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!
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 $v=0.$