# Let $ABC$ be a well-defined product of matrices. Suppose that $A,C$ are both invertible. Prove that $rank(ABC)=rank(B)$.

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 $rank(ABC)=rank(B)$.

• Which is your definition of "rank" of a matrix? – Agustí Roig Oct 22 '13 at 16:57
• @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 $C$. – agent154 Oct 22 '13 at 17:09
• Well, you could have defined "rank" by means of reduced matrices... Anyway, see my answer. – Agustí Roig Oct 22 '13 at 17:10
• @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 $B$... – 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

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

Anyway,

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