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Finding a basis for a subspace |
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| Nov4-06, 03:25 PM | #1 |
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Finding a basis for a subspace
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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| Nov4-06, 07:35 PM | #2 |
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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... |
| Nov4-06, 08:30 PM | #4 |
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Finding a basis for a subspace
ok i found that basis, now how do i find an orthonormal basis?????
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