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    http://www.maths.ox.ac.uk/system/fil...a_II_notes.pdf

    (Pg 11 Thm 3.9 in the above link)

    I am missing something here. We are given an orthonormal basis of W. How do we know that <w_i, w_i> does not equal zero always?
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    Because  \langle w_i ,w_i \rangle = \| w_i \|^2 , which is zero if and only if  w_i is zero, which cannot be the case since a basis for W cannot contain the zero vector (otherwise its elements are not linearly independent)
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    WLOG i = 1. We have that \langle , \rangle is non-degenerate on W, so \langle w_1, w \rangle is not zero for all w in W. Then write w = a_1w_1 + ... + a_nw_n for some scalars a_i (which we can do since the {w_i} form a basis), then \langle w_1, w \rangle = a_1 \langle w_1, w_1 \rangle using linearity and that \langle w_1, w_i \rangle = 0 for i not 1 (definition of orthogonality). So a_1 \langle w_1, w_1 \rangle is not always zero, so \langle w_1, w_1 \rangle \ne 0
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    (Original post by jb444)
    Because  \langle w_i ,w_i \rangle = \| w_i \|^2 , which is zero if and only if  w_i is zero, which cannot be the case since a basis for W cannot contain the zero vector (otherwise its elements are not linearly independent)
    The trouble is that \| x \| = \sqrt{\langle x,x \rangle} is usually by definition, so you're kinda begging the question.
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    Retracted my last post, just read the actual text that was cited and \langle,\rangle isn't assumed to be an inner product - it is merely a nondegenerate (at least on W) symmetric hermitian form.

    Therefore, you can't assume positive-definiteness.

    So that line of reasoning doesn't hold in any case and that is the point of posts three and four. I guess I should have checked the context before commenting!
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    Actually I didn't notice that either....I'll go change "inner product" to "form" above.



    ....or just delete them....
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    (Original post by jb444)
    Actually I didn't notice that either....I'll go change "inner product" to "form" above.



    ....or just delete them....
    Yeah, I was about to say your first post is only true for positive definite operators!



    (Original post by Glutamic Acid)
    WLOG i = 1. We have that \langle , \rangle is non-degenerate on W, so \langle w_1, w \rangle is not zero for all w in W. Then write w = a_1w_1 + ... + a_nw_n for some scalars a_i (which we can do since the {w_i} form a basis), then \langle w_1, w \rangle = a_1 \langle w_1, w_1 \rangle using linearity and that \langle w_1, w_i \rangle = 0 for i not 1 (definition of orthogonality). So a_1 \langle w_1, w_1 \rangle is not always zero, so \langle w_1, w_1 \rangle \ne 0
    Thanks!

    I think "all w in W" should be "some w in W" in:
    "We have that \langle , \rangle is non-degenerate on W, so \langle w_1, w \rangle is not zero for all (some) w in W."
    (This is the definition of <,> being non-deg on W I think?)
    Then for this w, a_1 =/= 0, as otherwise <w_1,w>=0. And then your argument follows. Would that be correct?

    I don't like how this thing in lectures was assumed true implicitly without mention, as I don't see it as obvious.
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    (Original post by twig)
    I think "all w in W" should be "some w in W" in:
    "We have that \langle , \rangle is non-degenerate on W, so \langle w_1, w \rangle is not zero for all (some) w in W."
    (This is the definition of <,> being non-deg on W I think?)
    Then for this w, a_1 =/= 0, as otherwise <w_1,w>=0. And then your argument follows. Would that be correct?

    I don't like how this thing in lectures was assumed true implicitly without mention, as I don't see it as obvious.
    Hah, I did mean 'all w in W' but that's ambiguous. But yes, <w_1,w> is not always zero, i.e. there is some w for which <w_1,w> is non-zero.
 
 
 
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