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Euge

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Euge

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- #2

Office_Shredder

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##x_1,...,x_N## and ##y_1,...,y_N## span ##N## dimensional spaces ##X## and ##Y##. Every element of ##Y## is in the span of all the x's and is almost orthogonal to the infinite dimensional space in the last paragraph, so is well approximated by an element in ##X##. Since ##X## and ##Y## have the same dimension, every element of ##X## is well approximated by an element of ##Y## as well.

If the y's do not span the space, there is some vector orthogonal to them. But that vector cannot be both almost orthogonal to ##X## and also almost orthogonal to the span of the other x's, which means it cannot be orthogonal to both ##Y## and the span of the other y's.

- #3

Paul Colby

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I've given a counter example. Could be missing something simple (as is usual for me).

Okay, I've edited this. It's more in line with what I was intending. It's still wrong because ##y_1 \notin H##.

The statement is false as the following counter example shows. Let ##z_n## be an orthonormal set of vectors where ##n\in\mathbb{N}##. Let, $$x_n = z_1,z_3,z_4,\cdots$$ Then let $$y_n = z_2, z_3,z_4, \cdots$$ Clearly, both sets ##x_n## and ##y_n## define orthogonal sets. ##\sum_1^\infty \|| x_n - y_n||^2 = 2 < \infty## since every term is 0 except for the first term which is 2.

The statement is false as the following counter example shows. Let ##z_n## be an orthonormal set of vectors where ##n\in\mathbb{N}##. Let, $$x_n = z_1,z_3,z_4,\cdots$$ Then let $$y_n = z_2, z_3,z_4, \cdots$$ Clearly, both sets ##x_n## and ##y_n## define orthogonal sets. ##\sum_1^\infty \|| x_n - y_n||^2 = 2 < \infty## since every term is 0 except for the first term which is 2.

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Office_Shredder

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- #5

Paul Colby

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I was right! I did miss something simple

- #6

mathman

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Some ##y_n=0##?

- #7

Paul Colby

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##y=0## isn’t normalSome ##y_n=0##?

- #8

Paul Colby

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For the case ##M < 1##, ##U## has an inverse on ##H##. Let ##X=1-U##. The sum, $$\sum_{n=0}^\infty X^n = \frac{1}{1-X} = U^{-1},$$ converges and defines the inverse operator on all of ##H##. The ##|y_n\rangle## therefore span ##H##.

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