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Let $A\colon H\to H$ be a compact, self-adjoint operator on a Hilbert space $H$. Then, the spectrum theorem says that for every $x\in H$

$$Ax = \sum_{k=1}^{\infty} \lambda_k\langle x,e_k \rangle e_k$$ where $\lambda_k$ are nonzero eigenvalues and $e_k$ are the corresponding orthonormal eigenvectors.

Assume that the range of A is dense.

Why does this imply that $\left\{e_k\right\}$ is a complete orthonormal System of $H$, i.e. an orthonormal Basis?

That $\left\{e_k\right\}$ is an orthonormal System is clear, but I do not see the completeness which would mean that the linear span of $\left\{e_k\right\}$ is dense in $H$.

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  • $\begingroup$ You asked this a couple of days ago, and accepted an answer. Why do you ask it again? $\endgroup$ – Daniel Fischer♦ May 10 '17 at 20:13
  • $\begingroup$ Because now I concentrate only on the dense range part. $\endgroup$ – Rhjg May 10 '17 at 20:14
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You assumed that the image of $A$ is dense. The image of $A$ is contained in the closure of the (linear) span of the $e_k$, from your formula, so $\{e_k\}$ is dense.

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  • $\begingroup$ Why is the Image of $A$ the closure of the linear span of the $e:k$ and what do you exactly mean with linear span of the $e_k$? Maybe I should ask this first. For me, linear span means the set of all linear combinations of the $e_k$. $\endgroup$ – Rhjg May 10 '17 at 20:16
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    $\begingroup$ Unless $H$ is finite-dimensional, the image of $A$ will not be closed. But the image of $A$ is contained in the closure of the span of the $e_k$. $\endgroup$ – Daniel Fischer♦ May 10 '17 at 20:20
  • $\begingroup$ Do you mean that $\text{Image}(A)=\left\{\sum_{k=1}^{\infty}\lambda_k\langle x,e_k\rangle e_k: x\in H\right\}\subset \text{span}\left\{e_k: k=1,2,...\right\}\subset \overline{\text{span}\left\{e_k: k=1,2,...\right\}}$? $\endgroup$ – Rhjg May 10 '17 at 20:24
  • $\begingroup$ @DanielFischer You're right -- thanks for pointing that out. Will edit. $\endgroup$ – Lorenzo Najt May 10 '17 at 20:24
  • $\begingroup$ Ah, maybe from $H=\overline{\text{Image}(A)}$ and my last comment one can conclude that $H=\overline{\text{Image}(A)}\subset\overline{\text{span}\left\{e_k: k=1,2,...\right\}}\subset H$ and hence we can replace the $\subset$ by $=$ and get $H=\overline{\text{span}\left\{e_k: k=1,2,...\right\}}$? $\endgroup$ – Rhjg May 10 '17 at 20:33

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