Eigen values?

**Beard** · April 22nd 2010, 06:37 AM

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.

**piglet** · April 22nd 2010, 07:54 AM

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

**jass10816** · April 22nd 2010, 08:20 AM

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.

**Beard** · April 23rd 2010, 06:42 AM

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?

**harish21** · April 23rd 2010, 06:53 AM

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

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