Do mathematicians prefer eigenvectors with purely integer entries? I was solving a trivial linear algebra question. 
Suppose we have $\begin{bmatrix} 1 & 3 \\ 5 & 3 \end{bmatrix}$, find all eigenvectors. 
Okay, so one of its eigenvecctor is $\begin{bmatrix} 1  \\ 5/3 \end{bmatrix}$
When I looked at the solution, the eigenvector was written as $\begin{bmatrix} 3  \\ 5 \end{bmatrix}$. At the beginning, it caused a lot of confusion and I spent hours trying to understand exactly what I am doing wrong. But now I realized people just love to "normalize" the eigenvector to purely integer (or non-fraction) entries.
Is this true? And why do people normalize these eigenvectors what is the point?
 A: You did nothing wrong: if $v$ is an eigenvector for the eigenvalue $\lambda$, then also $\alpha v$ is an eigenvector for every $\alpha\ne0$.
Since
$$
\begin{bmatrix}
3\\
5
\end{bmatrix}=
3\begin{bmatrix}
1\\
5/3
\end{bmatrix}
$$
your solution is as good as the book's.
If it were my solution, it would probably be
$$
\begin{bmatrix}
3/5\\
1
\end{bmatrix}
$$
because my method uses as free variables the ones with the higher index, so the final equation would be
$$
x_1=\frac{3}{5}x_2
$$
and the method is “set the free variable to $1$ and compute the bound variable”.
Why does your book choose $\left[\begin{smallmatrix}3\\5\end{smallmatrix}\right]$? I see no particular reason other than avoiding fractions. I'd much prefer following a standard method that doesn't put emphasis on integers. For instance, what would your book prefer between
$$
\begin{bmatrix}
\sqrt{2}-1\\
1
\end{bmatrix}
\qquad\text{and}\qquad
\begin{bmatrix}
1\\
\sqrt{2}+1
\end{bmatrix}
$$
where no “simple” choice is possible?
A: An eigenvector is simply a vector in an eigenspace, so it seems convenient to chose one with integer components when it is possible.
A: I think it's a matter of simplicity - coping with integers should just be easier.
The only form that a mathematician might prefer is the normalized eigenvector. With the usual norm definition, the solution for your example would be:
$$ \hat{x} = \frac{1}{\sqrt{34}}\begin{bmatrix}
3
\\ 
5
\end{bmatrix}$$
that is an eigenvector whose modulus is $1$.
