Weinstein–Aronszajn Identity

${\displaystyle \det(I_{m}+AB)=\det(I_{n}+BA),}$

where ${\displaystyle I_{k}}$ 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.

The identity may be proved as follows. Let ${\displaystyle M}$  be a matrix consisting of the four blocks ${\displaystyle I_{m}}$ , ${\displaystyle A}$ , ${\displaystyle B}$  and ${\displaystyle I_{n}}$ :

${\displaystyle M={\begin{pmatrix}I_{m}&A\\B&I_{n}\end{pmatrix}}.}$ 

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

${\displaystyle \det {\begin{pmatrix}I_{m}&A\\B&I_{n}\end{pmatrix}}=\det(I_{m})\det \left(I_{n}-BI_{m}^{-1}A\right)=\det(I_{n}-BA).}$ 

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

${\displaystyle \det {\begin{pmatrix}I_{m}&A\\B&I_{n}\end{pmatrix}}=\det(I_{n})\det \left(I_{m}-AI_{n}^{-1}B\right)=\det(I_{m}-AB).}$ 

Thus

${\displaystyle \det(I_{n}-BA)=\det(I_{m}-AB).}$ 

Substituting ${\displaystyle -A}$  for ${\displaystyle A}$  then gives the Weinstein–Aronszajn identity.

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

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

It follows that the non-zero eigenvalues of ${\displaystyle AB}$  and ${\displaystyle BA}$  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.