# Homework Help: Basis and dimention

1. Jan 15, 2009

### Dell

W is a vector space inside of R4 (1 0 1 0) and (1 2 1 1)
what is the basis and dimention of W

what i did was put them into a homogenic system and eliminated whatever i could

1 1 | 0
0 2 | 0
1 1 | 0
0 1 | 0

eventually got to

1 0 | 0
0 1 | 0

is (1,0) and (0,1) my basis
and 2 my dimention??
does this mean that i have a 2 dimention vector space inside a 4 dimentional space?

is there any way to check my answers??

2. Jan 15, 2009

### Staff: Mentor

This is not a clear description of the problem. Is W the subspace of R4 consisting of all linear combinations of the two vectors you showed?
Assuming that I have correctly interpreted your problem description, the two vectors--(1, 0, 1, 0) and (1, 2, 1, 1)--are a basis for the subspace W. These vectors are clearly linearly independent, and they span W, so they form a basis. Since there are two of them, the dimension (note spelling) is 2.
The work below is no help.
No and yes. Any vectors in R^4 or a subspace of it have to be 4-dimensional (have 4 components), so (1,0) and (0,1) could not possibly form a basis for any subspace of R^4.
Yes, the dimension of the subspace W (if I actually understand what it is) is 2, but there is very little connection between the work you showed and this result.

3. Jan 15, 2009

### sutupidmath

Just a few more words,aside from what was already said:

Any time you have, say a subspace W that consists of all linear combinations of some vectors, say a,b,c etc. in R^n, then all you need to do is to pick the ones that are lin. independent, since you already know that they span your space.

4. Jan 15, 2009

### Dell

so how do i prove that they are lin independant? by doing what i did in my 1st step , ie
1 1 | 0
0 2 | 0
1 1 | 0
0 1 | 0

eventually got to

1 0 | 0 x1=0
0 1 | 0 x2=0

??

5. Jan 15, 2009

### Staff: Mentor

That will work to show that the vectors are linearly independent.

With two vectors, it's easier to show that they are linearly independent by showing that neither is a multiple of the other. It's pretty obvious that no multiple of (1, 0, 1, 0) will get you to (1, 2, 1, 1). With three or more vectors, you have to do what you did.

In any case, something that always works is the definition of linear independence, namely, that the equation $a_1*x_1 + a_2*x_2 + ... + a_n*x_n = 0$ has only a single solution for the constants a_i.

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