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We have the following theorem

Let $T:X\rightarrow Y$ be a compact operator between the Hilbert spaces $X,Y$. Then there exist (possibly finite) orthonormal bases $\{e_1,e_2,\ldots \}$ and $\{f_1,f_2,\ldots \}$ of $X,Y$ and (possibly finite) numbers $s_1,s_2,\ldots$ such that $s_n \rightarrow 0$ (if there are countably many numbers), such that $ Tx=\sum\limits_{n=1}^{\infty}s_n \leftf_n $

This is the singular value decomposition for operators between Hilbert spaces.

My questions are: If $X$ and $Y$ are finite dimensional then the theorem, as stated above is true? If $X$ is finite dimensional, both orthonormal bases become finite, since then we work in $T(X)$ which is a finite dimensional subspace in the (possible infinite dimensional) space $Y$.

But what happens, if $X$ is infinite dimensional, but $Y$ is not? Are the orthonormal bases in that case still finite?

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    Several missing words make this question difficult to understand. Please proofread it sometime soon. Thanks!2012-06-28

1 Answers 1

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If one space is infinite-dimensional and the other is finite-dimensional, you have finite orthonormal sets, but obviously not an orthonormal basis of the infinite-dimensional one. Similarly if $X$ and $Y$ are finite-dimensional but with different dimensions, you don't get a basis of the larger one. Even in the case where $X=Y$, it's not at all clear to me that $\{e_i\}$ and $\{f_i\}$ can both be taken to be bases, rather than just orthonormal sets.