## Rank of the product of two matrices

Posted: September 9, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: , ,

Problem. Let $A, B$ be $m \times n$ and $n \times k$ matrices, respectively, with entries in some field. Prove that

$\text{rank}(AB) \geq \text{rank}(A)+\text{rank}(B) - n.$

Solution. Let $T_1,T_2$ be the corresponding linear transformations.

Claim. $\text{nul}(T_1T_2) \leq \text{nul}(T_1) + \text{nul}(T_2).$

Proof of the claim. Define $f: \ker (T_1T_2) \to \ker T_1$ by $f(x)=T_2(x)$ for all $x \in \ker(T_1T_2).$ See that $f$ is well-defined, linear  and $\ker f = \ker T_2.$ So, by the rank-nullity theorem

$\text{rank}(f)+ \text{nul}(f)=\dim \ker(T_1T_2)=\text{nul}(T_1T_2).$

But

$\text{rank}(f)=\dim \text{im}(f) \leq \dim \ker T_1=\text{nul}(T_1)$

and

$\text{nul}(f)=\dim \ker f=\dim \ker T_2 = \text{nul}(T_2). \Box$

Now applying the claim and the rank-nullity theorem gives

$\text{rank}(T_1T_2)+n=k- \text{nul}(T_1T_2)+n \geq n- \text{nul}(T_1)+ k- \text{nul}(T_2)=$ $\text{rank}(T_1) + \text{rank}(T_2). \ \Box$