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Finite dimensional real vector space

  1. Oct 5, 2005 #1
    Hi can someone assist me with the following question?

    Q. Let V be a finite dimensional real vector space with inner product < , > and let W be a subspace of V. Then the orthogonal complement of W is defined as follows.

    [tex]
    W^o = \{ v \in V: < v,w > = 0,w \in W\}
    [/tex]

    Prove the following:

    a) [tex]W^o[/tex] is a subspace of V.
    b) [tex]W \cap W^o = \left\{ {\mathop 0\limits^ \to } \right\}[/tex]
    c) [tex]\dim W + \dim W^o = \dim V[/tex]

    My working:

    I can do the first part but the others are a problem for me.

    b) W and W^o are both subspaces of V and so they both contain the zero vector. Then their intersection also contains the zero vector. Suppose the intersection contains some non-zero vector say f. Then we must have <f,f> = 0 for some non-zero vector f. But this contradicts some inner product property which says <f,f> = 0 iff f = zero vector. So from that I conclude that [tex]W \cap W^o = \left\{ {\mathop 0\limits^ \to } \right\}[/tex].

    c) I can't think of a way to do this one. I know that dim(V) >= dim(W), dim(W_0) because any linearly independent set in V has most k elements where k is the number of vectors in a basis for V.

    Can someone help me with part c or check my answer for part b? Any help appreciated.
     
  2. jcsd
  3. Oct 5, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Your answer to (b) is completely correct.

    To do (c), choose a basis for [tex]W[/tex] and a basis for [tex]W^0[/tex].
    Show that their union is a basis for [tex]V[/tex].
     
  4. Oct 5, 2005 #3
    Ok thanks for your help HallsofIvy, I'll give that a go.
     
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