rank theory

• March 6th 2011, 11:22 AM
situation
rank theory
literally have no clue on how to do this one... :'(

Prove that if A and B are n x n matrices and if rank(A) = n and rank(B) = n then rank(AB) = n.

any help would be much appreciated here as i have a test tomorrow!
• March 6th 2011, 11:57 AM
TheEmptySet
Quote:

Originally Posted by situation
literally have no clue on how to do this one... :'(

Prove that if A and B are n x n matrices and if rank(A) = n and rank(B) = n then rank(AB) = n.

any help would be much appreciated here as i have a test tomorrow!

I don't know what tools you have exactly but here is a way.

A square matrix has full rank if and only if it has non zero determinant. (only the zero vector can be in the nullspace) We also know that

$\det(AB)=\det(A)\det(B)$
• March 6th 2011, 12:24 PM
HallsofIvy
Another way- the "rank" of a linear transformation, A, from vector space V to itself is the dimension of A(V). Here, A is given by an n by n matrix so it is from $R^n$ to $R^n$. Since the rank of A is n, $A(R^n)= R^n$. Similarly, the rank of B is n so that $B(R^n)= R^n$. Putting those together, AB maps $R^n$ into all of $R^n$: $AB(R^n)= R^n$ and so AB has rank n.
• March 6th 2011, 01:31 PM
situation
Quote:

Originally Posted by HallsofIvy
Another way- the "rank" of a linear transformation, A, from vector space V to itself is the dimension of A(V). Here, A is given by an n by n matrix so it is from $R^n$ to $R^n$. Since the rank of A is n, $A(R^n)= R^n$. Similarly, the rank of B is n so that $B(R^n)= R^n$. Putting those together, AB maps $R^n$ into all of $R^n$: $AB(R^n)= R^n$ and so AB has rank n.

please could you just briefly explain what is meant by the dimension, a linear transformation, and a vector space? i have so pretty awful lecture notes which just arent making sense to me in working out what your explanation means.

thank you both very much!