Find Basis for Subspaces & Dimension of W,U,W+U,W\capU

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SUMMARY

This discussion focuses on finding bases and dimensions for the subspaces W and U, defined as W=sp{(1 0 2 0), (1 1 1 1), (1 0 0 0)} and U=sp{(1 0 1 1), (1 2 1 2), (0 0 1 0)}. Both W and U are confirmed to be linearly independent with a dimension of 3. The basis for W+U is determined to be sp{(1 0 2 0), (1 1 1 1), (1 0 0 0), (1 0 1 1)} with a dimension of 4. For the intersection W∩U, the method involves setting up a homogeneous system to find vectors in W that satisfy conditions derived from both subspaces.

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Dell
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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)}

and am asked to find
a basis for each one and their dimension
a basis for W+U
a basis for W[tex]\cap[/tex]U
-----------------------------------------------
for the basis i found that they are both linearly independent therefore my basis is the span given and the dimension is 3 for both of them
------------------------------------------------
how do i find a basis for W+U? can i take the 6 vectors given and check which are independent and which are independent, take the independent 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
------------------------------------------------
for W[tex]\cap[/tex]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
 
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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.
 

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