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

$$T$$ 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?

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.

• In point 2, you wrote T:R^n-> R^n, although T:R^n-> R^m. And i dont understand your point 3.Please explain – Parul Sep 12 '16 at 20:21

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