Finding the Eigenvectors Consider the matrix given below:
$\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}.$
The eigenvalues for this matrix are
$\dfrac{1+\sqrt 5}{2},\dfrac{1-\sqrt 5}{2}.$
I am facing trouble finding the eigenvectors. Please help.
 A: You have an eigenvalue $\phi=\dfrac{1+\sqrt5}2$.
An eigenvector will satisfy $A\begin{bmatrix}x \\y \end{bmatrix}=\phi\begin{bmatrix}x \\y \end{bmatrix}$; i.e., $\begin{bmatrix}x+y\\x\end{bmatrix}=\begin{bmatrix}\phi x \\ \phi y \end{bmatrix}.$ 
Solutions of this will be multiples of each other.  
We could take $y=1$.  Then $x=\phi y=\phi$.  So an eigenvector is $\begin{bmatrix}\phi  \\  1 \end{bmatrix}.$
All the eigenvectors for this eigenvalue are given by $\begin{bmatrix}c\phi  \\  c\end{bmatrix}.$
Approach for the other eigenvalue is similar.
A: Find eigenvalues from the characteristic polynomial
$\lambda^2-\lambda-1=(\lambda+(\sqrt(5)-1)/2)*(\lambda-(\sqrt(5)+1)/2)$
$\lambda_1=(-\sqrt(5)+1)/2$
$\lambda_2=(\sqrt(5)+1)/2$

For every λ we find its own vector(s):
$\lambda_1=(-\sqrt(5)+1)/2$
$A-\lambda_1I=\left(\begin{matrix}
\frac{\sqrt{5}+1}{2} & 1 \\
1 & \frac{\sqrt{5}-1}{2}
\end{matrix}\right)$
$Av=\lambda v$ 1
$ \Rightarrow (A-\lambda I)v=0$
So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination: 2
$\left(\begin{matrix}
\frac{\sqrt{5}+1}{2} & 1 & 0 \\
1 & \frac{\sqrt{5}-1}{2} & 0
\end{matrix}\right)$
$\begin{matrix}
x_1 & +\frac{\sqrt{5}-1}{2}*x_2 & = & 0
\end{matrix}$
General Solution: $X=\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2}*x_2 \\
x_2
\end{matrix}\right)$
Let $x_2=1, v_1=\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2} \\
1
\end{matrix}\right)$

$\lambda_2=(\sqrt(5)+1)/2$
$A-\lambda_2I=\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2} & 1 \\
1 & \frac{-\sqrt{5}-1}{2}
\end{matrix}\right)$
$(A-\lambda I)v=0$ 1
So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination: 2
$\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2} & 1 & 0 \\
1 & \frac{-\sqrt{5}-1}{2} & 0
\end{matrix}\right)$
$\begin{matrix}
x_1 & -\frac{\sqrt{5}+1}{2}*x_2 & = & 0
\end{matrix}$
General Solution: $X=\left(\begin{matrix}
\frac{\sqrt{5}+1}{2}*x_2 \\
x_2
\end{matrix}\right)$
Let $x_2=1, v_2=\left(\begin{matrix}
\frac{\sqrt{5}+1}{2} \\
1
\end{matrix}\right)$
