Is it acceptable to have a fraction in an eigenvector?

By definition, an eigenvalue $\lambda$ and one corresponding eigenvector $v$ must satisfy the following equation:

$$Av = \lambda v.$$

Now, consider the vector

$$w = \alpha v,$$

where $\alpha$ is a non null number.

Then, notice that:

$$Aw = A(\alpha v) = \alpha (Av) = \alpha (\lambda v) = \lambda (\alpha v) =\lambda w.$$

Therefore:

$$Aw = \lambda w,$$

and hence $w$ is another eigenvector associated to the eigenvalue $\lambda$.

In general, it is not true that there is only one eigenvector associated to the eigenvalue $\lambda$. Instead, there is a linear subspace, also known as the eigenspace associated to $\lambda$. Informally, there are infinitely many eigenvectors to $\lambda$, which belong to a certain eigenspace. Given one eigenvector (say $v$), then all the multiples of $v$ except for $0$ (i.e. $w = \alpha v$ with $\alpha \neq 0$) are also eigenvectors.

There are matrices with eigenvectors that have irrational components, so there is no rule that your eigenvector must be free of fractions or even radical expressions. As an example:

$$\begin{bmatrix}1 & 2\\1 & 1\end{bmatrix}$$

Which has eigenvalues of $1 \pm \sqrt{2}$ and eigenvectors of $\begin{bmatrix}\pm\sqrt{2} \\ 1\end{bmatrix}$

(Additionally, because of the unsolvability of the quintic, there are even eigenvectors that cannot be expressed in elementary functions.)

Because of the form of the equations that you solve to get the eigenvectors, you have infinite solutions to the eigenvectors. Additionally, an eigenvector is only really valuable as a direction. So if any eigenvector can be said to be the "correct" or "most special" one, it's the one that has a norm of 1, or a norm of the associated eigenvalue. However, these can be actually a bit ugly to express.

For our example they are: norm 1, $\begin{bmatrix}\pm\sqrt{\frac{2}{3}} \\ \sqrt{\frac1{3}}\end{bmatrix}$ norm $\lambda$: $\frac1{3}\begin{bmatrix}\mp 2\sqrt{3} \pm\sqrt{6} \\ \sqrt{3} \pm \sqrt{6} \end{bmatrix}$.

That being said, it is, of course, nicer to work with whole numbers when the opportunity arises.

$$\begin{bmatrix}-5 & 5\\4 & 3\end{bmatrix}\begin{bmatrix}\frac12\\1 \end{bmatrix}=\begin{bmatrix}\frac52\\5\end{bmatrix}=5\begin{bmatrix}\frac12\\1 \end{bmatrix}$$

so that $\begin{bmatrix}\frac12\\1 \end{bmatrix}$ is undisputably an Eigenvector associated to the Eigenvalue $5$.