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

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