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I need to be able to find the eigenvalues and eigenvectors of a matrix (sticking to $2 \times 2$ matrices at the moment).

I'm fine with finding the eigenvalues, but my answers for the eigenvectors don't match up with those on my problem sheet. However the proportion between my values seem to be the same, but I'm wondering why, and especially how, the writer of the problem sheet came up with his answers.

Matrix $A$: $$A = \left(\begin{array}{rr} 2 & -1\\ -1 & 1 \end{array}\right).$$

Eigenvalues found: $2.618$ and $0.382$ as $(A - \lambda I)X = 0$ $$\left(\begin{array}{cc} 2 - 2.618 & -1\\ -1 & 1-2.618 \end{array}\right)\left(\begin{array}{c} x\\ y \end{array}\right) = \left(\begin{array}{c} 0\\0 \end{array}\right).$$ so: $$\begin{array}{rcccl} -0.618x & - & y & = & 0\\\ -x & - & 1.618y & = & 0. \end{array}.$$ So: \begin{align*} x &= -1.618y\ y &= -0.618x. \end{align*} so this should give an eigenvector of: $$\left(\begin{array}{c} 1\\-1.618\end{array}\right)\quad\text{or}\quad \left(\begin{array}{c} 0.618\\ -1 \end{array}\right).$$ Or can the signs be reversed? $$\left(\begin{array}{c} -1\\ 1.618 \end{array}\right)\quad \text{and}\quad \left(\begin{array}{c} -0.618\\ 1\end{array}\right)\quad???$$ The eigenvector given in my answers sheet for the eigenvalue $2.618$ is: $$\left(\begin{array}{c} -0.85\\ 0.53 \end{array}\right)$$ which if you look at the proportion and take into account the fact that these numbers have been rounded seem to be at least proportional to my answers, but what I'd like to know is:

How come the answer above was given on the question sheet? Does the sign actually matter? I suspect yes, but can't see why or how one chooses.

Many thanks in advance.

Joe

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Your examples have $y=-1.618x$, not the other way around. Any non-zero multiple (including -1) of an eigenvector is an eigenvector, so you can flip the signs on both components (but not only one). As you say, the book answer is proportional to yours once you get x and y right. I don't know how they chose the normalization.

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    $\begingroup$ Nitpick: any nonzero multiple of an eigenvector is an eigenvector. $\endgroup$ – Arturo Magidin Feb 17 '11 at 16:20
  • $\begingroup$ @Arturo: good point. I'll add it. $\endgroup$ – Ross Millikan Feb 17 '11 at 18:07

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