Finding a basis for a subspace

  • Thread starter JasonJo
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  • #1
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Let U be a proper subspace of R^4 and let it be given by the equations:

1) x1+x2+x3+x4=0
2) x1-x2+2x3+x4=0

how do i find a basis for this subspace???

I got that (0,1,2,0) is one of the basis vectors since x2=2x3, therefore whatever we pick for x2, x3 will be twice that value.

i also got that x4=-x1-1.5x3, but does this require two more basis vectors or one?

ie, i'm asking, it seems that every x value can be determined once x1 and x2 are determined, therefore it should have 2 basis vectors, but i cant quite put it into that form
 
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Answers and Replies

  • #2
I think that you have 2 leading variables and 2 dependent variables (parameters), i.e. your solution subspace is in dimension 2, with 2 vectors generating it.
reduced row echelon form gives you
x1 + 1.5x3 + x4 = 0
x2 - 0.5x3 = 0

from the first equation you have
x1 = - 1.5x3 - x4
from the second one you have
x2 = 0.5x3

and x3, x4 are parameters, so you can let x3 = t and x4 = s

Now

(x1,x2,x3,x4) = (-1.5t+s, 0.5t, t, s) =
t(-1.5, 0.5, 1, 0) + s(1, 0, 0, 1)

so the vectors you need to find are:

(-1.5, 0.5, 1, 0) and (1, 0, 0, 1).

I hope I didn't mess up with the numbers...
 
  • #3
Office_Shredder
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JasonJo said:
I got that (0,1,2,0) is one of the basis vectors since x2=2x3, therefore whatever we pick for x2, x3 will be twice that value.

It's 2*x2 = x3 actually
 
  • #4
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ok i found that basis, now how do i find an orthonormal basis?????
 

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