Let V be an inner product space, and let U and W be subspaces of V. Show that

and

here, U* denotes the orthogonal complement of U.

having trouble even finding where to start this one.

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- Apr 26th 2008, 12:49 PMmathisthebestpuzzlePlease Help!!!
Let V be an inner product space, and let U and W be subspaces of V. Show that

and

here, U* denotes the orthogonal complement of U.

having trouble even finding where to start this one. - Apr 27th 2008, 02:49 PMmathisthebestpuzzle
i know i need to prove subspaces both ways.

x the perp of the intersection of U and W what does that tell me?

i'm stuck. - Apr 27th 2008, 07:15 PMmathmathmathhey
hey have you been able to figure out this problem. I also have the same problem just wondering if u found a way to do it or maybe we can but our ideas together to solve it.

- Apr 27th 2008, 11:16 PMDanshader
- Apr 28th 2008, 02:06 AMOpalg
See here.

- Apr 28th 2008, 08:45 PMmathisthebestpuzzle
the link sends me to a login for live journal? thank you though

- Apr 29th 2008, 12:02 AMOpalg
Sorry, I hadn't noticed that LiveJournal page was "friends only". Here's a copy of the relevant comment. The "perp" symbol ⊥ (denoting an orthogonal complement) has come out on the line instead of as a superscript, which makes it a bit hard to read.

You need to know that U⊥⊥ = U. Also, if G and H are subspaces with G⊆H, then H⊥⊆G⊥.

If x = y+z with y∈U⊥ and z∈W⊥ then it is easy to see that y and z both lie in (U∩W)⊥, hence so does x. Therefore U⊥+W⊥⊆(U∩W)⊥.

For the converse inclusion, if x∈(U⊥+W⊥)⊥ then x∈U⊥⊥ = U, and similarly x∈W⊥⊥ = W. Thus (U⊥+W⊥)⊥⊆U∩W. Take the perp of each side to see that (U∩W)⊥⊆U⊥+W⊥.

Thus (U∩W)⊥ = U⊥+W⊥. You get the other identity by taking the perp of both sides and replacing U with U⊥ and W with W⊥.