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$A$ 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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – user1551
    Jul 18 '19 at 14:25
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We have

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

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

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