# Dimensions of general vector spaces

• Apr 3rd 2011, 06:43 PM
ugkwan
Dimensions of general vector spaces
The concepts are starting to snowball. I have been studying and doing the homework assigned, hoping that as the concept accumulate it would create a pool of rules that made more sense, but this hasn't worked out. This particular question hit a nerve, because I am unable to make sense of why a linearly independent set can have zero basis and zero dimensions. I kinda understand that there is only the trivial answer to each skalar multiple inorder for the answer to be zero. I do not understand how could a set that is proven to be linear independent, and span R3, have no dimension or basis. Can someone please take a min to explain what is going on?

Find a basis for the solution space of the homogeneous linear system, and find the dimension of that space.

(xn) where n is the subset

2(x1)+(x2)+3(x3)=0
(x1) +5(x3)=0
(x2)+(x3)=0

I augmented the matrix to reduced row echelon form:

1 0 0 0
0 1 0 0
0 0 1 0
• Apr 3rd 2011, 07:17 PM
tonio
Quote:

Originally Posted by ugkwan
The concepts are starting to snowball. I have been studying and doing the homework assigned, hoping that as the concept accumulate it would create a pool of rules that made more sense, but this hasn't worked out. This particular question hit a nerve, because I am unable to make sense of why a linearly independent set can have zero basis and zero dimensions. I kinda understand that there is only the trivial answer to each skalar multiple inorder for the answer to be zero. I do not understand how could a set that is proven to be linear independent, and span R3, have no dimension or basis. Can someone please take a min to explain what is going on?

Find a basis for the solution space of the homogeneous linear system, and find the dimension of that space.

(xn) where n is the subset

2(x1)+(x2)+3(x3)=0
(x1) +5(x3)=0
(x2)+(x3)=0

I augmented the matrix to reduced row echelon form:

1 0 0 0
0 1 0 0
0 0 1 0

Why work with the augmented one if the last column is all 0's?!? Anyway, the above reduced form show the only

solution to the original homogeneoussystem is the trivial one, so the solution is the zero subspace whose basis is

the empty set and whose dimension is zero.

Tonio
• Apr 3rd 2011, 09:27 PM
ugkwan
That clears it up. So the subspace is the zero vector, thats why the basis is zero, and dimension is zero.

You can tell I am sincerely not getting this subject, but I am trying even if I end up with horrible nose bleeds staring down these questions that seem so foreign to me.

Can you also explain why I shouldn't have used augmented matrix operations when the question is homogenous to begin with?
• Apr 4th 2011, 03:07 AM
Ackbeet
Quote:

Originally Posted by ugkwan
That clears it up. So the subspace is the zero vector, thats why the basis is zero, and dimension is zero.

You can tell I am sincerely not getting this subject, but I am trying even if I end up with horrible nose bleeds staring down these questions that seem so foreign to me.

Can you also explain why I shouldn't have used augmented matrix operations when the question is homogenous to begin with?

I think tonio was getting at the fact that, for a homogeneous system, adding the augmented matrix extra column doesn't gain you any information, since it will be all zeros, and any row operations performed on that column will not change the zeros. They will stay zero. So all you've gained is more work for yourself in writing it out.