# linear dependence of matrix columns

• Feb 26th 2011, 03:24 PM
situation
linear dependence of matrix columns
if i have a m x n matrix (call this matrix A for arguments sake) and an n x p matrix (call this matrix B) how would i prove that, if the columns of B are linearly dependent, then so are the columns of AB?

i have been looking at the definition for linear dependance but I have been struggling to put this in to context here... (Headbang)

any thoughts would be greatly appreciated.

Cheers
• Feb 26th 2011, 09:24 PM
LoblawsLawBlog
You could use some facts about rank, namely that the rank of a matrix is the maximum number of independent columns and rank AB<= rank B.

This might be overkill because depending on which definition of rank you use, it could be hard to prove either one of these. Sometimes rank is defined as the largest number of independent columns, but then I don't think showing that rank AB<=rank B is too easy. If rank is defined as the size of a largest minor with nonzero determinant, then rank AB<=rank B follows quickly but the first fact might not. Maybe someone can come up with a simpler solution.
• Feb 27th 2011, 05:11 AM
HallsofIvy
The columns of a matrix are dependent if and only multiplication by that matrix is not "one to one".

If the columns of B are dependent, then multiplication by B is not one to one- there exist vectors u and v, \$\displaystyle u\ne v\$ such that Bu= Bv. From that, AB(u)= A(Bu)= A(Bv)= AB(v) so that AB is not one to one and so its columns are dependent.
• Feb 27th 2011, 05:35 AM
situation
Quote:

Originally Posted by HallsofIvy
The columns of a matrix are dependent if and only multiplication by that matrix is not "one to one".

If the columns of B are dependent, then multiplication by B is not one to one- there exist vectors u and v, \$\displaystyle u\ne v\$ such that Bu= Bv. From that, AB(u)= A(Bu)= A(Bv)= AB(v) so that AB is not one to one and so its columns are dependent.

thanks for that just unsure what you mean by matrix multiplication being "one to one". please could you elaborate slightly?

thank you very much! :)
• Feb 27th 2011, 09:29 AM
LoblawsLawBlog
Thanks, HallsofIvy. I knew something like that would work but I always get the relationship between dependent rows/columns and onto/one to one mixed up. Last night I thought it was columns/onto... should have checked a book :)

Situation, I don't know if there's much more to say about one to one than what HallsofIvy wrote. If B is one to one, it means that if x and y are different vectors, then Bx and By are also different vectors (as Fraleigh would say, "two to two" might be a better description). Equivalently, if Bx=By, then it must be the case that x=y.