# What can be said about $\text{Rank}(C)?$ [duplicate]

Assume that $A\in\mathbb{R}^{n\times k}, \ B\in\mathbb{R}^{k\times m}$ and that $\text{Rank}(A)=r, \ \text{Rank}(B)=s.$ What can be said about $\text{rank}(C),$ where $C=AB?$

The only thing I can quickly see here is that $C\in\mathbb{R}^{n\times m}.$ I also know that $\text{Rank}(C)=\text{Dim}(V(C)).$ But what is the dimension of $C$ then?

The exam has a really messy solution where the answers are that the rank of $C$ is sometimes $=s$ and sometimes $=r$. I don't understand any of it.

## marked as duplicate by Arnaud Mortier, Chris Custer, Leucippus, B. Mehta, Xander HendersonMay 25 '18 at 2:29

Hint: To show $\text{Rank } AB \leq \text{Rank } B$, consider the function $f(x) = Ax$. Then, $AB = (f(b_1),...,f(b_k))$, where $b_i$ is the $i$-th column of $B$. If $f(b_i)$ cannot be represented by the other vectors in $AB$ (i.e. there exist no scalars $z_1,...,z_{i-1},z_{i+1},...,z_k$ such that $f(b_i) = z_1f(b_1)+...+z_{i-1}f(b_{i-1})+z_{i+1}f(b_{i+1})+z_kf(b_k)$), then what can be said about $b_i$?
To show $\text{Rank } AB \leq \text{Rank } A$, make use of $\text{Rank} A = \text{Rank} A^T$ and the previously shown proposition.