Rank of matrix AB when A and B have full rank Define $A$ as $m\times n$ matrix with rank $n$, and $B$ as $n\times p$ matrix with rank $p$. Calculate the rank of matrix $C=AB$.
--edit--
Rank of a matrix is the number of linear independent rows.
 A: The conditions imply (as pointed out in YACP's +1 answer) that $m\ge n\ge p$. Another way of reaching the desired conclusion is that consequently the underlying linear mappings (the linear mappings that have the given matrices w.r.t to natural bases)
$A:\mathbb{R}^n\rightarrow\mathbb{R}^m$ and $B:\mathbb{R}^p\rightarrow\mathbb{R}^n$ are both injective. Therefore the composition $AB$ is also injective. Therefore its rows are linearly independent. Therefore $AB$ has full rank, too.
A: Suppose $M$ is an $r\times s$ matrix, $\varphi_M$ denotes a linear operator related to $M$, i.e.
\begin{align}
\varphi_M:k^s&\to k^r\\
x&\mapsto Mx
\end{align}
Since $\operatorname{rk}(A)=n$, we have $\dim(\operatorname{im}(\varphi_A))=n$. By Rank-nullity theorem, $\dim(\ker(\varphi_A))=0$, therefore $\varphi_A$ is injective. Likewise, $\varphi_B$ is injective, hence $\varphi_C=\varphi_{AB}=\varphi_A\circ\varphi_B$ is injective, thus $\dim(\ker(\varphi_C))=0$, and $\dim(\operatorname{im}(\varphi_C))=p$, so $\operatorname{rk}(C)=p$.
A: The conditions from the hypothesis imply that $m\ge n\ge p$. One knows that $\operatorname{rank}(AB)\le \min(m,p)=p$. On the other side, from Sylvester Rank Inequality we get $p=n+p-n\le \operatorname{rank}(AB)$, so $\operatorname{rank}(AB)=p$.
