Homework Help: A Simple Linear Algebra Problem

1. Feb 26, 2010

maherelharake

1. The problem statement, all variables and given/known data
If A is a 4x2 matrix, explain why the rows of A must be linearly dependent.

2. Relevant equations

3. The attempt at a solution
I put...
Since the rank of the matrix is either 0, 1, or 2, I can conclude that the nullility is either 2, 1, or 0. So since there are 4 vectors in a 2 dimensional space, at most two are independent. Therefore the other rows must be dependent. '
Is this anywhere close? Thanks in advance.

2. Feb 26, 2010

Tedjn

What you've said is correct, although the last two sentences are what you really want. R2 is 2-dimensional, so any 4 vectors must have dependencies. Of course, however, just from the question itself we can't tell what your professor would expect you to use; that would depend on what he's teaching in class.

3. Feb 26, 2010

maherelharake

So I don't have to add anything else to that? It makes sense, but it seems like that is more of a Chapter 2 explanation (we are in Chapter 5), and I was trying to explain it in a way that deals with our current material. Thanks

4. Feb 26, 2010

Tedjn

In the subject there would be theorems proved that say something to the effect that any set of n+1 vectors in an n dimensional vector space are never linearly independent. These considerations come up when you prove that every basis of a finite dimensional vector space has the same number of vectors; this number is defined to be the dimension. If you've covered this, then I don't see why you can't use the fact that R2 is 2-dimensional.

On the other hand, with the current material, you might say that the column and row rank of a matrix are equivalent (which is what you hinted at), and that the row rank can be at most 2. By definition, row rank is the dimension of the rowspace; if it is less than or equal to 2, then 4 vectors certainly cannot be linearly independent. Essentially, these two arguments are almost equivalent, and they both rely on the concept of a unique dimension. I personally don't know another way to do this problem.

5. Feb 26, 2010

maherelharake

Ok thanks a lot. I will keep what I had, but add a little more explanation.