Eigen values ....again :(

**Beard** · April 26th 2010, 05:53 AM

Hi,
Here is the question that I am stuck on
$B = \left[\begin{array}{cc}1&1\\1&2\end{array}\right]$
Construct the matrix X that reduces B to diagonal form through the transformation
$D = X^{-1}BX.$ Confirm that D is diagonal, that the diagonal elements of D are the
eigenvalues of B and that det(B) = det(D).

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

I then get to the point where $\lambda^{2} - 3\lambda + 1$ and therefore my eigen values are $\lambda = \frac{3 \pm \sqrt{5}}{2}$ and my eigen vectors as $\left[\begin{array}{c}-\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right]$ and $\left[\begin{array}{c}\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right]$ .

The matrix X is made from the eigen values of B. So $det(X) =\left[\begin{array}{cc}-\frac{(1 + \sqrt{5})}{2}&\frac{(1 + \sqrt{5})}{2}\\1&1\end{array}\right]$ . I get as being $-1 -\sqrt{5}$

Now I have to find the cofactor of X and then transpose of the cofactor in order to eventually get the inverse of X.

$X_{cof} = \left[\begin{array}{cc}1&-1\\ \frac{(-1 -\sqrt{5})}{2}&\frac{-1 -\sqrt{5})}{2} <br /> \end{array}\right]$

$X_{cof}T = \left[\begin{array}{cc}1&\frac{-1 -\sqrt{5})}{2}\\ -1&\frac{-1 -\sqrt{5})}{2} <br /> \end{array}\right]$

$X^{-1} = \frac{1}{det(X)} \cdot X_{cof}T$ which I get as $\left[\begin{array}{cc}\frac{1 -\sqrt{5}}{4}&\frac{1}{2}\\ \frac{\sqrt{5} - 1}{4}&\frac{1}{2}\end{array}\right]$

I now have all the components required to find D.

So when I multiply them all together i get $\left[\begin{array}{cc}0&\frac{\sqrt{5} - 1}{2}+1\\ 0&1 +\sqrt{5}\end{array}\right]$

Which is evidently wrong. Can anyone see roughly where i could have gone wrong and then i will post a more detailed process of what i was doing at that stage.

Thanks for reading and any help that come with it

**Opalg** · April 26th 2010, 11:28 AM

Originally Posted by Beard

... my eigen values are $\lambda = \frac{3 \pm \sqrt{5}}{2}$ and my eigen vectors as $\left[\begin{array}{c}-\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right]$ and $\left[\begin{array}{c}\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right]$ .

I agree with the eigenvalues and the first eigenvector. But I think you'll find that the second eigenvector is $\begin{bmatrix}\frac{-1+\sqrt5}2\\1\end{bmatrix}$ , not $\begin{bmatrix}\frac{1+\sqrt5}2\\1\end{bmatrix}$ .

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