Is it acceptable to have a fraction in an eigenvector? The professor teaching a class I am taking wants me to find the eigenvalues and the eigenvectors for the following matrix below. 
$$\begin{bmatrix}-5 & 5\\4 & 3\end{bmatrix}$$
I have succeeded in getting the eigenvalues, which are $\lambda= \{ 5,-7 \}$.
When finding the eigenvector for $\lambda= 5$, I get $\begin{bmatrix}1/2\\1 \end{bmatrix}$. However, the correct answer is $\begin{bmatrix}1\\2 \end{bmatrix}$ .
I have tried doing this question using multiple online matrix calculators. One of which gives me  $\begin{bmatrix}1/2\\1 \end{bmatrix}$, and the other gives me $\begin{bmatrix}1\\2 \end{bmatrix}$.
The online calculator that gave me $\begin{bmatrix}1\\2 \end{bmatrix}$ explains, that y=2, hence $\begin{bmatrix}1/2·2\\1·2 \end{bmatrix} = \begin{bmatrix}1\\2 \end{bmatrix}$. 
What I do not understand is, why is y must equal to 2?Is it because there cannot be a fraction in an eigenvector?
 A: 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.
A: $$\begin{bmatrix}-5 & 5\\4 & 3\end{bmatrix}\begin{bmatrix}\frac12\\1 \end{bmatrix}=\begin{bmatrix}\frac52\\5\end{bmatrix}=5\begin{bmatrix}\frac12\\1 \end{bmatrix}$$
so that $\begin{bmatrix}\frac12\\1 \end{bmatrix}$ is undisputably an Eigenvector associated to the Eigenvalue $5$.
A: 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 $\alpha \neq 0$.
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 $\lambda$. In other words, 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.
A: Calculations with whole numbers are easier than fractions, so people often take eigenvectors with whole numbers for convenience.
For example, $(1/2,1)$ and $(1,2)$ both normalize to the same vector, but the calculations give you 
$$\left(\frac{1/2}{\sqrt{5/4}},\frac1{\sqrt{5/4}}\right)\text{ and }\left(\frac1{\sqrt5},\frac2{\sqrt5}\right)$$
