Theorem: $r(B)\ge r(AB)$

Question

Theorem: $r(B)\ge r(AB)$, where $r(A)$ denotes the rank of $A$

I would appreciate any corrections on the proof below and any alternate proofs. Many thanks for your time.

Proof

Step 1. By the theorem that any linear map $T:F^n→F^m$ is of the form $T(v)=Av$ for some matrix $A_{m\times n}$, we take the following linear maps along with matrix representations:

$T:F^n→F^m$, represented by $A_{m\times n}$

$S:F^p→F^n$, represented by $B_{n\times p}$

$TS:F^p→F^m$, represented by $(AB)_{m\times p}$

2. We will assume $T$ is onto to give the image of $T$ and $TS$ maximal dimension, therefore $\dim(\operatorname{Im}T)=m$ and $\dim(\operatorname{Im}TS)=m$

3. By definition $r(A)=\dim(\operatorname{Im}A)$, therefore $r(B)=\dim(\operatorname{Im}B)=n$ and $r(AB)=\dim(\operatorname{Im}AB)=m$. To show $r(B)\ge r(AB)$ we need to show $n\ge m$

4. The rank-nullity theorem states $\dim V=\dim(\operatorname{Im}T)+\dim(\operatorname{Ker}T)$. We take $k=\dim(\operatorname{Ker}T)$ and apply rank-nullity to $T$ from step 1, this gives $n=m+k$. We apply rank-nullity to $T$ as both n and m are included in the inequality we are attempting to prove $n\ge m$

5. By definition of dimension, $k$ cannot be negative. By the theorem that a kernel is a subspace and the definition of a subspace of a vector space, $k\ge 0$

6. As $k\ge 0$, $n\ge m$. This is true in all cases as by the assumption in step 2, m has maximal dimension. By step 3, $r(B)=n$ and $r(AB)=m$, therefore $r(B)\ge r(AB)$

Q.E.D.

Answer 1

For any $x\in \Bbb R^m$ we have, $$x^tAB=(x^tA)B\in \{y^tB:y\in \Bbb R^n\}.$$ So that, $$\{x^tAB:x\in \Bbb R^m\}\subseteq \{y^tB:y\in \Bbb R^n\}$$ $$\implies r(AB)=\text{dim}\big(\{x^tAB:x\in \Bbb R^m\}\big)\leq \text{dim}\big(\{y^tB:y\in \Bbb R^n\}\big)=r(B).$$