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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 $C=AB$.


Rank of a matrix is the number of linear independent rows.

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answered in question. A*B is just a matrix multiplication – Jonny Jan 7 '13 at 10:33
up vote 6 down vote accepted

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 $\operatorname{rank}(AB)=p$.

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You are right, my mistake. I deleted my post. – Sunny88 Jan 7 '13 at 11:01

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.

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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 $\operatorname{rk}(C)=p$.

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