Sylvester's determinant theorem

Statement

Let $K$ be a commutative ring, let $A$ be an $m \times n$ matrix and $B$ an $n \times m$ matrix over $K$. Then $$\det(I_m + AB) = \det(I_n + BA),$$ where $I_m$ and $I_n$ are identity matrices and $\det$ is the determinant.

Remarks

The two sides are determinants of different sizes, $m$ and $n$. For a column $u$ and a row $v^T$, it gives $\det(I_m + u v^T) = 1 + v^T u$. The statement is also called the Weinstein–Aronszajn identity.

Explanations

L1 human

The products $AB$ and $BA$ can have very different sizes. Still, adding the identity to them gives the same determinant. This can turn a large determinant into a small one.