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Let $A$ be a $2\times2$ matrix: $$ \begin{bmatrix} 1 & 1\\ 1 & 0\\ \end{bmatrix}. $$ I found the eigen values $$\lambda_1=1-\sqrt{5} \\\text{and} \\ \lambda_2=1+\sqrt{5}.$$ But for some reason when I try to find the eigenvectors I keep getting that $$\begin{bmatrix} 1-(1-\sqrt{5}) & 1 \\ 1 & -1+\sqrt{5}\\ \end{bmatrix}$$ reduces to $$\begin{bmatrix} \sqrt{5} & 1 \\ 0 & -4+\sqrt{5}\\ \end{bmatrix}$$ but then that would mean my eigenvector would be $$V_1 = \begin{pmatrix} 0 \\ 0\\ \end{pmatrix}.$$ Which cant be true.

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    $\begingroup$ The roots should be halved. You forgot the "over $2a$" portion of the quadratic formula. $\endgroup$ – Herng Yi Feb 12 '13 at 2:25
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$$\lambda_1=\frac{1}{2}\sqrt{5}+\frac{1}{2}$$ $$\lambda_2=\frac{1}{2}-\frac{1}{2}\sqrt{5}$$

Because

$$ \begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix} $$

$$A-\lambda I= \begin{bmatrix} 1-\lambda & 1 \\ 1 & -\lambda \\ \end{bmatrix} $$ $$\det(A-\lambda I)=(1-\lambda)(-\lambda)-1\cdot1$$ $$=\lambda^2-\lambda-1$$

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