As the title says, both $A$ and $B$ are $ntimes n$ matrices, I want to prove the Null spaces $Null (A)$ = $Null(BA)$.

I do not know if the statement is correct,

and I am not sure whether my derivation as follows is correct. Is there a new method to prove or any counterexamples？

Proof: $forall u in Null(A)$, then $Au = 0$. Multiply $B$ from left, we obtain $BAu=0$, which means any vector $u in Null(A)$ is a vector in $Null (BA)$.

Similarly, $forall v in Null(BA)$, we have $BAv = 0$. Multiply $B^{-1}$ from left, we obtain $B^{-1}BAv=0$, ie. $Av=0$, which means any vector $v in Null(BA)$ is a vector in $Null (A)$.

Therefore, we obtain $Null(A) = Null (BA)$, ie., $Ax=0$ and $BAx=0$ have the same solution.

Mathematics Asked on November 18, 2021

1 AnswersGiven that $B$ is invertible, we have that $0 = BAv = B(Av)$ if and only if $Av = 0$ because $B$ is injective. But this precisely that $ker A = ker BA.$

Answered by Dylan C. Beck on November 18, 2021

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