In mathematics, the Weinstein–Aronszajn identity states that if A {\displaystyle A} and B {\displaystyle B} are matrices of size m × n and n × m respectively (either or both of which may be infinite) then, provided A B {\displaystyle AB} (and hence, also B A {\displaystyle BA} ) is of trace class,
det ( I m + A B ) = det ( I n + B A ) , {\displaystyle \det(I_{m}+AB)=\det(I_{n}+BA),}
where I k {\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.
Proof The identity may be proved as follows. Let M {\displaystyle M} be a matrix consisting of the four blocks I m {\displaystyle I_{m}} , A {\displaystyle A} , B {\displaystyle B} and I n {\displaystyle I_{n}} :
M = ( I m A B I n ) . {\displaystyle M={\begin{pmatrix}I_{m}&A\\B&I_{n}\end{pmatrix}}.}
Because Im is invertible, the formula for the determinant of a block matrix gives
det ( I m A B I n ) = det ( I m ) det ( I n − B I m − 1 A ) = det ( I n − B A ) . {\displaystyle \det \!{\begin{pmatrix}I_{m}&A\\B&I_{n}\end{pmatrix}}=\det(I_{m})\det(I_{n}-BI_{m}^{-1}A)=\det(I_{n}-BA).}
Because In is invertible, the formula for the determinant of a block matrix gives
det ( I m A B I n ) = det ( I n ) det ( I m − A I n − 1 B ) = det ( I m − A B ) . {\displaystyle \det \!{\begin{pmatrix}I_{m}&A\\B&I_{n}\end{pmatrix}}=\det(I_{n})\det(I_{m}-AI_{n}^{-1}B)=\det(I_{m}-AB).}
Thus
det ( I n − B A ) = det ( I m − A B ) . {\displaystyle \det(I_{n}-BA)=\det(I_{m}-AB).}
Substituting − A {\displaystyle -A} for A {\displaystyle A} then gives the Weinstein–Aronszajn identity.
Applications Let λ ∈ R ∖ { 0 } {\displaystyle \lambda \in \mathbb {R} \setminus \{0\}} . The identity can be used to show the somewhat more general statement that
det ( A B − λ I m ) = ( − λ ) m − n det ( B A − λ I n ) . {\displaystyle \det(AB-\lambda I_{m})=(-\lambda )^{m-n}\det(BA-\lambda I_{n}).}
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