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I'm working on a problem that requires the following operator, $A^TA$, to have an orthonormal set of eigenfunctions.

Note $A:H_1 \mapsto H_2$, where $H_1$ and $H_2$ are separable Hilbert spaces. Furthermore, $A$ is also continuous, linear and injective. In addition, $A^T$ denotes the adjoint of $A$.

Could anyone point me towards a theorem or something that might help? Thanks.

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  • $\begingroup$ Does $A^T$ denote the adjoint of $A$? Then $A^TA$ is a positive definite self-adjoint operator. $\endgroup$ – Daniel Fischer Jul 25 '15 at 13:19
  • $\begingroup$ I realize this, but how can I use this property to prove the above? $\endgroup$ – user127159 Jul 25 '15 at 13:36
  • $\begingroup$ The spectral theorem is pretty useful in that situation. $\endgroup$ – Daniel Fischer Jul 25 '15 at 13:44
  • $\begingroup$ Yes, but the spectral theorem requires compactness, and I'm sure that in this case, compactness may be too strong an assumption. $\endgroup$ – user127159 Jul 25 '15 at 13:56
  • $\begingroup$ No, there is a version of the spectral theorem that works for all normal (even unbounded) operators. But indeed it does not imply that the there is an orthornomal basis of eigenvectors. One sufficient conditions is that $A^T A$ has compact resolvent, i.e. $(A^T A+1)^{-1}$ is compact. $\endgroup$ – MaoWao Jul 26 '15 at 15:33

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