Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 10F (Linear Algebra)

If $A$ is an invertible matrix, thenThus $AB$ and $BA$ are similar matrices, and similar matrices have the same characteristic polynomial.

For all but finitely many $s\in \mathbb{C}$, the matrix $A-sI$ is invertible. Part (a), applied to $A-sI$ and $B$, then givesBoth sides are polynomials in the two variables $s,t$. Since they agree for infinitely many values of $s$, their coefficients as polynomials in $s$ agree. Setting $s=0$ yieldseven when $A$ is singular.

TakeThen $AB=B\ne 0$ and $(AB{)}^2=0$, whereas $BA=0$. Hence their minimal polynomials are

For every nonnegative integer $k$,Consequently, if $m_{AB}(AB)=0$, thenBy the divisibility characterization of the minimal polynomial,Interchanging $A$ and $B$ also gives $m_{AB}(t)\mid t,m_{BA}(t)$. Therefore the Minimal polynomials of AB and BA can differ only in the exponent of the factor $t$, and that exponent differs by at most one.

If $AB$ is diagonalizable, its minimal polynomial has no repeated roots. The preceding divisibility says that every nonzero Jordan block of $BA$ has size one and every zero-eigenvalue block has size at most two. Squaring $BA$ preserves the one-dimensional nonzero blocks and sends every size-at-most-two nilpotent block to zero. Thus the Jordan normal form of $(BA{)}^2$ is diagonal, so