# Linear independance

1. Jan 22, 2009

### Dell

given these 4 vectors X1..4 in vector space W

W=sp{(X1,X2,X3X4)$$\in$$r4| X1+2X3+X4=0}

Find the basis of W,

what i did was--
for these to be the basis of W they need to span W and be independant of one another,-
since X1+2X3+X4=0 it is clear that these 3 are not independant, since X1+2X3=-X4 etc..

does this mean that dimW=3 and i can take any combination of 3 of the 4 x's for my basis as long as X2 is one of them?

2. Jan 22, 2009

### Staff: Mentor

x1, x2, x3, and x4 aren't vectors; they are coordinates of an arbitrary vector in R4. The problem could just have easily been stated in terms of x, y, z, and w.

From the definition of the set W, I can tell by inspection that (1, 1, 0, -3) is in the set, and I got this by substituting values into the equation x1 + 2x2 + x4 = 0.

You can characterize all of the vectors in W just from the equation and three other obvious equations:
x1 = - 2x2 - x4
x2 = 1x2
x3 = 1x3
x4 = 1x4

If you look at this system of equations for a while, you might see some vectors lurking in there. You might even be able to convince yourself that they are linearly independent and span W.

3. Jan 22, 2009

### Dell

thanks, so the dimention would be 1
and the basis would be that vector (made up of)
x1 = - 2x2 - x4
x2 = 1x2
x3 = 1x3
x4 = 1x4

4. Jan 22, 2009

### Staff: Mentor

No, the dimension is not 1. Notice that for each vector (x1, x2, x3, x4) there are three parameters that can be set. Does that tell you something?

5. Jan 22, 2009

### Dell

right, x2 x3 and x4 are all independant,
dim=3

6. Jan 22, 2009

### Staff: Mentor

No on x2, x3, and x4. Yes on the dimension.

As I said earlier, x2, x3, and x4 are not vectors, so it doesn't make any sense to describe them as linearly dependent or linearly independent.

You should be able to get three lin. independent vectors out of the system of equations I wrote in a previous post.