1
$\begingroup$

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 $H$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.