Suppose that A and B are 2x2 matrices, A is invertible, and AB = 2BA. Prove that B is not invertible.

My first thought is that I need to find what B is in terms of the terms of A, and proving B's invertibility from there, but I don't know how to start that..

Im really stumped. Any help/tips/advice would be much appreciated!

  • $\begingroup$ If $B$ were invertible then its inverse would be unique and by the equality given that would imply $A=2A$. In turn this implies $A=0$ but $0$ is not invertible, contradicting the hypothesis. $\endgroup$
    – Guest
    Mar 27, 2015 at 12:19
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    $\begingroup$ @Guest False. The equality would imply $B^{-1}AB=2A$, not $A=2A$. $\endgroup$
    – 5xum
    Mar 27, 2015 at 12:26

3 Answers 3


Taking determinant, we have $$ \det(A)\det(B)=2^2\det(B)\det(A)\implies \det(A)\det(B)=0\implies \det(B)=0. $$ The last implication above uses $\det(A)\neq 0$.


Hint : Take $\det$ on both sides...

  • 1
    $\begingroup$ I think you should write a hint as a comment, not as an answer. $\endgroup$
    – Mike
    Mar 27, 2015 at 12:22
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    $\begingroup$ @Mike Hints are very commonly written as answers on this site. The point is that often times, the OP only needs a small hint and is then able to solve the problem himself. In such a case, it is better if the hint is posted as an answer, as it can then be accepted. $\endgroup$
    – 5xum
    Mar 27, 2015 at 12:24
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    $\begingroup$ I feel that rather than providing a complete solution till the end, giving a little help as a hint is more beneficial to the OP. $\endgroup$
    – Arpan
    Mar 27, 2015 at 12:24
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    $\begingroup$ @Mike Also the OP did ask for "any help/tips/advice", rather than a full solution! $\endgroup$
    – Jason C
    Mar 27, 2015 at 17:32

All answers so far suggest the result is false in characteristic$~3$. Indeed (and I see this is already in a comment by user1551) over a field of characteristic$~3$ one has $$ A=\pmatrix{0&1\\1&0}, B=\pmatrix{1&0\\0&-1} \quad\text{then } AB=\pmatrix{0&-1\\1&0\\}=-\pmatrix{0&1\\-1&0\\}=-BA=2BA. $$ Probably not intended, but the question does not specify where the matrix entries live.

I may add another example to show that while (outside characteristic $3$) "$B$ singular" may be strengthened to "$B$ nilpotent" (as the answer by 'math' indicates), one cannot strengthen it to "$B=0$": $$ A=\pmatrix{2&0\\0&1}, B=\pmatrix{0&1\\0&0} \quad\text{then } AB=\pmatrix{0&2\\0&0\\}=2\pmatrix{0&1\\0&0\\}=2BA. $$

Finally (a pet subject of mine), while certain other matrix sizes will allow counterexamples in other odd characteristics, there is one size where counterexamples exist independently of the characteristic: for $0\times0$ matrices, $AB=2BA$ holds "always" (there is just one case), but $A,B$ are both invertible.

  • $\begingroup$ this is neat! $~~$ $\endgroup$
    – 6005
    Mar 28, 2015 at 11:00

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