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# Linear Algebra Orthogonal Projections Watch

1. 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?
2. Because , which is zero if and only if is zero, which cannot be the case since a basis for W cannot contain the zero vector (otherwise its elements are not linearly independent)
3. WLOG i = 1. We have that is non-degenerate on W, so is not zero for all w in W. Then write for some scalars a_i (which we can do since the {w_i} form a basis), then using linearity and that for i not 1 (definition of orthogonality). So is not always zero, so
4. (Original post by jb444)
Because , which is zero if and only if 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 is usually by definition, so you're kinda begging the question.
5. Retracted my last post, just read the actual text that was cited and 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!
6. Actually I didn't notice that either....I'll go change "inner product" to "form" above.

....or just delete them....
7. (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 is non-degenerate on W, so is not zero for all w in W. Then write for some scalars a_i (which we can do since the {w_i} form a basis), then using linearity and that for i not 1 (definition of orthogonality). So is not always zero, so
Thanks!

I think "all w in W" should be "some w in W" in:
"We have that is non-degenerate on W, so 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.
8. (Original post by twig)
I think "all w in W" should be "some w in W" in:
"We have that is non-degenerate on W, so 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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Updated: April 6, 2013
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