Finding Fibonacci matrix's eigenvectors. But I'm algebraic disaster. stucking at calculation.

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Likewise title, I was calculating eigenvectors of Fibonacci matrix. I set the Fibonacci matrix to be $$ \begin{pmatrix} f_{n+1}\\ f_n \end{pmatrix} = \begin{pmatrix} 1&1\\ 1&0 \end{pmatrix} \begin{pmatrix} f_n\\ f_{n-1} \end{pmatrix} $$ $$ \begin{pmatrix} 1&1\\ 1&0 \end{pmatrix} =M $$ And by determinant computation, I get the eigenvalues $\lambda = (1\pm\sqrt5)/2$ (I set plus one to $\phi$ and minus one to $\psi$).

And I chose to find the eigenvectors through brute computation searching the kernel space of matrix $(M-\lambda I)$. $$ \begin{pmatrix} 1-\phi & 1 \\ 1&-\phi \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix} $$ Which is expressed as above. I though this computation would give me solution for $x$ and $y$. However, instead I got $$ (1-\phi)x+y=0 $$ $$x-\phi y=0 $$ which led to $$(1-\phi+1/\phi)x=0$$ which gave me nothing but faint.

I think I'd faced this kind of error during algebra I'm not sure last time experience was also about Fibonacci or not. However I remember at that time I missed some operation which seems adding no information but still making equation solvable (such as dividing lefthand-side to righthand-side, ...).

I know that there are eigenvectors and How they look like. What I am interested in is how to achieve. I searched a bit and I found a youtube course. but he just guessed out the vector like mathmagician which is not what im looking for. https://www.youtube.com/watch?v=1wPEB2XDMJo (It's around 19:50~) Please help me Hyper math bois..

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If $\lambda$ is any of the eigenvalues, it is clear that $(\lambda,1)$ is a solution of the equation $x-\lambda x=0$. Is it also a solution of the equation $(1-\lambda)x+y=0$? Yes, because this is the same thing as asserting that$$(1-\lambda)\lambda+1=0,$$or$$\lambda^2-\lambda-1=0.\tag1$$And the eigenvalues are the roots of the equation $(1)$.

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When you get to

$(1-\phi+(1/\phi))x=0$

you're right on target. The coefficient of $x$ on the left side is zero because your eigenvalue equation is $\phi^2-\phi-1=0$ and the coefficients just $1-\phi+(1/\phi)=-(\phi^2-\phi-1)/\phi$. So your equation really says "Put anything you want for $x$", then once you've chosen you solve for $y$.