1. ## Determine eigenvector

Hey guys,

i am supposed to determine the eigenvalues and eigenvectors of:
$\begin{pmatrix}1 & 1 \\1 & 0\\ \end{pmatrix}$

I calculated the eigenvalues with the formula:

$det(\lambda I - A) = 0$

to be $\lambda_1 = \frac{1}{2}(1 + \sqrt 5)$ and $\lambda_2 = \frac{1}{2}(1 - \sqrt 5)$

To get the eigenvectors i need to solve the following equation, right?

$\begin{pmatrix}\lambda - 1 & -1 \\-1 & \lambda\\ \end{pmatrix}\cdot\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$

Now i get x = 0 and y = 0, but thats just not right says Wolfram Alpha and the exercise text.

What is wrong?

Thanks, Inf

2. Originally Posted by Inf
To get the eigenvectors i need to solve the following equation, right?

$\begin{pmatrix}\lambda - 1 & -1 \\-1 & \lambda\\ \end{pmatrix}\cdot\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$

No, you have to solve the systems $(A-\lambda_iI)\begin{pmatrix}x \\ y \end{pmatrix}=\begin{pmatrix}0\\ 0 \end{pmatrix} \quad (i=1,2)$

3. To find eigenvectors you solve simultaneous equations for:
$\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \begin{matrix} x \\ y \end{matrix} = \lambda \begin{matrix} x \\ y \ \end{matrix}$

4. if λ is an eigenvalue for A, then Av = λv, for some eigenvector v. which means that

(A - λI)v = 0. this is the system you have to solve (for each value of λ). so for λ1 = (1/2)(1 + √5) this would be:

[(1/2)(1 - √5) ...........1........][x]....[0]
[.......1.......... (-1/2)(1 + √5)][y] = [0], which leads to:

y = (-1/2)(1 - √5)x.

if we choose (as apparently wolfram|alpha) did, y = 1, this makes x = (1/2)(1 + √5).

you should be able to compute the 2nd eigenvector, now.

5. Thank you guys!

I thought both equations have to be solved at the same time.

Now I did it like you told me and get this:

$p_1 = \begin{pmatrix}\frac{1 +\sqrt 5}{2}\\1\\\end{pmatrix}$
$p_2 = \begin{pmatrix}\frac{1 -\sqrt 5}{2}\\1\\\end{pmatrix}$
$P = \begin{pmatrix}\frac{1 +\sqrt 5}{2} & \frac{1 -\sqrt 5}{2}\\1 &1\\\end{pmatrix}$

And this perfectly solves the equation: $P^{-1} \cdot A \cdot P = D$

Thank you guys!
Greetings, Inf