Geometric interpretations theorems about rank Consider the following theorems about rank:

*

*If $A$ is a $k \times m$-matrix, $B$ a $m \times n$ matrix and $\operatorname{rank}(A)=m$, then $\operatorname{rank}(AB)=\operatorname{rank}(B)$

*$\operatorname{rank}(AB) \geq \operatorname{rank}(A) + \operatorname{rank}(B)-m$
I know that the "geometric" interpretation of a matrix is that it's a linear map from one vectorspace to another vectorspace and that the rank is by definition the amount of vectors in the basis of the image of the linear map=dimension column space of the matrix that belongs to that linear map.
Does someone know if there exists a geometric intepretation for these theorems? I am familiar with the proofs of both of them. Thanks in advance
 A: Both of these theorems concern the result of applying $A$ to the range (i.e. the set of outputs) of $B$.

If A is a $k \times m$ matrix, $B$ $m \times n$, and $\operatorname{rank}(A)=m$, then $\operatorname{rank}(AB)=\operatorname{rank}(B)$

For a set $S \subset \Bbb R^m$, let $A(S) \subset \Bbb R^k$ denote the set $A(S) = \{Ax : x \in S\}$. In particular, note that $\operatorname{rank}(A) = \dim(A(\Bbb R^m))$.
The key here is to understand that $\operatorname{rank}(A) = m$ means that $A$ is an embedding of $\Bbb R^m$ into $\Bbb R^k$. More specifically, it maps the canonical vectors ("axis directions") of $\Bbb R^k$ to the (linearly independent) columns of $A$. Thus, for any subspace $U$ of $\Bbb R^m$, $\dim(A(U)) = \dim(U)$. Thus, we have
$$
\operatorname{rank}(AB) = \dim AB(\Bbb R^n) = \dim A(B(\Bbb R^n)) = \dim(B(\Bbb R^n)) = \operatorname{rank}(B).
$$

$\operatorname{rank}(AB) \geq \operatorname{rank}(A) + \operatorname{rank}(B) - m$

Here, it's useful to have a geometric understanding of the kernel (AKA nullspace) of a linear transformation. The kernel of $A$ (denoted by $\ker(A)$) consists of all vectors that are mapped (or if you prefer, "squished") to the zero $0$.  The rank nullity theorem tells us that when $A$ has input space $\Bbb R^m$ (i.e. has $m$ columns), we have
$$
\operatorname{rank}(A) + \dim \ker (A) = m.
$$
In other words, the dimension of the input space is "conserved" in some sense. If the outputs of $A$ span a $\operatorname{rank}(A) \leq m$-dimensional subspace, then there must have been $m - \operatorname{rank}(A)$ "dimensions of space" that were mapped to zero.
With that established, we can now interpret the inequality as follows: we can rearrange the inequality to get
$$
\operatorname{rank}(AB) \geq \operatorname{rank}(B) - (m - \operatorname{rank}(A)) = \operatorname{rank}(B) - \dim \ker(A).
$$
In other words: the greatest amount by which multiplication with $A$ can reduce the rank of $B$ is $\dim \ker(A)$.

We can leverage our understanding into a proof if we apply the reasoning behind the rank-nullity theorem and considering what $A$ does to the outputs of $B$.  Let $A|_{B(\Bbb R^n)}$ denote the linear map whereby $A$ is only applied to the range of $B$.  Noting that
$$
A|_{B(\Bbb R^n)}(B(\Bbb R^n)) = A(B(\Bbb R^n)) = (AB)(\Bbb R^n),
$$
we see that $\operatorname{rank}(A|_{B(\Bbb R^n)}) = \operatorname{rank}(AB)$. By the rank nullity theorem, we have
$$
\operatorname{rank}(AB) + \dim \ker (A|_{B(\Bbb R^n)}) =\\
\operatorname{rank}(A|_{B(\Bbb R^n)}) + \dim \ker (A|_{B(\Bbb R^n)}) =
 \\ \dim B(\Bbb R^n) = \operatorname{rank}(B).
$$
Now, we want to find an upper bound for $\dim \ker (A|_{B(\Bbb R^n)})$. By definition, $\ker (A|_{B(\Bbb R^n)})$ consists of the vectors that are in the range of $B$ that are also mapped to $0$. In other words, it is the intersection of the spaces $B(\Bbb R^n)$ and $\ker(A)$.  With that, we can see that $\ker (A|_{B(\Bbb R^n)}) \subset \ker(A)$, so that
$$
\dim \ker (A|_{B(\Bbb R^n)}) \leq \dim \ker (A) = m - \operatorname{rank}(A).
$$
Putting all this together, we have
$$
\operatorname{rank}(B) = 
\operatorname{rank}(AB) + \dim \ker (A|_{B(\Bbb R^n)}) 
\\ \leq \operatorname{rank}(AB) + \dim \ker (A)
\\ = \operatorname{rank}(AB) + m - \operatorname{rank}(A).
$$
Rearranging the inequality $\operatorname{rank}(B) \leq \operatorname{rank}(AB) + m - \operatorname{rank}(A)$ yields the desired result.
