Let $(e\_j)\_{j\in\mathbb N}$ be an orthonormal basis of a Hilbert space $V$. Let $T:V\to V$ be continuous, selfadjoint linear operator. 
Assume that for all $i,j\in\mathbb N$ the number $\langle Te\_i,e\_j\rangle$ is rational.
Let $\lambda\in\mathbb C$. Suppose there exists an eigenvector $v\in V$ with $Tv=\lambda v$.
The question is this: Does there exist an eigenvector $w$ for the eigenvalue $\lambda$ such that $\langle w,e\_j\rangle$ lies in the field ${\mathbb Q}(\lambda)$ for every $j\in\mathbb N$?


This is true for finite dimensional spaces, even without the self-adjointness, since $\lambda$ is a zero of the characteristic polynomial.


If it turns out to be false as stated, you may add further conditions on $T$, like the one that it preserves the linear span of the basis vectors.



Algebraicity of Eigenvectors in a Hilbert space
The answer is no.


**Update.** Here is an explicit example (which also preserves the $\mathbb Q$-span of the basis). Let $V=\ell\_2$ over $\mathbb R$ and $(e\_j)$ the standard basis. Define $T:V\to V$ by
$$
\begin{aligned}
 (Tx)\_1 &= x\_1+3x\_2+x\_3 \\
 (Tx)\_2 &= 3x\_1+10x\_2+6x\_3+x\_4 \\
 (Tx)\_n &= x\_{n-2}+6x\_{n-1}+11x\_n+6x\_{n+1}+x\_{n+2}, \qquad n>2 .
\end{aligned}
$$
Let $\alpha$ be the root of $p(x)=x^2+3x+1$ such that $|\alpha|<1$. Then the vector $v\in\ell\_2$ defined by $v\_n=\alpha^n$ satisfies $Tv=0$, so it is an eigenvector for $\lambda=0$. On the other hand, no nonzero rational vector $w$ satisfies $Tw=0$ because the relation $Tw=0$ implies the recurrence relation
$$
 w\_{n+4}=-6w\_{n+3}-11w\_{n+2}-6w\_{n+1}-w\_n ,
$$
which does not have rational solutions tending to 0.


**Below is the original answer.**


Let $V=\ell\_2$ over $\mathbb R$ and $(e\_j)$ the standard basis, so that $\langle x,e\_j\rangle=x\_j$ for every $x\in\ell\_2$. Let $v\in V$ be a unit vector with irrational ratios of coordinates (for definiteness, take a geometric progression like $(c\pi^{-n})\_{n\in\mathbb N}$). Then there exist a continuous self-adjoint $T:V\to V$ with rational coordinates such that $\ker T=\langle v\rangle$. So we have $\lambda=0$ but no $\lambda$-eigenvector is rational.


I represent $T$ by its (infinite) matrix $(t\_{ij})\_{i,j\in\mathbb N}$, $t\_{ij}=\langle Te\_i,e\_j\rangle$. This matrix should be symmetric, and continuity of the resulting operator should be taken care of.


Begin with a self-adjoint $A$ such that $\ker A=\langle v\rangle$ with irrational components $a\_{ij}$. For example, let $A$ be the projection to the orthogonal complement of $v$, so $a\_{ij}=\delta\_{ij}-v\_iv\_j$. I am going to approximate the matrix $(a\_{ij})$ by a rational matrix $(t\_{ij})$ such that the kernel stays the same.


Since $Av=0$, each row of our matrix is orthogonal to $v$. Approximate the first row $a\_i=(a\_{1i})$ by a rational vector $t\_1=(t\_{1i})$ such that $\langle t\_1,v\rangle=0$ and $|t\_{1i}-a\_{1i}|$ is bounded by a rapidly decaying geometric progression (see below for details). Replace the first row and the first column of the matrix by this approximation. Then adjust the diagonal elements $a\_{ii}$ so that the rows remain orthogonal to $v$, namely change $a\_{ii}$ to $a\_{ii}-(a\_{1i}-t\_{1i})v\_1v\_i^{-1}$. Recall that $v\_i$ is a not-so-fast decaying geometric progression, so this adjustment is a small change.


Now remove the first row and the first column from the matrix, and the first component from $v$. Apply the same procedure to the truncated data: namely approximate the first remaining row and the first remaining column by a rational vector whose scalar product with $v$ is the same as before, then adjust the diagonal elements so that the scalar products of all rows with $v$ are preserved. Repeat ad infinitum. Note that every element of the original matrix is changed only finitely many times, and $v$ belongs to the kernel of the matrix after each step.


The approximations are controlled and the adjustments are bounded in terms of approximations, so the sum $\sum (a\_{ij}-t\_{ij})^2$ can be made arbitrarily small. This implies that $T$ is continuous and $\|T-A\|$ is small (say, less than 1/2), so we have $Tv=0$ and $\|Tw\|\ge\frac12\|w\|$ for all $w$ orthogonal to $w$. Therefore $\ker T=\langle v\rangle$. However $v$ cannot be rescaled to a rational vector.


It remains to show that every vector $w\in\ell\_2$ can be approximated by a rational vector $w'$ such that $\langle w',v\rangle=\langle w,v\rangle$ and $w'\_i-w\_i$ is bounded by a small, fast decaying progression. Let $w'\_1=w\_1+\varepsilon\_1$ be a rational approximation of $w\_1$, then let $w'\_2=w\_2-\varepsilon\_1v\_1v\_2^{-1}+\varepsilon\_2$ be a rational approximation of $w\_2-\varepsilon\_1v\_1v\_2^{-1}$, then let $w'\_3=w\_3-\varepsilon\_2v\_2v\_3^{-1}+\varepsilon\_3$ be a rational approximation of $w\_3-\varepsilon\_2v\_2v\_3^{-1}$, and so on. The $\varepsilon\_i$ at each step can be chosen arbitrarily small, and the resulting vector $w'$ satisfies $\langle w'-w,v\rangle=0$.


