Weinstein–Aronszajn identity

The identity may be proved as follows.^[1] 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.^[2]