look similar to the characteristic polynomials of $AB$ and $BA$, we obtain that they are equal up to a factor of
$(-\lambda)^{m - n}$ where, without loss of generality, $m \geq n$. In other words, $AB$ and $BA$ have the same eigenvalues counting multiplicities except
for the zeros.
Given $A \in \mathscr M_{m \times n} (\mathbf F), B \in \mathscr M_{n \times m} (\mathbf F)$, with $m \geq n$,
$$
\chi_{AB} = \det (AB - \lambda \mathbf 1) = (-\lambda)^{m-n} \det (BA - \lambda \mathbf 1) = (-\lambda)^{m-n} \chi_{BA}.
$$
$$
\begin{align*}
\chi_{AB} &= \det (AB - \lambda \mathbf 1) = (-\lambda)^m \det \left(\mathbf 1 - \frac{1}{\lambda} AB\right)\\
&= (-\lambda)^m \det \left(\mathbf 1 - \frac{1}{\lambda} BA\right) = (-\lambda)^{m - n} \det (BA - \lambda \mathbf 1) = (-\lambda)^{m - n} \chi_{BA}.
\end{align*}
$$
Because the essence of mathematics is solving problems, let's apply this new tool to one:
Let $U \in \mathscr M_n (\mathbf C)$ be a unitary matrix, block-partitioned as follows:
$$
U = \begin{pmatrix}
U_{11} & U_{12}\\
U_{21} & U_{22}
\end{pmatrix},
$$
where $U_{11}$ is $k\times k$. Prove that $|\det U_{11}| = |\det U_{22}|$.
Recall that the definition of being unitary states that $U\adj U = \adj U U = \mathbf 1$, so multiplying the two previous matrices must give
$$
\begin{align*}
U \adj U &= \begin{pmatrix}
U_{11} & U_{12}\\
U_{21} & U_{22}
\end{pmatrix} \begin{pmatrix}
\adj U_{11} & \adj U_{21}\\
\adj U_{12} & \adj U_{22}
\end{pmatrix} = \begin{pmatrix}
U_{11} \adj U_{11} + U_{12} \adj U_{12} & U_{11} \adj U_{21} + U_{12} \adj U_{12}\\
U_{21} \adj U_{11} + U_{22} \adj U_{12} & U_{21} \adj U_{21} + U_{22} \adj U_{22}
\end{pmatrix} = \begin{pmatrix}
\mathbf 1 & 0\\
0 & \mathbf 1
\end{pmatrix},\\
\adj U U &= \begin{pmatrix}
\adj U_{11} & \adj U_{21}\\
\adj U_{12} & \adj U_{22}
\end{pmatrix} \begin{pmatrix}
U_{11} & U_{12}\\
U_{21} & U_{22}
\end{pmatrix} = \begin{pmatrix}
\adj U_{11} U_{11} + \adj U_{21} U_{21} & \adj U_{11} U_{12} + \adj U_{21} U_{22}\\
\adj U_{12} U_{11} + \adj U_{22} U_{21} & \adj U_{12} U_{12} + \adj U_{22} U_{22}
\end{pmatrix} = \begin{pmatrix}
\mathbf 1 & 0\\
0 & \mathbf 1
\end{pmatrix};
\end{align*}
$$
from this we get the following equalities:
$$
\begin{align}
U_{11} \adj U_{11} &= \mathbf 1 - U_{12} \adj U_{12} \implies |\det U_{11}|^2 = \det(\mathbf 1 - U_{12} \adj U_{12}),\\
\adj U_{11} U_{11} &= \mathbf 1 - \adj U_{21} U_{21} \implies |\det U_{11}|^2 = \det(\mathbf 1 - \adj U_{21} U_{21}),\\
U_{22} \adj U_{22} &= \mathbf 1 - U_{21} \adj U_{21} \implies |\det U_{22}|^2 = \det(\mathbf 1 - U_{21} \adj U_{21}),\\
\adj U_{22} U_{22} &= \mathbf 1 - \adj U_{12} U_{12} \implies |\det U_{22}|^2 = \det(\mathbf 1 - \adj U_{12} U_{12}).
\end{align}
$$
Finally, using Sylvester's theorem for determinants on $(1)$ and $(4)$ finishes the problem off:
$$
|\det U_{11}|^2 = \det(\mathbf 1 - U_{12} \adj U_{12}) = \det(\mathbf 1 - \adj U_{12} U_{12}) = |\det U_{22}|^2
\implies |\det U_{11}| = |\det U_{22},|
$$
where the last step is possible because the modulus of a complex number is always a positive real.
To conclude this post, I would like to cite "The Usefulness of Formulas in Solving Matrix Problems" (Stanescu) in
Crux Mathematicorum vol. 51, no. 9, as a good source of some lesser-known determinant
identities.