Not sure where to go with this... I've done the following, though:

$A$ and $C$ must be square matrices if they are invertible. Let $A$ be an $m\times m$ matrix, and let $C$ be an $n\times n$ matrix. Let $B$ be a $m\times n$ matrix so that $ABC$ is well defined.

$A$ has a rank of $m$ and $C$ has a rank of $n$. $rank(AB)\leq m$ and $rank(AB)\leq rank(B)$ Also, $rank(BC)\leq n$ and $rank(BC)\leq rank(B)$.

From here, I have no idea what to do. I don't even know if I'm on the right track. The best I can figure is that $rank(ABC)\leq \min(rank(A),rank(B),rank(C))$, but I have no idea how to show that $rank(ABC)=rank(B)$.

  • $\begingroup$ Which is your definition of "rank" of a matrix? $\endgroup$ – Agustí Roig Oct 22 '13 at 16:57
  • $\begingroup$ @AgustíRoig What do you mean? "rank" is the dimension of the column space... so if $A$ and $C$ are invertible, and $A$ is an $m\times m$ matrix, then the columns of $A$ are linearly independent, meaning the dimensions of the column space is $m$, thus the rank of $A$ is $m$... simmilarly for $C$. $\endgroup$ – agent154 Oct 22 '13 at 17:09
  • $\begingroup$ Well, you could have defined "rank" by means of reduced matrices... Anyway, see my answer. $\endgroup$ – Agustí Roig Oct 22 '13 at 17:10
  • $\begingroup$ @AgustíRoig Ahh.. I think I see how I might do this, then... If $A$ and $C$ are invertible, they can be reduced to $I$... and $I$ has the same rank as $A$ and $C$. So if I have the product $I_mBI_n$, then I get $B$... $\endgroup$ – agent154 Oct 22 '13 at 17:14

Hopefully this isn't flawed... but it seems like it should be correct...

If $A$ and $C$ are invertible, they must be square matrices. Let $A$ be an $m\times m$ matrix, $B$ be an $m\times n$ matrix, and $C$ be an $n\times n$ matrix. $rank(A)=m$ and $rank(C)=n$.

Since $A$ and $C$ are invertible, they can be reduced via Gaussian Elimination to their respective identity matrices $I_{m}$ and $I_{n}$ respectively, both of which have the same rank: $rank(I_{m})=m$ and $rank(I_{n})=n$. Consider the new matrix product \begin{align*} I_{m}BI_{n}=B \end{align*} which has the same rank as \begin{align*} ABC \end{align*} Therefore, $rank(ABC)=rank(I_{m}BI_{n})=rank(B)$.


So, if $b_1, \dots , b_n$ are the columns of $B$,

$$ \mathrm{rank}\ B = \mathrm{dim}\ \mathbf{span} (b_1, \dots , b_n) \ . $$

Also you can see that the columns of $AB$ are

$$ Ab_1, \dots , Ab_n \ . $$

Hence, if $A$ is invertible, multiplication by $A$ is an isomorphism and

$$ \mathrm{dim}\ \mathbf{span} (Ab_1, \dots , Ab_n) = \mathrm{dim}\ \mathbf{span} (b_1, \dots , b_n) \ . $$

Alternatively you can show easily, just with the definition of linearly independent vectors and a short computation that, if $A$ is invertible, then

$$ b_1, \dots , b_n \quad \text{are l.i.}\qquad \Longleftrightarrow \qquad Ab_1, \dots , Ab_n \quad \text{are l.i.} \ . $$


$$ \mathrm{rank}\ AB = \mathrm{rank}\ B \ . $$


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.