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I just searched a handout written by the professor in Cornell. However, I feel that there is some problem in it. enter image description here

First, if $V$ is infinite-dimensional, then if $W$ is infinite-dimensional, does $W$ necessarily have an orthonormal basis(related question)? Even if $W$ has an orthonormal basis, say $\beta$, then how can we "extend the orthnormal basis for $W$ to $V$"? Although we can choose "one" of a vector $v\in V-W$, and we know that $v\cup\beta$ is linearly independent, how can we do the Gram-Schmidt process, turning the $v\cup\beta$ to an orthonormal set? Here the $\beta$ is infinite! Does my doubt make sense? And can this proof be modified to be correct?

Eric
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  • This is true if and only if $f|_U$ is non degenerate. – Alberto Andrenucci Sep 12 '17 at 17:01
  • At some point in your course did your instructor say "from now on all vector spaces are finite-dimensional"? – Randall Sep 12 '17 at 17:07
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    The course is MATH 2310 at Cornell, which is introductory linear algebra. Almost certainly will all vector spaces be finite-dimensional. – Michael L. Sep 12 '17 at 17:09
  • Thank you all. Suppose if $V$ is infinite-dimesional and $U$ is finite-dimensional. Is the proof valid? It seems that the theorem itself is valid, but the proof involving take orthonormal basis for "$W$" which is again can be infinite-dimensional right? – Eric Sep 12 '17 at 17:10

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This is not valid for infinite-dimensional spaces. If $U$ is dense in $V$ then $U^\perp=\{0\}$, so a dense but proper subspace gives a counterexample.

Angina Seng
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  • Thanks. "This is not valid for infinite-dimensional spaces." Do you refers to $U$ or $V$? – Eric Sep 12 '17 at 17:04