I have a doubt, in a Hilbert space it is known that if I have $ \{x_k\}_{k=1}^n$ vectors in a finite space and these n elements are orthogonal then are linear independent. Is it true for a space of infinite dimension when if $(x_k, x)=0$ $\implies $ $x=0$?
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Does this answer your question? Linear independence in vector spaces of infinite dimension – Kumar Nov 08 '20 at 02:10
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2As long as the vectors are non-zero, orthogonality implies linear independence in ANY inner product space. – balddraz Nov 08 '20 at 02:49
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It is apparent that the space is an inner product space, say $H$ and that the given vectors, say $\{e_k\}_{k\in I}$ for some indexing infinite set $I$, are mutually orthogonal. To that end, then yes, it is true that $H$ is infinite-dimensional ( what, however, is generally not true —in contrast to the $n$-dimensional case—is that $H=span\{\hat{e}_k\}_{k\in I}$, where $\hat{e}_k$ is the unit vector in the direction of $e_k$; even $span\{\hat{e}_k\}_{k\in I}$ may not be isomorphic to $H$.)
Jack LeGrüß
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