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$A$ is a $m \times n$ matrix and $B$ is a $m \times m$ matrix.
I have that if a vector $x$ is in the kernel of $A$, then $Ax=0$, so that vector $x$ would be also in the kernel of $BA$. Then, from the rank nullity theorem I get that the nullity of $A$ is $n-rank(A)$ and the nullity of $BA$ is $m-rank(BA)$

Dan Klymenko
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Irene
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  • @MohammadZuhairKhan I'm sorry I wrote wrong the question. Is BA, not AB. So far I've gotten that if x is in the nullspace of A then it has to be in the nullspace of BA. But I don't know if that is correct – Irene Jan 04 '18 at 15:48
  • @JoséCarlosSantos it is the other way around, rank(BA) not rank(AB) and that changes a lot right? – Irene Jan 04 '18 at 15:52
  • @Irene Since $\operatorname{rank}(AB)=\operatorname{rank}\bigl((AB)^T\bigr)=\operatorname{rank}(B^TA^T)$, it changes nothing. – José Carlos Santos Jan 04 '18 at 15:53
  • @JoséCarlosSantos I would say that the fact the rank of a matrix is equal to the rank of its transpose is non-trivial in this context – Ben Grossmann Jan 04 '18 at 15:55

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You have already shown that if $x$ is in the nullspace of $A$, then $x$ is in the nullspace of $BA$. Now, using the invertibility of $B$, conclude that if $x$ is in the nullspace of $BA$, then it is also in the nullspace of $A$.

All together, we see that the matrices $A$ and $BA$ have the same nullspace. By the rank-nullity theorem, they must also have the same rank.

Ben Grossmann
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  • I've edited the question to say the dimensions of the matrices, so, having that dimensions could I apply what you said about the rank nullity theorem? – Irene Jan 04 '18 at 16:04
  • Yes, there is no issue there; the rank-nullity theorem applies for non-square matrices as well. – Ben Grossmann Jan 04 '18 at 16:08
  • Yes, I know that but wouldn't they be null(A) = n-rank(A) and null(BA) = m - rank (BA) and if n and m are different... – Irene Jan 04 '18 at 16:10
  • Note that the matrix $BA$ has the same size as $A$. The formula should be $$ \operatorname{null}(BA) = n - \operatorname{rank}(BA) $$ – Ben Grossmann Jan 04 '18 at 16:13