Finding Basis of W: Linear Independence of X1,X2,X3,X4

[Dell]

given these 4 vectors X^1..4 in vector space W

W=sp{(X₁,X₂,X₃X₄)\inr⁴| 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?

 

[Mark44]

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.

 

[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

 

[Mark44]

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

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?

 

[Dell]

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

 

[Mark44]

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

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.