# Math Help - Orthogonal complements

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

$(U \cap W)* = U*+W*$

and

$(U+W)=U* \cap W*$

here, U* denotes the orthogonal complement of U.

having trouble even finding where to start this one.

2. i know i need to prove subspaces both ways.

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

i'm stuck.

3. ## hey

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.

4. Originally Posted by mathisthebestpuzzle
$(U \cap W)* = U*+W*$

and

$(U+W)=U* \cap W*$ ====> shouldn't it be $(U+W)*=U* \cap W*$
well this is basically De Morgans theorem, the way i would prove it is through drawing a Venn diagram and shade out the required region.

5. Originally Posted by mathisthebestpuzzle
Let V be an inner product space, and let U and W be subspaces of V. Show that

$(U \cap W)* = U*+W*$

and

$(U+W)=U* \cap W*$

here, U* denotes the orthogonal complement of U.

having trouble even finding where to start this one.
See here.

6. the link sends me to a login for live journal? thank you though

7. Originally Posted by mathisthebestpuzzle
the link sends me to a login for live journal? thank you though
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⊥.