rank theory

**situation** · March 6th 2011, 11:22 AM

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!

**TheEmptySet** · March 6th 2011, 11:57 AM

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

$\det(AB)=\det(A)\det(B)$

**HallsofIvy** · March 6th 2011, 12:24 PM

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.

**situation** · March 6th 2011, 01:31 PM

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!

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