Eigenvalues of compact self adjoint operator that is cyclic Let $T: H \to H$ be an arbitrary compact self adjoint operator, and $H$ a Hilbert space. I am asked to prove that $T$ being cyclic is equivalent to all of the eigenvalues of $T$ having multiplicity one.
I believe it might be easier to check the $\Rightarrow$ implication first. I know that $T$ has the property that there is an othonormal basis $\{e_i\}_{i \in I}$ of $H$ consisting of eigenvectors of $T$. I also know that eigenvalues are real and of finite multiplicity. We must check that if $Te_i = \lambda_i e_i$ and $Te_j = \lambda_j e_j$ then $\lambda_i \neq \lambda_j$ for all $i \neq j$.
Since $T$ is cyclic, there is some $v \in H$ such that $e_i=\sum_k \mu_kT^{n_k}v$ and similarly $e_j=\sum_k \theta_sT^{n_s}v$. Can someone point out the contradiction that it will follow if $\lambda_i=\lambda_j$ using these facts? I have tried using self adjointness to reach some contradiction using the inner products of some expressions, but I am not clear about how to express my goal in terms of it. What other way would you guys recommend? Thank you!
 A: In the first part we will use the fact that for a self-adjoint operator if  $Tx=\lambda x$ and $y\perp x$ then $Ty\perp x.$
Let $v$ is a cyclic vector of $T.$ Assume for a contradiction that an eigenvalue $\lambda$ is of multiplicity greater than $1.$ Therefore there are two vectors $e$ and $f$ such that $$
Te=\lambda e,\ Tf=\lambda f,\ e\perp f,\ \|e\|=\|f\|=1$$
Let $$u=\langle v,e\rangle \,e+\langle v,f\rangle \,f $$
Then $$v-u\perp e,\quad v-u\perp f$$ Moreover $u\neq0$ as otherwise $T^nv\perp e,f$ for any $n\ge 0.$
Let $$w=\langle f,v\rangle \,e-\langle e,v\rangle \,f $$ Then $w\neq 0$ ( because $u\neq 0)$ and $$Tw=\lambda w,\qquad v-u\perp w, \quad u\perp w$$
We have $$T^nv=T^n(v-u)+T^nu$$ As $T$ is self-adjoint we get
$$T^n(v-u)\perp w, \qquad T^nu\perp  w$$ Thus $T^nv\perp w$ for any $n\ge 0,$ i.e. $v$ is not a cyclic vector.
Conversely, assume that all eigenvalues of $T$ are single.
Let $$ v=\sum_{k=1}^\infty 2^{-k} e_k$$ We claim that $v$ is a cyclic vector for $T.$ Indeed, we have $$T^nv=\sum_{k=1}^\infty 2^{-k} \lambda_k^ne_k$$ Assume that there is $v\in \mathcal{H}$ orthogonal to $T^nv$ for any $n\ge 0.$ Then $$\sum_{k=1}^\infty 2^{-k}\,\langle v,e_k\rangle\,\lambda_k^n =0, \qquad n\ge 0 \qquad (*)$$
Consider the set $$ K=\{\lambda_k\,:\, k\ge 0\}\cup \{0\}$$ Then $K$ is compact. By the Stone-Weierstrass theorem the polynomials are uniformly dense in $C(K),$ the space of continuous (at $0$) complex-valued  functions on $K.$ Let $f(\lambda_k)=\langle v,e_k\rangle $ and $f(0)=0.$ Then $f\in C(K).$ By $(*)$ we have $$\sum_{k=1}^\infty f(\lambda_k)\,\lambda_k^n\,2^{-k}=0,\qquad n\ge 0$$ Hence $$\sum_{k=1}^\infty f(\lambda_k)\,p(\lambda_k)\,2^{-k}=0$$ for any polynomial $p(x).$ Take a sequence $p_m(x)$ of polynomials convergent to $\bar{f}$ uniformly on $K.$ Then $$\sum_{k=1}^\infty |f(\lambda_k)|^2\,2^{-k}=\lim_{m\to \infty}\,\sum_{k=1}^\infty f(\lambda_k)\,p_m(\lambda_k)\,2^{-k}=0$$ Thus $f(\lambda_k)=0,$ i.e. $\langle v,e_k\rangle =0$ for $k\ge 0.$ Hence $v=0.$
