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Do I just pick a vector arbitrarily say a vector v = (x,y,z) and go from there?

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- Thread starter 7sqr
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- #1

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Do I just pick a vector arbitrarily say a vector v = (x,y,z) and go from there?

- #2

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pick vectors that are of the form (a,0,0), and otherwise arbitrary.

- #3

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let u, v ∈ W

u=(a1,0,0) v= (a2,0,0)

u+v = (a1+a2, 0,0) ∈ W

∴u+v ∈ W

am i anywhere close to doing that right?

- #4

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Yes, that's the right way to do it.

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Thank you guys, I really appreciate the help.

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This is true, but all you have to do to see it is to write ##(a,0,0)=a(1,0,0,)##.

- #8

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This is true, but all you have to do to see it is to write (a,0,0)=a(1,0,0,)(a,0,0)=a(1,0,0,).

Yes, in this case it is easy. But if the specification had been more complicated, having a standard recipe is not a bad idea.

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Svein thanks for that, it clears up the concept a little more.

- #10

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What Svein described is how to find the orthogonal complement of the orthogonal complement of the set W. This is

- #11

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Well, in the [itex]l_{2}[/itex] space (with the standard scalar product), this algorithm is used in proofs of completeness...

What Svein described is how to find the orthogonal complement of the orthogonal complement of the set W. This isalwaysa subspace, even if W isn't. If you know this, you can find out if ##W## is a subspace by checking if ##W^{\perp\perp}=W##. This is rarely (never?) the easiest way to do it.

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