Understanding the method to find Eigenvectors For a matrix:
$
   $[A]$  =   \begin{bmatrix}
        5 & 4  \\
        1 & 2  \\
        \end{bmatrix}
$
I have the eigenvalues: $\lambda = 6, 1$
Now for each value I need to find eigenvectors:
When $\lambda = 6$
$$
\begin{bmatrix}
  5-6 & 4  \\   
  1 & 2-6  \\
  \end{bmatrix} \begin{bmatrix}
   x \\ y
   \end{bmatrix}
= \begin{bmatrix} 0 \\ 0 \end{bmatrix}
$$
$$
\begin{bmatrix}
  -1 & 4  \\   
  1 & -4  \\
  \end{bmatrix} \begin{bmatrix}
   x \\ y
   \end{bmatrix}
= \begin{bmatrix} 0 \\ 0 \end{bmatrix}
$$
Operating $R_2 \rightarrow R_2 + R_1$
$$
\begin{bmatrix}
  -1 & 4  \\   
  0 & 0  \\
  \end{bmatrix} \begin{bmatrix}
   x \\ y
   \end{bmatrix}
= \begin{bmatrix} 0 \\ 0 \end{bmatrix}
$$
So, the rank is 1 which is less than the unknowns
Take, $y = k$
$-x + 4y = 0$
$x = 4y = 4k$ 
$X = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4k \\ k \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \end{bmatrix} $ because $(k = 1)$ which are the eigenvectors.
Now, my questions are:  


*

*Why did we take $y = k$? Can we take $x$ as $k$?

*And, why $k$ was taken to be $1$? why can it not be any other value?

 A: Recall that $\lambda$ is an eigenvalue of $A$ if $\det(A-\lambda I) = 0$. This means $A - \lambda I$ is singular and therefore $$(A - \lambda I)v = 0$$ has a non-trivial solution $v \neq 0$ called an eigenvector of $A$ with respect to $\lambda$. 
With this, for any $t \in \mathbb{F}$, we have
$$(A-\lambda I) (tv) = t (A-\lambda I) v = 0$$
which shows $tv$ is also an eigenvector of $A$ with respect to $\lambda$. The same argument also tells the fact that homogenous system with any singular coefficient matrices have infinitely many solutions. In particular, if we are only interested in finding one eigenvector, we can at least set one variable as any number (which fix a particular $t$).
So back to your questions: Yes we can also take $x = k$. The only reason is that then we will have $y = \frac{1}{4}k$ which has a non-so-beautiful fraction. In this path you can in fact take $k = 4$ to get the same result.
A: An eigenvector $\vec r$ of a matrix $A$ is defined as vector which fulfils the equation
$$ A\vec r = \lambda \vec r$$
with the eigenvalue $\lambda$.
So only the direction of the vector matters. Not its lenght. Thus $\vec r'=k\vec r$ is also an eigenvalue of $A$ because.
$$ A\vec r' = k\cdot A\vec r = k\cdot \lambda \vec r=\lambda\vec r'$$
So you can use any value for $k$
A: An eigenspace is a linear subspace, so there is never just one eigenvector: you can always at least takes its nonzero multiples as well (sometimes the dimension is more than$~1$, and your eigenvectors are arbitrary combinations of more than one chosen eigenvector). The linear system you solve to find eigenvectors is homogeneous, so that its solutions form a subspace; if your eigenvalue was correctly computed, you are certain to find nonzero solutions, so there is always at least one free parameter to choose a general eigenvector; in the example this is your $k$ (you might as well continue calling it $y$, but maybe they wanted to mark the fact that $y$ was promoted from being an unknown to be a parameter). However, in the end one often wants a basis of eigenvectors, and from the parametric description of the eigenspace this is obtained by taking the parameters one by one (often, like here, there is only one parameter to take), make that parameter nonzero and any remaining parameters zero. This is why they made $k=1$ in the end. Once you get a hang of it, you might just identify which unknowns are your parameters, and without any name change directly make one of them nonzero and the others zero.
Given a linear system brought to echelon form, you can decide which columns are your pivot columns, and the remaining columns will be solution parameters. Since in the example your remaining equation $-x+4y=0$ has both coefficients nonzero, you can choose your pivot either in the first or second column, and correspondingly your parameter will be $y$ or $x$. So indeed, in the example you could have chose $x$ to be the parameter, and set it to any nonzero value of your choosing, after which solve for$~y$.
