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I'm having some trouble with the following question:

Let $A, B \in \mathbb{R}^{n \times n}$ and let $A$ be invertible. Is it true that in this case $rank(BA)=rank(B)$?

I think that this statement is correct, but I'm unable to prove it.

My thoughts so far:

If $B$ is also invertible the statement clearly holds, since $GL_n(\mathbb{R})$ is a group.

For $B$ not invertible we immediately have the inequality $rank(BA) \leq rank(B)$ because the columns of $BA$ are linear combinations of the columns of $B$.

Now I've tried to prove the other inequality by contradiction, i.e. assuming that $rank(BA)<rank(B)$ and showing that this cannot be. But I can't complete this step.

Thanks in advance for any help!

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  • $\begingroup$ You will find a proof here (+1 for good presentation of the question) $\endgroup$ – AlexR May 19 '14 at 7:40
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Hint. If $A$ is invertible then the set of vectors $A{\bf x}$, where ${\bf x}\in{\Bbb R}^n$, is the whole of ${\Bbb R}^n$. So $${\rm im}(BA)=\{BA{\bf x}\mid {\bf x}\in{\Bbb R}^n\}=\{B{\bf y}\mid {\bf y}\in{\Bbb R}^n\}={\rm im}(B)\ .$$

This is more or less the whole answer, but see if you can provide the reason for each step.

Note: depending on the notation you are using in your course, ${\rm im}(B)$ is the same as ${\rm col}(B)$ or ${\rm CS}(B)$.

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  • $\begingroup$ Thank you for your answer! I like this proof much better than the proof referred to above! $\endgroup$ – Tom Bombadil May 19 '14 at 7:50
  • $\begingroup$ Thank you! BTW note that almost exactly the same argument shows easily that for any $A$ of suitable size we have ${\rm rank}(BA)\le{\rm rank}(B)$. $\endgroup$ – David May 19 '14 at 7:57
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Hint: For any two matrices $A$ and $B$ the inequality $$ \DeclareMathOperator{rk}{rk}\rk(BA)\leq\rk(B) $$ holds since the columns of $BA$ are linear combinations of the columns of $B$. Since $\rk(X)=\rk(X^\top)$ we then also have $$ \rk(BA)=\rk((BA)^\top)=\rk(A^\top B^\top)\leq\rk(A^\top)=\rk(A) $$ Can you use this to finish your problem?

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