# How to find eigenvectors

Good afternoon,

I have some problem finding the eigenvectors for matrix $$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$$

I found eigen values $$\mu=\frac{\sqrt{5}+1}{2}$$ and $$\lambda=\frac{-\sqrt{5}+1}{2}$$. But when I try to find eigenvectors I am blocked with the system : \begin{align*} &\begin{cases} y = \mu x\\ x + y = \mu y \end{cases} \\ \iff &\begin{cases} y = \mu x\\ x +(1-\mu) \mu x = 0 \end{cases}\\ \iff &\begin{cases} y = \mu x\\ x(-\mu^2 + \mu +1)=0 \end{cases} \end{align*}

But $$(-\mu^2 + \mu +1)=0$$ so I don't know how to continue. I wonder if x could be any value (1 for example) then I deduce y, but it doesn't work. Did I make a mistake? Thanks you for your help .

• You say it doesn't work. Could you be note detailed as to how you can tell it didn't work? Dec 4, 2021 at 21:57

You made no mistake, and the part that you said "It doesn't work" has to do with finding $$x$$. You would have expected that an $$x$$ could be solved for from the second equation: $$x\cdot 0 = 0$$. This equation has infinitely many solutions $$x$$. And the eigenvector corresponding to this eigenvalue is of the form $$(c,\mu c), c\neq 0$$.
Since $$-\mu^2+\mu+1=0$$, the second equation is just $$0=0$$. All that remains is the equation $$y=\mu x$$. So, an eigenvector corresponding to the eigenvalue $$\mu$$ is $$(1,\mu)$$. (I took $$x=1$$, as you suggested. Any number different from $$0$$ will do.)
You can check that it works by computing$$\begin{bmatrix}0&1\\1&1\end{bmatrix}.(1,\mu)-\mu(1,\mu).$$It is equal to $$(0,1+\mu-\mu^2)=(0,0)$$. So, the method works.
• @Arthur I have added more details. But it seemed to me that the OP got confused by the fact that the second equation became $0=0$. Dec 4, 2021 at 22:03
• Actually I was confused with 0x=0 and that $1+\mu=\mu^2$ but everything was finally right.Thanks you very much Dec 5, 2021 at 8:55