Does the Null spaces of matrix $ntimes n$ matrix $A$ and matrix $BA$ equal to each other if the matrix $B$ is invertible?

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

Given 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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