Linear Algebra, $IM(AB)=IM(A)$ for non square matrices.

Question

For non-square matrices $A \in \mathbb{R}^{m \times n}, B\in \mathbb{R}^{n\times p}$ with both having rank = n. (m>n and m>p)

Do we have $Im(AB)=Im(A)$? I guess it should be true, but can not find a way to proof it.

What seems clear to me is the following: By Sylvester's Rank Formula it holds rank(AB)$\geq$ rank(A)$+$rank(B) $- K = K + K - K.$ And on the other hand trivially rank(AB) $\leq$ rank(A) since $Im(AB)\subset Im(A)$, i.e. rank(AB) = rank(A) which indicates that images of $AB$ and $A$ could be equal.

Answer 1

Clearly Im(AB)$\subset$ Im(A).

On the other hand, let $z\in$ Im(A). Then there is $y\in \mathbb{R}^n$ with $Ay=z$. Since $B:\mathbb{R}^p \to \mathbb{R}^n$ is surjective we find $x \in \mathbb{R}^p$ with $Bx = y$, i.e. $ABx=Ay=z$ which shows that $z\in $Im(AB).