Finding the Eigenvectors

Question

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.

Answer 1

Find eigenvalues from the characteristic polynomial

$\lambda^2-\lambda-1=(\lambda+(\sqrt(5)-1)/2)*(\lambda-(\sqrt(5)+1)/2)$

$\lambda_1=(-\sqrt(5)+1)/2$

$\lambda_2=(\sqrt(5)+1)/2$

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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 $x_2=1, v_1=\left(\begin{matrix} \frac{-\sqrt{5}+1}{2} \\ 1 \end{matrix}\right)$

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$\lambda_2=(\sqrt(5)+1)/2$

$A-\lambda_2I=\left(\begin{matrix} \frac{-\sqrt{5}+1}{2} & 1 \\ 1 & \frac{-\sqrt{5}-1}{2} \end{matrix}\right)$

$(A-\lambda I)v=0$ 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 $x_2=1, v_2=\left(\begin{matrix} \frac{\sqrt{5}+1}{2} \\ 1 \end{matrix}\right)$

Answer 2

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.