Let $A$ be an $m \times n$ matrix with rank $m$ and $B$ be an $n \times p$ matrix with rank $n$. Determine the rank of $AB$. Justify your answer.
So to begin after doing some specialized examples I worked out that the rank of the matrix $AB$ will be $m$. Now I'm having difficulties proving it.
First off in the textbook I'm using (Linear Algebra, Friedberg, Insel, et al), the definition of the rank is the following:
If $A \in M_{m \times n}(\mathbb{F})$, we define the rank of $A$, denoted $rank(A)$, to be the rank of the linear transformation $L_{A}:\mathbb{F}^{n} \to \mathbb{F}^{m}$, where $L_{A}$ is the left multiplication transformation, $L_{A}(x) = Ax$ for $x \in \mathbb{F}^{n}$. And as a result $L_{B}: \mathbb{F}^{p} \to \mathbb{F}^{n}$ is the left multiplication transformation, $L_{B}(x) = Bx$ for $x \in \mathbb{F}^{p}$
Since I deduced that the rank of $AB = m$, that means I have to show that $rank(AB) = rank (A)$. THis means:
$$rank(L_{AB}) = rank(L_{A}) \\ \Rightarrow\ dim(R(L_{AB})) = dim(R(L_{A})),\\ \text{where $R$ is the range of a linear transformation}$$
So it means I have to show that:
$$dim(R(L_{A}L_{B})) = m$$
So form here to me I deduced that somehow I have to show that $R(L_{A}L_{B}) = R(L_{A})$
And this is where I'm stuck. There was a proof I saw that used the idea of $L_{B}(\mathbb{F^{p}}) = \mathbb{F^{n}}$ but that seemed very much off because for that to be the case, it must be that $L_{B}$ is invertible, which it may not necessarily be. Perhaps my approach is wrong, but I feel I'm close to what I need to accomplish but can't seem to see what last steps I need to take. Some assistance would be nice.
Best Answer
Here is a simple proof:
$B$ is surjective, so Im $AB =$ Im $A$, so rank$(AB) =$ rank$(A) = m$.
To see $B$ is surjective, just note that $B$ maps to a $n$-dimensional space and has rank $n$.