# Subspaces and basis

1. Jan 22, 2009

### Dell

i am given these 2 groups
W=sp{(1 0 2 0) (1 1 1 1) (1 0 0 0)}
U=sp{(1 0 1 1) (1 2 1 2) (0 0 1 0)}

a basis for each one and their dimention
a basis for W+U
a basis for W$$\cap$$U
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for the basis i found that they are both linearly independant therefore my basis is the span given and the dimention is 3 for both of them
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how do i find a basis for W+U? can i take the 6 vectors given and check which are dependant and which are independant, take the independant ones in which case i get

w+u=sp{(1 0 2 0) (1 1 1 1) (1 0 0 0) (1 0 1 1)}
dim(W+U)=4
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for W$$\cap$$U am i looking for all the vectors in W which are perpendicular to U? how would i do this?
i know how to find one vector perpendicular to a subspace but how do i find a basis for a group perpendicular to another group

2. Jan 22, 2009

### Dell

for vectors in W which intersect U can i set up a parameter vector (a b c d) then compare it to the basis of W to get a homogenic system

1 1 1 | a
0 1 0 | b
2 1 0 | c
0 1 0 | d

then i get d-b=0

then do the same for u

1 1 0 | a
0 2 0 | b
1 1 1 | c
1 2 0 | d

and i get b-2(d-a)=0

then any vector that adheres to these 2 conditions is an answer.