Sylvester's Determinant Identity

(I_m + AB) = (I_n + BA) — swap to the smaller side.

The idea

Sylvester's determinant identity relates the determinants of the two products of a pair of rectangular matrices, taken in either order.

Theorem (Sylvester's determinant identity).

Let $A$ be an $m \times n$ matrix and $B$ an $n \times m$ matrix. Then $\det(I_m + AB) = \det(I_n + BA).$

The two sides involve matrices of different sizes: $AB$ is $m \times m$ while $BA$ is $n \times n$. The identity says their determinants, each shifted by the identity, agree exactly nonetheless. The figure records the fact behind this: the two products have the same nonzero eigenvalues, with the same multiplicities, and the larger one pads its spectrum with zeros.

The identity matters most when $m$ and $n$ differ greatly. For a tall $A$ of size $1000 \times 3$ and a wide $B$ of size $3 \times 1000$, one side is a $1000 \times 1000$ determinant and the other is $3 \times 3$. The identity lets us always compute the smaller side.

Ways to work on it

• Walkthrough. Verify Sylvester on a concrete 2 × 3 / 3 × 2 pair.
• Practice. Apply the rank-1 case: (I_2 + ab^ ) = 1 + b^ a.
• Hardest. What Sylvester's identity says about traces, characteristic polynomials, and small perturbations.

Not sure where to start? Take the ten-question placement test.