Eigenvectors of this matrix I have a matrix A
$$A = \begin{pmatrix}
0 & 1 \\
-1 & 1 
\end{pmatrix}
$$
Of which I have calculated the eigenvalues to be $\lambda = e^{\pm i \frac{\pi}{3}}$
Hence we have $\lambda_1 = e^{i \frac{\pi}{3}}$ and $\lambda_2 = e^{-i \frac{\pi}{3}}$
I have then tried to find the eigenvectors as follows:
$$
 \begin{pmatrix}
0 & 1 \\
-1 & 1 
\end{pmatrix}
 \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix}
=
 \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix}
\lambda
$$
The way I know how to go now is to 'guess' a value for $x_1$ as the length is irrelevant and the second term $x_2$ will come out as proportional and then I can just normalise it, however I run into some issues. Starting with $\lambda_1$ I guess $x_1$ to be 1:
$$
 \begin{pmatrix}
0 & 1 \\
-1 & 1 
\end{pmatrix}
 \begin{pmatrix}
1 \\
x_2
\end{pmatrix}
=
 \begin{pmatrix}
\lambda_1 \\
\lambda_1 x_2
\end{pmatrix}
$$
Which when solving gives me the following equations:
$$
0(1) +1(x_2) = \lambda_1
$$
$$
-1(1) + 1(x_2) = \lambda_1{x_2}
$$
Solving the first gives me:
$$
x_2 = \lambda_1
$$
But the second gives me:
$$
x_2 = \frac{1}{1-\lambda_1}
$$
Which has left me confused because I was under the impression that these should be the same - especially since the way I calculated the eigenvalues should mean that these are linearly coupled. I have seen some stuff online about finding RREF of the matrix however I have not been taught this and I am assuming there is an easier way or I have gone wrong somewhere. Is one of these correct? If so, why is the other not the same? Any help would be appreciated.
 A: They are the same though. Since $\lambda_1 = e^{i\frac \pi 3} = \frac 12 + i\frac {\sqrt{3}}{2}$, we have
$$\frac 1{1 - \lambda_1} = \frac 1{\frac 12 - i\frac {\sqrt{3}}{2}} = \frac 1{\frac 12 - i\frac {\sqrt{3}}{2}} \frac{\frac 12 + i\frac {\sqrt{3}}{2}}{\frac 12 + i\frac {\sqrt{3}}{2}} = \lambda_1.$$
As a side note, be careful with the method of taking $x_1 = 1$, it may fail when the first coordinates of the eigenvectors are 0. It is safer to solve the linear system resulting from the equation for the eigenvectors:
$$\begin{cases}
x_2 &= \lambda x_1,\\
-x_1 + x_2 &= \lambda x_2.
\end{cases}$$
A: Your two answers for $x_2$ are equivalent: start by converting $\lambda_1$ to Cartesian using Euler's formula. $$\lambda_1 = e^{i\frac{\pi}{3}} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
Now let's simplify your other form for $x_2$.
$$x_2 = \frac{1}{1-\lambda_1} = \frac{1}{1 - {\frac{1}{2} + i\frac{\sqrt{3}}{2}}} = \frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}}$$
Multiplying the numerator and denominator by the conjugate we get:
$$\frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} \cdot \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = \lambda_1$$
Furthermore, notice that the original equation $\lambda_1 = \frac{1}{1 - \lambda_1}$ is equivalent to $\lambda_1(1 - \lambda_1) - 1=0$, and this is likely the same equation you solved to determine your eigenvalues, $\lambda_{1,2}$.
