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Subspaces and perpendiculuar subspaces

  • Thread starter jumbogala
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  • #1
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Homework Statement


How do you show that M double perp is a subset of M?

Homework Equations





The Attempt at a Solution


My prof told me to try proving that M is a subset of M perp perp, then to use the facts that if M is a subspace of Rn then T(X) = projU(X) for all X in Rn.

I'm not sure how to go about that. I know logically that it's a subset, but I don't know how to prove it.

I'm thinking that once I prove it, maybe I can show that the dimension of U and U perp perp are equal, so the spaces are equal too?

Can anyone help get me started? Thanks :)
 
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Answers and Replies

  • #2
Office_Shredder
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It follows pretty much from the definition of perpendicularity. Try it out and see what happens
 
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  • #3
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Okay well I tried this:
Say X is a vector in M.

Then Y is a vector in M perp if Y ● X = 0.

Z is a vector in M perp perp if Z ● Y = 0.

But I can show that every vector X in M is also in M perp perp:
X ● Y = 0 because Y ● X = 0.

Therefore M is a subset of M perp perp. Is that part right?
 
  • #4
Office_Shredder
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Yeah, that's all there is to it for that part
 
  • #5
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Okay now for the next part.
T(X) = projM(X) = Y
dim(M) + dim (M perp) = n

S(Y) = projM perp(Y) = Z
dim(M perp) + dim(M perp perp) = n

Then n - dim(M) = n - dim(M perp perp)
dim (M) = dim(M perp perp)

Since M is a subset of M perp perp then they are equal?
 
  • #6
Office_Shredder
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I'm not sure precisely what your projection notation is supposed to say but the argument basically goes like how you posted:

For all subspaces U, dim(U)+dim(U perp)=n

So dim(M)+dim(M perp)=n

M perp is a subspace also, so dim(M perp)+dim(M perp perp)= n

And then subtract like you did to finish it off
 
  • #7
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Hmm okay, so the projection thing isn't actually necessary?

Thanks for your help, by the way!
 

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