Eigenvectors of this matrix

Question

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.

Answer 1

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: $$\begin{cases} x_2 &= \lambda x_1,\\ -x_1 + x_2 &= \lambda x_2. \end{cases}$$