# cyclic vector exists for symmetric operator iff there no repeated eigenvalues

Considering a symmetric operator $A$ acting on a finite dimensional Hilbert space $H$, we say $x\in H$ is a cyclic vector for $A$ if the set of finite linear combinations of $\{A^n x:n=0,1,2,...\}$ is equal to $H$. I am looking for a proof of whether $A$ must have a cyclic vector iff $A$ has no repeated eigenvalues. Hints are welcomed.

-
It would be helpful if you comment on what have you attempted and tried so far. – Javier Álvarez Dec 14 '12 at 12:55
Up to isomorphism you can assume that the Hilbert space is $\mathbb{C}^n$ and the operator is represented by a diagonal matrix $$\begin{bmatrix} \lambda_1 && \\ & \ddots & \\ &&\lambda_n \end{bmatrix}$$This way we are reduced to a problem in linear algebra. Have you tried this already? – Giuseppe Negro Dec 14 '12 at 13:09
If $H$ is of finite dimension, then "dense" is a strange term to use; $H$ itself is the only dense subspace. – Marc van Leeuwen Dec 14 '12 at 13:10
Is your Hilbert space real or complex? When you say that $A$ is symmetric do you mean that $A$ is self-adjoint or that some matrix representing is equal to its (regular, non-conjugate) transpose? – user38355 Dec 14 '12 at 15:39

I will assume that we are talking a real Hilbert space. Everything works the same in the complex case if we use "hermitian" instead of "symmetric".

Note that, from your definition, the condition "$A$ has a cyclic vector $x$" is equivalent to $$\tag{1} H=\{p(A)x:\ p\in\mathbb R[t]\}.$$ If we fix an orthonormal basis for $H$, the fact that $A$ is symmetric guarantees that $A$ is diagonalizable, i.e. $A=VDV^T$, with $V$ orthogonal and $D$ diagonal containing the eigenvalues of $A$ (counting multiplicities) in its diagonal.

As $p(A)=Vp(D)V^T$, condition $(1)$ translates into $$\tag{2} H=\{p(D)y:\ p\in\mathbb R[t]\},$$ where $y=V^Tx$.

The key observation is that $\{p(D):\ p\in\mathbb{R}\}$ is a vector space (since $\mathbb R[t]$ is), and that $$\tag{3} \dim\{p(D):\ p\in\mathbb{R}\}=\text{the number of distinct eigenvalues of }A.$$ This can be seen from the fact that $p(D)$ is the diagonal matrix with diagonal $p(D_{11}),\ldots,p(D_{nn})$, and that given distinct numbers $\lambda_1,\ldots,\lambda_k$, the set $\{(p(\lambda_1),\ldots,p(\lambda_k)):\ p\in\mathbb R[t]\}$ equals $\mathbb R^k$

Now, if we have equality in $(2)$, then the set on the right has dimension $n$, and by $(3)$ we get that $A$ has $n$ distinct eigenvalues.

Conversely, if $A$ has $n$ distinct eigenvalues, then choosing $y$ to be the vector with all its coordinates equal to one (i.e. the sum of the unit eigenvectors for $A$), we get that the set on the right-hand-side on $(2)$ has dimension $n$, and so it equals all of $H$.

-