Weinstein–Aronszajn identity

In mathematics, the Weinstein–Aronszajn identity states that if $A$ and $B$ are matrices of size m × n and n × m respectively (either or both of which may be infinite) then, provided $AB$ (and hence, also $BA$) is of trace class,

$\det (I_m+AB)=\det (I_n+BA),$

where $I_k$ is the k × k identity matrix.

It is closely related to the matrix determinant lemma and its generalization. It is the determinant analogue of the Woodbury matrix identity for matrix inverses.

Proof

The identity may be proved as follows. Let $M$ be a matrix consisting of the four blocks $I_m$, $A$, $B$ and $I_n$:

$M=\left(\begin{array}{cc}{I}_m& A\\ B& I_n\end{array}\right).$

Because I_m is invertible, the formula for the determinant of a block matrix gives

$\det \left(\begin{array}{cc}{I}_m& A\\ B& I_n\end{array}\right)=\det (I_m)\det (I_n-B{I}_m^{-1}A)=\det (I_n-BA).$

Because I_n is invertible, the formula for the determinant of a block matrix gives

$\det \left(\begin{array}{cc}{I}_m& A\\ B& I_n\end{array}\right)=\det (I_n)\det (I_m-A{I}_n^{-1}B)=\det (I_m-AB).$

Thus

$\det (I_n-BA)=\det (I_m-AB).$

Substituting $-A$ for $A$ then gives the Weinstein–Aronszajn identity.

Applications

Let $\lambda \in \mathbb{ℝ}\setminus \{0\}$. The identity can be used to show the somewhat more general statement that

$\det (AB-\lambda I_m)=(-\lambda {)}^{m-n}\det (BA-\lambda I_n).$

It follows that the non-zero eigenvalues of $AB$ and $BA$ are the same.

This identity is useful in developing a Bayes estimator for multivariate Gaussian distributions.

The identity also finds applications in random matrix theory by relating determinants of large matrices to determinants of smaller ones.