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!

  • $\begingroup$ You will find a proof here (+1 for good presentation of the question) $\endgroup$
    – AlexR
    May 19 '14 at 7:40

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 ${\rm CS}(B)$.

  • $\begingroup$ Thank you for your answer! I like this proof much better than the proof referred to above! $\endgroup$ 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

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.