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 $L_{B}: mathbb{F}^{m} to mathbb{F}^{p}$ is the left multiplication transformation, $L_{B}(x) = Bx$ for $x in mathbb{F}^{m}$

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.