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.
cyclic vector exists for symmetric operator iff there no repeated eigenvalues
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functional-analysis
operator-theory
hilbert-spaces
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1It would be helpful if you comment on what have you attempted and tried so far. – 2012-12-14
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0Up 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? – 2012-12-14
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1If $H$ is of finite dimension, then "dense" is a strange term to use; $H$ itself is the only dense subspace. – 2012-12-14
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1Is 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? – 2012-12-14