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Orthogonal spanning set

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Let V be an inner product space. Show that if w is orthogonal to each of the vectors
    u1,u2,...,ur, then it is orthogonal to every vector in the span{u1,u2,...,ur}.


    2. Relevant equations



    3. The attempt at a solution
    Not sure how to show this, if w is orthogonal to every vector u1,u2,...,ur, wouldn't w already be orthogonal to the spanning set?

    The spanning set is a subspace of the subspace of the inner product space of V.

    Am I suppose to show it symbolically?
     
  2. jcsd
  3. Apr 14, 2014 #2

    jbunniii

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    Two things to consider:

    (1) What is the definition of orthogonality?
    (2) What does a general element of ##\text{span}\{u_1, \ldots,u_r\}## look like?
     
  4. Apr 14, 2014 #3
    So for orthogonality two vectors in an inner product space are orthogonal if (in this case) the vectors <u, w> = 0.

    A spanning set is just the set of all linear combinations of the vectors in V.

    So for the spanning set w = k1u1 + k2u2,...,krur

    which becomes

    0 = k1u1 + k2u2,...,krur

    so the only difference between w being orthogonal with u1,u2,...,ur is that there is some scalar k thats been thrown into the mix?
     
  5. Apr 14, 2014 #4

    jbunniii

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    OK.

    Well, the right hand side is a general element of the span, so that's good. But we want to check whether ##w## is orthogonal to this element, we don't want to set ##w## equal to this element. So to avoid confusion, let's give the general element a new name, say ##v = k_1 u_1 + \ldots + k_r u_r##. Now you want to check whether ##w## is orthogonal to ##v##. So what is ##\langle w, v \rangle##?
     
  6. Apr 14, 2014 #5
    ##\langle w, v \rangle## = ku1w+ku2w+...+kurw =0

    then after saying that, if k was the zero vector it proves that to be true
     
  7. Apr 14, 2014 #6

    jbunniii

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    No, you don't need ##k## to be the zero vector. Let's take a step back.
    $$\langle w,v \rangle = \langle w, k_1 u_1 + \ldots + k_r u_r\rangle = ?$$
    What properties of the inner product do you know that can help to simplify this? The goal is to get an expression involving ##\langle w, u_i\rangle## for ##i=1, \ldots, r##.
     
  8. Apr 14, 2014 #7
    Properties of inner products say the scalar k can be brought out as such k<w,u>

    so we could say k<w, ui> = 0 for k ε R and for i = 1, 2, ... r

    then at this point the value of k doesnt matter because i have already said that <w, ui> = 0 for i = 1, 2, ... r.

    Please correct me if Im wrong, I think im starting to see it.

    I
     
  9. Apr 14, 2014 #8

    jbunniii

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    Yes, that's right. You also need one more fact, namely:
    $$\langle w, k_1 u_1 + \ldots + k_r u_r \rangle = \langle w, k_1 u_1 \rangle + \ldots + \langle w, k_r u_r \rangle$$
    Now you can apply your step of sliding the ##k_i##'s out of the inner product, and use the fact that ##\langle w, u_i \rangle = 0## as you said.

    One small detail: in case you happen to be working with complex numbers, the ##k_i##'s get conjugated if you move them from the second position in the inner product. So:
    $$\langle w, k_i u_i \rangle = \overline{k_i} \langle w, u_i\rangle$$
    But if you move them from the first position then there is no conjugate:
    $$\langle k_i u_i, w \rangle = k_i \langle u_i, w \rangle$$
    If the ##k_i##'s are real numbers, then it doesn't make any difference, since the conjugate of ##k_i## is just ##k_i##.
     
  10. Apr 14, 2014 #9
    This is a basic linear algebra course so we only work with real values. But thank you so much for all your help, I truly understand this now!
     
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