The Characteristic Polynomial of a Product

Ralph Howard

This is a note to prove the most notorious of all qualifying exam questions:

Theorem Let A and B be n × n matrices over a field F. Then the characteristic polynomials of AB and BA are the same.

Remark 1 If one of the two matrices, say A, is invertible then A^{−1}(AB)A = BA. Thus AB and BA are similar which certainly implies AB and BA have the same characteristic polynomial. However if A and B are both nonsingular then AB and BA do not have to be similar. For example if

A = [ \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ] , B = [ \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ]

then AB = B ≠ 0, but BA = 0 so that AB and BA are not similar.

Remark 2 I leaned of the following proof from Tom Markham who said it is originally due to Paul Halmos.

Proof: For any square matrix let C let char_{C}(λ) := det(λI − C) be the characteristic polynomial of C. Let r be the rank of A. Then by doing row and column reduction there are invertible n × n matrices P and Q so that

A = P [ \begin{matrix}I & 0 \\ 0 & 0 \end{matrix} ] Q

where I is the r × r identity matrix. Now express B as

B = Q^{−1} [ \begin{matrix}B_{11} & B_{12} \\ B_{21} & B_{22} \end{matrix} ] P^{−1}

Where B_{11} is r × r, B_{22} is (n − r) × (n − r) etc. Then

AB = P [ \begin{matrix}I & 0 \\ 0 & 0 \end{matrix} ] QQ^{−1} [ \begin{matrix}B_{11} & B_{12} \\ B_{21} & B_{22} \end{matrix} ] P^{−1} = P [ \begin{matrix}B_{11} & B_{12} \\ 0 & 0 \end{matrix} ] P^{−1}

so that char_{AB}(λ) = λ^{n−r} char_{B_{11}}(λ). Likewise

BA = Q^{−1} [ \begin{matrix}B_{11} & B_{12} \\ B_{21} & B_{22} \end{matrix} ] P^{−1}P [ \begin{matrix}I & 0 \\ 0 & 0 \end{matrix} ] Q = Q^{−1} [ \begin{matrix}B_{11} & 0 \\ B_{21} & 0 \end{matrix} ] Q

so that char_{BA}(λ) = λ^{n−r} char_{B_{11}}(λ). This completes the proof. □