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Math Help - orthonomal set

  1. #1
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    orthonomal set

    Hi,
    how can one shows that an orthonomal set in an inner product space is automatically an independent set?
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  2. #2
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    Quote Originally Posted by mtmath View Post
    Hi,
    how can one shows that an orthonomal set in an inner product space is automatically an independent set?

    Let \{v_1,\ldots,v_n\} be n orthonormal vectors , and suppose \sum^n_{k=1}a_kv_k=0\,,\,\,a_i\in\mathbb{F}= the linear space's definition field. If we denote by <,> the inner product, we get:

    \forall j=1,2,\ldots,n\,,\,\,0=<0,v_j>=<\sum^n_{k=1}a_kv_k  ,vj>=\sum^n_{k=1}a_k<v_k,vj>= \sum^n_{k=1}a_i\delta_{k,j}=a_j ...and we're done.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Let \{v_1,\ldots,v_n\} be n orthonormal vectors , and suppose \sum^n_{k=1}a_kv_k=0\,,\,\,a_i\in\mathbb{F}= the linear space's definition field. If we denote by <,> the inner product, we get:

    \forall j=1,2,\ldots,n\,,\,\,0=<0,v_j>=<\sum^n_{k=1}a_kv_k  ,vj>=\sum^n_{k=1}a_k<v_k,vj>= \sum^n_{k=1}a_i\delta_{k,j}=a_j ...and we're done.

    Tonio
    thanks a lot Tonio,

    so this simply tells that the summation equals zero when all a's are zeros, hence the so mention set is not linearly dependant?
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  4. #4
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    Yes, and since it is "not linearly dependent", it is linearly independent.
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  5. #5
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    Quote Originally Posted by mtmath View Post
    thanks a lot Tonio,

    so this simply tells that the summation equals zero when all a's are zeros, hence the so mention set is not linearly dependant?

    Well, any such summation is zero when all the coefficients are zero......what you wanted to prove, and we achieved, is that only then the sum is zero, or in other words: if the sum is zero then all the coefficients equal zero. This, according to definition, means the vectors are lin. independient .

    Tonio
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