1. ## Eigen values?

Hi,

I am totally lost in how I am meant to solve this.

I have to find the eigen values of the matrix

$\left[\begin{array}{cc}1&1\\1&2\end{array}\right]$

I then find $det(B - \lambda I) = \left[\begin{array}{cc}1 - \lambda &1\\1&2 - \lambda \end{array}\right]$

$= (1 - \lambda )(2 - \lambda ) - (1\cdot1)$

From there it starts to get messy, I just want to know if I am doing anything wrong.

2. Originally Posted by Beard
Hi,

I am totally lost in how I am meant to solve this.

I have to find the eigen values of the matrix

$\left[\begin{array}{cc}1&1\\1&2\end{array}\right]$

I then find $det(B - \lambda I) = \left[\begin{array}{cc}1 - \lambda &1\\1&2 - \lambda \end{array}\right]$

$= (1 - \lambda )(2 - \lambda ) - (1\cdot1)$

From there it starts to get messy, I just want to know if I am doing anything wrong.
Explains very clearly how to do it here Eigenvalue algorithm - Wikipedia, the free encyclopedia

3. You want the determinant to be zero so that you will have a nontrivial solution. Just expand your determinant function and set it equal to zero. You will only have to solve a quadratic equation.

4. Carrying on with your method I get the eigen values to be $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$.

For my set of eigen vectors I get $\left[\begin{array}{c}-\frac{(1 + \sqrt{5})}{2}\\1 \end{array}\right]$ and the second to be $\left[\begin{array}{c}\frac{1 + \sqrt{5}}{2}\\1 \end{array}\right]$

Can someone confirm this?

5. Originally Posted by Beard
Carrying on with your method I get the eigen values to be $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$.

For my set of eigen vectors I get $\left[\begin{array}{c}-\frac{(1 + \sqrt{5})}{2}\\1 \end{array}\right]$ and the second to be $\left[\begin{array}{c}\frac{1 + \sqrt{5}}{2}\\1 \end{array}\right]$

Can someone confirm this?

correct