How does solving this system yield to such results?

Question

$$\begin{cases} (1-\lambda_i)x+y = 0 \\x-\lambda_iy = 0 \end{cases} \iff \begin{pmatrix}x\\y\end{pmatrix}\in\left\langle\begin{pmatrix}\lambda_i\\1\end{pmatrix} \right\rangle$$ I have tried to solve it in various manners and none yields to that result for x and y.

Answer 1

Transform your system into the form eigenvalue-eigenvector :

$$\underbrace{\begin{pmatrix}1&1\\1&0\end{pmatrix}}_M\begin{pmatrix}x\\y\end{pmatrix}=\lambda_i\begin{pmatrix}x\\y\end{pmatrix}$$

Matrix $M$ has characteristic equation

$$\lambda^2-\lambda-1=0,\tag{1}$$

Thus, its eigenvalues are

$$\lambda_1=\underbrace{\frac{1+\sqrt{5}}{2}}_{\text{Golden ratio} \ \Phi}\ \ \text{and} \ \ \lambda_2=\frac{1-\sqrt{5}}{2}=1-\Phi$$

These eigenvalues being distinct, the dimensions of their eigenspaces is 1. Thus you can drop out the brackets ($\langle$ and $\rangle$) : you have simplified your issue into the plain checking that :

$$\begin{pmatrix}\lambda_i\\1\end{pmatrix} \ \text{is an eigenvector associated with} \ \lambda_i$$

for $i=1,2$.

Let us check it for $i=1$, i.e., for $\lambda_i=\Phi$ :

We must have (coming back to your initial system) :

$$\begin{cases} (1-\Phi)\Phi+1 = 0 \\ \Phi-\Phi \times 1 = 0 \end{cases}$$

The second equation is clearly verified. The first one is equivalent to $\Phi^2-\Phi-1=0$ which is true because $\Phi$ is a root of equation (1).

Answer 2

It appears you are trying to find eigenvalues and eigenvectors of a matrix, specifically $$\begin{pmatrix}1&1\\1&0\end{pmatrix}$$ The second equation is enough to show that the vector is proportional to your right side for any given eigenvalue $\lambda_i$. Plugging that into the first equation gives $$1+\lambda_i-\lambda_i^2=0$$ which you can solve to find $$\lambda_i=\frac {1\pm \sqrt 5}2$$