# Linear algebra

1. Jan 26, 2009

### Dell

in this question i am given 2 subspaces of R4

W={(x1,x2,x3,x4)$$\in$$R4|x1-x4=0}

U=sp{(1,2,0,-1),(2,3,1,-1),(1,-1,1,-1)}

1] a basis and dimention for W
2] a basis for W+U
3] a basis for W$$\cap$$U

1]
since i only have limitations on x1 and x4 i call x2=t x3=s x1=x4=q

therefore W=sp{(1,0,0,1),(0,1,0,0),(0,0,1,0)} and dimW=3

2]
to find a basis for W+U, i look for linearly independant vectors that span the space, so i set up a matrix to see which are combinations of the others,
i got W+U=sp{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)}

these 4 span the whole of R4 so does this mean that W+U = R4
3]
to find a basis of W$$\cap$$U, i found a homogenic system for each subspace and compared the 2.
taking a random vector (a b c d) in the subspace, i get
for W==> -4a+b+3c-sd=0
for U==> a-d=0
from the combination i get -6a+b+3c=0
a=t b=6t-3s c=s d=t

and so i get W$$\cap$$U=sp{(1,6,0,1)(0,-3,1,0)} dim=2
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does this all look okay,
also, if i am asked to find a basis and i write the span, is it the same thing, or must i write just one possible basis in the span

Last edited: Jan 27, 2009