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Rank of the product of two matrices

aukie

New member
Jul 19, 2012
1
Hello

Both of the below theorems are listed as properties 6 and 7 on the wikipedia page for the rank of a matrix.

I want to prove the following,

If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A).

Apparently this is a corollary to the theorem
If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ).

which I know how to prove. But I can't prove the first theorem. Any ideas? I would especially like to see how it is a corollary to the second theorem which the author in the book I am reading claims. Thanks for reading
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
I want to prove the following,

If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A).

Apparently this is a corollary to the theorem
If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ).
You want to prove that if A is an M by n matrix and B is an n by n matrix of rank n, then rank(AB) = rank(A). But an n by n matrix of rank n is necessarily invertible. So $B$ has an inverse $B^{-1}$. It follows from the theorem that $\text{rank}(A) = \text{rank}((AB)B^{-1}) \leqslant \text{rank}(AB).$ The reverse inequality $\text{rank}(AB)\leqslant \text{rank}(A)$ follows directly from the theorem. Hence $\text{rank}(AB)= \text{rank}(A).$