0
$\begingroup$

Say we have two square, non-singular matrices A and B, which are not equal to each other. And such that both AB and BA is defined. So will the only solutions to the equation AB = BA be that either one of them is the Identity matrix or A is the inverse of B or B is the inverse of A. But my teacher had said that the last two are true only when AB = BA = I (Identity matrix). But how can AB = BA be equal to something other than the Identity matrix ? Is it possible that : $$AB = BA \neq I$$

$\endgroup$
4
  • 2
    $\begingroup$ Simplest case for instance: $3 \times 2 = 2 \times 3 = 6$. Extend it to diagonal matrices when $n > 1$. $\endgroup$
    – Zhanxiong
    Nov 9, 2020 at 16:02
  • $\begingroup$ For, say, $B=A^3+17A^2+\frac12A+9I$ we have $BA=AB$ (and $A^3+17A^2+\frac12A+9I\notin\{ A^{-1}, I, A\}$ more often than not). $\endgroup$
    – user239203
    Nov 9, 2020 at 16:07
  • $\begingroup$ In general, if $B$ is any polynomial in $A$ with real coefficients, then $AB=BA$. Also, if both $A$ and $B$ are any diagonal matrices, then $AB=BA$. $\endgroup$
    – Somos
    Nov 9, 2020 at 16:08
  • 1
    $\begingroup$ Let $A=B=\pmatrix{1&0\\ 0&0}$. Then $AB=BA$ and none of $A,B,AB$ or $BA$ is equal to any scalar multiple of $I$. $\endgroup$
    – user1551
    Nov 9, 2020 at 17:48

1 Answer 1

2
$\begingroup$

any set of matrix that are simultaneously diagonalizable (meaning they have the same eigenvectors) verify your equality

$\endgroup$
4
  • $\begingroup$ As a high schooler, who has just started learning about matrices. I am not familiar with eigenvectors. $\endgroup$ Nov 9, 2020 at 16:01
  • $\begingroup$ Could you please explain what they are ? $\endgroup$ Nov 9, 2020 at 16:02
  • 1
    $\begingroup$ it means that if there exist a matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are diagonal matrices $\endgroup$
    – NHL
    Nov 9, 2020 at 16:02
  • 1
    $\begingroup$ @KoustubhJain So for example, $AB = BA$ will always work if $$ A = \pmatrix{p&0\\0&q}, \quad B = \pmatrix{s & 0 \\ 0 & t} $$ for numbers $p,q,s,t$. $\endgroup$ Nov 9, 2020 at 18:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .