# Proving $\operatorname{rank}A =\operatorname{rank}B$ when $AB = 2A + 3B$

$$A$$ and $$B$$ are two square matrices such that $$AB = 2A + 3B$$. Show that $$\operatorname{rank}A =\operatorname{rank}B$$.

I managed to prove that the matrices $$A-3I$$ and $$B-2I$$ are invertible and that $$AB=BA$$.

Also if $$A$$ is invertible then $$B$$ is invertible because otherwise determinat of $$2A$$ would be $$0$$ which is false.

I don't know what to do when $$A$$ is not invertible.

Let $$x \in ker(B)$$, then $$0=ABx=2Ax+3Bx=2Ax$$, hence $$x \in ker(A).$$

Thus $$ker(B) \subset ker(A).$$

Similar arguments give: $$ker(A) \subset ker(B).$$

Conclusion: $$ker(B) =ker(A).$$

The rank - nullity -theorem gives now the result.

• We have $x^TA=0\Rightarrow x^TB=0$, but that amounts to $\ker(A^T)\subset\ker(B^T)$, not $\ker(A)\subset\ker(B)$. We can still get the result using the rank-nullity theorem, though. Jul 18 '19 at 14:25

We have

$$A(B-2I) = 3B$$

Since $$B-2I$$ is invertible, it follows that $$\operatorname{rank} A = \operatorname{rank}(3B) = \operatorname{rank} B$$.