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Is it possible to find a selfadjoint compact operator $T$ on a Hilbert space such that there is a vector $v$ which does not belong to any finite dimensional invariant subspace (for $T$) ?

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This will not be a full solution, but here is an idea which you could try to extend and verify, perhaps it will get you going. Let us take the Hilbert space $X = l^2(\mathbb{N})$ of square-summable sequences of complex numbers, and define the operator $T:X \to X$ which takes a sequence $(x_n)$ to $(w_nx_n)$, for some fixed choice of complex numbers $w_n$. To ensure that $T$ is self-adjoint we will need to have the $w_n$ be real numbers. To ensure compactness we will need the $w_n$ to tend to zero as $n \to \infty$. Let us fix the choice $(w_n) = (1/n)$.

Take for example the vector $x = (1/n)$. The motivation for choosing this one is that infinitely many (actually all) of its coordinates are non-zero. Then certainly $x \in X$, and $T^kx = (1/n^{k+1})$. I strongly believe that this choice of $x$ will produce a counter-example. If $x$ were to be contained in a finite-dimensional $T$-invariant subspace, then the subspace generated by the vectors $x, Tx, T^2x, \ldots$ will be finite-dimensional. In particular, for some $k \geq 0$ I should be able to express $T^kx$ as a linear combination of the vectors $x, Tx, T^2x, \ldots, T^{k-1}x$. Such an expression will give me a solution to a certain infinite system of linear equations. Now check the link below.

https://en.wikipedia.org/wiki/Vandermonde_matrix

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    $\begingroup$ Thank you for your answer. I am not seeing the direct link with the Vandermonde matrix, but I can conclude to a contradiction thanks to your linear combination. Since $T^k x = \sum_{i=0}^{k-1} \lambda_i T^i x$ implies the following impossible equality of sequences : $\sum_{i=1}^{k} \lambda_i n^{k-i+1}=1$. $\endgroup$ – user488906 Oct 16 '17 at 15:43
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As long as the dimension of the Hilbert space is larger than one, this is possible:

On $\mathbb R^2$, take the operator $$ T=\pmatrix{1&0\\0&2}. $$ Then the vector $\pmatrix{1\\1}$ is not in an invariant subspace of $T$.

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  • $\begingroup$ well yes it is in $\mathbb{R}^2$, which is invariant by $T$ and finite dimensional. In other words : the only way to find an example here is on an infinite dimensional Hilbert space. $\endgroup$ – user488906 Oct 16 '17 at 14:02

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