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Let $A$ be a $3 \times 2$ matrix, and $B$ be a $2\times 3$ matrix. If $(AB)^2 = 0$, then how can I show that $(BA)^2 = 0$?

I have done some calculation but it's too much. I was wondering if there is a better way:

$ A = \begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] a_{31} & a_{32} \end{bmatrix} , B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\[0.3em] b_{21} & b_{22} & b_{32} \end{bmatrix} $

then we have:

$ AB = \begin{bmatrix} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} & a_{11}b_{13}+a_{12}b_{23} \\[0.3em] a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} & a_{21}b_{13}+a_{22}b_{32} \\[0.3em] a_{31}b_{11}+a_{32}b_{21} & a_{31}b_{12}+a_{32}b_{22} & a_{31}b_{13}+a_{32}b_{23} \end{bmatrix} $

$ BA = \begin{bmatrix} b_{11}a_{11}+b_{12}a_{21}+b_{13}a_{31} & b_{11}a_{12}+b_{12}a_{22}+b_{13}a_{32} \\[0.3em] b_{21}a_{11}+b_{22}a_{21}+b_{23}a_{31} & b_{21}a_{12}+b_{22}a_{22}+b_{23}a_{32} \end{bmatrix} $

The elements of AB and BA are some how similar, maybe by calculation I get to something. But I really hope there is a better way.

Martin Sleziak
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Shadi Taghian
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  • There is one way that I calculate every element of AB and then show that the elements of BA are somehow similar, but I was actually looking for a better way. – Shadi Taghian May 28 '16 at 08:54
  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – anonymous May 28 '16 at 08:55
  • BTW there is a similar question about $2\times2$ matrices: http://math.stackexchange.com/questions/1698033/prove-or-disprove-for-2-times-2-matrices-a-and-b-if-ab2-o-then-b – Martin Sleziak May 29 '16 at 07:39
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    Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak May 29 '16 at 07:39
  • In this case I have tried to improve the title in order to better describe the question. Good title is useful for other users (so that they know more about the question from the title). But it is also for you, because the SE software then can generated better list of related questions (see the sidebar on the right) and also a list of questions with similar titles which is displayed to you while asking the question. (In fact, by typing the new title in the ask question dialog I found the question I linked in the above comment.) – Martin Sleziak May 29 '16 at 07:42

1 Answers1

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Notice that $AB$ is a $3\times3$ matrix and $BA$ is a $2\times2$ matrix.

  • If $(AB)^2=0$ then $(BA)^3=A(BA)^2B=0$.
  • Therefore the minimal polynomial of the matrix $BA$ divides $x^3$.
  • As $BA$ is $2\times2$ matrix, the degree of this polynomial is at most two.
  • So the minimal polynomial of $AB$ is either $x$ or $x^2$ and thus $$(BA)^2=0.$$
Martin Sleziak
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