If $A$ and $B$ are matrices in $\textsf{M}_{n \times n}(\mathbb{R})$, prove or refute that $AB$ and $BA$ have the same minimal polynomial

Question

Help please, I have a doubt.

If $A$ and $B$ are matrices in $\textsf{M}_{n \times n}(\mathbb{R})$, prove or refute that $AB$ and $BA$ have the same minimal polynomial.

I know they share the same eigenvalues, but I do not know how to use this fact to prove the proposition, I would like to know any tip

Answer 1

The claim is not true. Let $E_{i,j}$ be the matrix unit with $(i, j)-$ entry 1, $0$ elsewhere. Then $E_{12}E_{3,1}$ is $0$, min poly $x$ while $E_{3,1}E_{1,2}$ is not zero.The claim is true if one of the matrices is invertible because in that case the products are similar matrices. For $2\times 2$-matrices take $E_{12}$ and $E_{11}.$

Answer 2

$$\begin{bmatrix}0&0\\ 1&0\end{bmatrix}\begin{bmatrix}1&0\\ 0&0\end{bmatrix}=\begin{bmatrix}0&0\\ 1&0\end{bmatrix}\\ \begin{bmatrix}1&0\\ 0&0\end{bmatrix}\begin{bmatrix}0&0\\ 1&0\end{bmatrix}=\begin{bmatrix}0&0\\ 0&0\end{bmatrix}$$

Answer 3

it should be said that the claim is 'almost' true.

i.e. with $C:=BA$
$AC^kB= (AB)^{k+1}=(AB)(AB)^{k}$
so if you take $p$, the minimal polynomial for $C$, then
$\mathbf 0 = p\big(C\big) = A p\big(C\big)B = AB\cdot p\big(AB\big)$

and re-running the argument with $C':=AB$ you get
$\mathbf 0 = f\big(C'\big) = B f\big(C'\big)A = BA\cdot f\big(BA\big)$

This implies that minimal polynomials $p$ and $f$ are the same with the possible exception that one may be 'zero-inflated', i.e. the above allows for $p(x)=x\cdot f(x)$ or $f(x)=x\cdot p(x)$.

My preferred example of different minimal polynomials is the 2x2 case there $A:=\mathbf{11}^T$ and $B:=\mathbf{11}^TD$ where $D$ is diagonal with a +1 and -1 on it (reflection)

Then
$AB=\mathbf{11}^T\mathbf{11}^TD = 2 \cdot \mathbf{11}^TD\neq \mathbf 0$
which has determinant zero and trace zero, so is nilpotent

$BA=\mathbf{11}^TD\mathbf{11}^T = \mathbf 0$
still is nilpotent but being real symmetric is diagonalizable and hence the zero matrix.