Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking for the simplest possible example of square matrices $A$ and $B$ such that

  • $A$ is similar to $B$,
  • $AB$ is not similar to $BA$.

Such an example should exist, but I would like to find the "smallest" one. If either $A$ or $B$ is invertible, then $AB$ will be similar to $BA$, so one needs to look at singular matrices.

share|cite|improve this question
up vote 12 down vote accepted

$A = \begin{bmatrix}1 & 0\\ 1 & 0\end{bmatrix},\ B = \begin{bmatrix}0 & 0\\1 & 1\end{bmatrix}$

Let $P = \begin{bmatrix}1 & -1\\-3 & 1\end{bmatrix}$. Then
$PAP^{-1} = \begin{bmatrix}1 & -1\\-3 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\ 1 & 0\end{bmatrix}\begin{bmatrix}-1/2 & -1/2\\-3/2 & -1/2\end{bmatrix} = B$.

$AB = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix},\ BA = \begin{bmatrix}0 & 0\\2 & 0\end{bmatrix}$

$AB$ and $BA$ are not similar as $\text{rank}(AB) = 0$ but $\text{rank}(BA) = 1$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.