$A:V\to W, B:W\to Z$ be linear map on finite dimensional vector spaces.
I need to show $\text{rank}(BA)\le \text{rank}(A)$.
I thought like this: suppose $\text{rank}(A)=m$, suppose $\text{rank}(BA)=m+1$, so there exists linearly independent vectors $z_1,\dots,z_{m+1}\in Z$ and $v_1,\dots,v_{m+1}\in V$ such that $BA(v_i)=z_i$,now $A(v_1),\dots, A(v_{m+1})$ must be linearly independent, but this is a contradiction to the fact that $\text{rank}(A)=m$, could anyone help me how to proceed now?
If $\mathrm{rank}(A)=m$, this means that the image of $A$ has dimension $m$, so every set of $\geq m$ vectors in the image is linearly dependent. But $B$, as a linear map, maps linearly dependent vectors to linaerly dependent vectors (it preserves the linear structure). So for the image of $B$ to be of rank $\geq m$, it must be used on a space of dimension at least this dimension $\geq m$ (linear maps can only lower the dimension). So $\mathrm{rank}(AB)=\dim(B[\mathrm{im}(A)])\leq \dim(\mathrm{im}(A))=m$
$BA(V)$ is clearly spanned by the images of a basis of $A(V)$, so a basis of $BA(V)$ cannot have more elements than a basis of $A(V)$.
