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Dimension vs Orthogonality

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data
    If {u1, u2,...,um} are nonzero pairwise orthogonal vectors of a subspace W of dimension n, prove that [tex]m \leq n[/tex].

    3. The attempt at a solution

    I look at all my notes but I still can't understand what this qurstion asks or what definitions I need to be using for this... I'm stuck and appreciate some guidance so I can get started. Thank you.
  2. jcsd
  3. Apr 8, 2009 #2

    matt grime

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    Well, what is n? It's the dimension. So we're asking if a set of mutually orthogonal vectors can have cardinality greater then n. Can it? Hmm, what is dimension? It is the cardinality (size) of a maximally linearly independent set. Does that help? (I.e. you should now think: if I can show that the set is ________ then by what's gone before it must have at most n elements.)
  4. Apr 9, 2009 #3
    Do you mean I have to prove that the set is linearly independent? A set of mutually orthogonal non-zero vectors is always linearly independent.

    OK so the vectors {u1, u2,...,um} are in Rm and they are orthogonal (i.e uj · ui=0 if i =/= j).

    Let's take a linear combination of the vectors in this set that gives the zero vector:

    k1u1+k2u2+...+kmum = 0

    I just need to show that there's only one value for all the constants, k1 = k2 = … = km = 0.

    If I take the dot product of both sides of the equation with u1:

    u1 · (k1u1+k2u2+...+kmum) = 0 · u1

    Factoring out the constant.

    k1(u1 · u1) + k2(u1 · u2) +...+km(u1 · um) = 0

    k1(u1)2 + k2(0) +...+km(0) = 0

    k1=0 & all scalar coefficients are zeros and therefore vectors are independent.

    I showed that the vectors in the set are linearly independent but I can't see exactly how this is useful in showing that [tex]m \leq n[/tex]...
    Could you please provide me with more explanation of what I need to do next?
  5. Apr 9, 2009 #4

    matt grime

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    I stated the definition of dimension in my first post: it is the maximum size of *any* set of linearly independent vectors. You have a set of m linearly independent vectors. You know that *by definition* the maximum size a set of linearly independent vectors can have is n.
  6. Apr 9, 2009 #5
    It proves that [tex]m \leq n[/tex], end of proof? Is what I've done sufficient to write as a proof or do I need to also further prove the definition of dimension that you stated?
  7. Apr 10, 2009 #6

    matt grime

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    Yes it obviously proves m<=n. Was that really a question? Do I need to explain it more?

    What I used above is the definition of dimension. What other one do you have?
  8. Apr 10, 2009 #7
    OK, thank you very much for your help Matt. :biggrin:
  9. Apr 11, 2009 #8

    matt grime

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    Seriously, are you using a different definition? You could be using the minimal size of a spanning set as the definition, so you probably ought to show that this agrees with maximal size of a linearly independent set. Or whatever definition you're using (I can't think of another elementary one).
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