0
$\begingroup$

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 .

$\endgroup$
1
  • $\begingroup$ You say it doesn't work. Could you be note detailed as to how you can tell it didn't work? $\endgroup$
    – Arthur
    Dec 4, 2021 at 21:57

2 Answers 2

0
$\begingroup$

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 $(c,\mu c), c\neq 0$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ @Arthur I have added more details. But it seemed to me that the OP got confused by the fact that the second equation became $0=0$. $\endgroup$ Dec 4, 2021 at 22:03
  • $\begingroup$ Then addressing the inherent homogeneity of eigenvector equations in general could also be a good idea. $\endgroup$
    – Arthur
    Dec 4, 2021 at 22:05
  • $\begingroup$ Actually I was confused with 0x=0 and that $1+\mu=\mu^2$ but everything was finally right.Thanks you very much $\endgroup$ Dec 5, 2021 at 8:55
  • $\begingroup$ If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ Dec 5, 2021 at 9:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .