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Given two $4 \times 4$ matrices $A$ and $B$. Prove that if $AB=0$ then $\operatorname{rank}(BA)<3$, i.e. $\operatorname{rank}(BA)$ is $2$ or less.

Now, i can definitely see the reason behind the conclusion, i tried many things to do in order to simplify AB its just too hard, so i tried to simplify the question.

If $AB = 0$ then $BA$ will have at least 2 zero rows.

From there i tried to get into conclusion that both matrixes must have at least 2 sets of 2 rows that they are must be similar (1 is multiplicity of the other) and its was a very long deduction.

I am a little hopeless and need help, thanks.

user1551
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omer s
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2 Answers2

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Assuming that by $p(BA)$ you mean $rank(BA)$ we approach by contradiction

Suppose that $rank(BA)\geq 3$

Since $rank(BA)\leq \min\{rank(A),rank(B)\}$ this implies that both $rank(A)$ and $rank(B)\geq 3$ further implying $nullity(A)\leq 1$ and $nullity(B)\leq 1$

Now, this would imply that as $rank(B)\geq 3$ there are at least three linearly independent nonzero outputs of $Bx$, at most one of which can be in the kernel of $A$, implying the remaining two are not in the kernel of $A$. By using one of those vectors which maps to the image of $B$ which is not in the kernel of $A$ we show that there exists a vector $x$ such that $ABx\neq 0$ hence $AB\neq 0$.

By contradiction then, if $AB=0$ then $rank(BA)<3$

Community
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JMoravitz
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Another simple proof.

$(BA)^2 = B(AB)A = 0$ => Im(BA) is subset of Ker(BA) => $dim(Im(BA)) \le dim(Ker(BA))$=> $2dim(Im(BA)) \le 4$ => $dim(Im(BA)) \le 2$

kotomord
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