Derivation of formula $\mathrm{rank}(A)+\mathrm{rank}(B)-m\le\mathrm{rank}(AB)\le\min\{\mathrm{rank}(A), \mathrm{rank}(B)\}$

Question

$T$ is a linear transformation $T:\Bbb R^n\longrightarrow\Bbb R^m$ and $U:\Bbb R^m\longrightarrow\Bbb R^n$ where $m≠n$ such that $TU$ is bijective. Then where did this formula
$\mathrm{rank}(T)+\mathrm{rank}(U) -m \le \mathrm{rank}(TU) \le \min\{\mathrm{rank}(T), \mathrm{rank}(U)\}$ come from? Can someone please give its proof?

Answer 1

Hints:

1. Note that for any linear map $$T:\mathbb{R}^n\rightarrow \mathbb{R}^m$$ the rank of $T$ is less or equal than $\min(n,m)$.

2. A linear map $$T:\mathbb{R}^n\rightarrow \mathbb{R}^n$$ is bijective iff its rank is $n$.

3. $m\geq n$, due to $1$.

4. The second inequality is standard.

Answer 2

This inequality is known as Sylvester's Rank inequality. The related properties are provided in the above link.