How to prove and interpret $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$?

Question

Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$

I proved $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ by interpreting $AB$ as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information.

How do you manage $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$? Is there a nice interpretation like the previous one?

Answer 1

Yes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get "more stuff" in the former case, as the former is bigger than the latter.

Answer 2

Prove first that if $f:X\to Y$ and $g:Y\to Z$ are functions between finite sets, then $|g(f(X))| \leq \min \{ |f(X)|, |g(Y)| \}.$

Then use the same idea.

Answer 3

Once you have proved $\operatorname{rank}(AB) \le \operatorname{rank}(A)$, you can obtain the other inequality by using transposition and the fact that it doesn't change the rank (see e.g. this question).

Specifically, letting $C=A^T$ and $D=B^T$, we have that $\operatorname{rank}(DC) \le \operatorname{rank}(D) \implies \operatorname{rank}(C^TD^T)\le \operatorname{rank} (D^T)$, which is $\operatorname{rank}(AB) \le \operatorname{rank}(B)$.

Answer 4

Each column of $AB$ is a linear combination of columns of $A$ so $\text{Range}(AB)\subseteq \text{Range} (A)$ equivalently $rank(AB)\leq rank(A).$

Similarly, Each row of $AB$ is a linear combination of rows of $B$ so $rowspace(AB)\subseteq rowspace(B)$.

Since rowspace=columnspace. Thus $rank(AB)\leq rank(B)$.

From both of the inequality we deduce that $rank(AB)\leq \min\{rank A, rank B\}$.