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What I am trying to figure out is how they got the negative sign in the red circle, because from my calculations (blue circle) I get the first one to negative but the second one, that is the red circled one, is positive; I don't think normalizing would change the sign. What am I doing wrong? I just don't see how the red-circled one can be negative and not the one above it. Can someone shed some light on what I am doing wrong, or thinking for that matter?

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Note that eigenvectors are not unique. Both $(-1,1)$ and $(\frac1{\sqrt2}, -\frac1{\sqrt2})$ are eigenvectors of your matrix to $\lambda = 3$.

In fact if $Ax = \lambda x$, then $Ay = \lambda y$ for any scalar multiple $y$ of $x$. Here $y = -\frac1{\sqrt2}x$ if $y$ is "their" eigenvector and $x$ is "yours".

The convention by wich they arrived at their eigenvectors is to enforce $x_1 \ge 0$ and $\|x\| = 1$ for every eigenvector $x$.

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  • $\begingroup$ So, from what I understand, it's safe to assume that even eigenvalue of 1 has an eigenvector where both elements are negative? $\endgroup$ – Dick Armstrong Aug 17 '15 at 19:49
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    $\begingroup$ @DickArmstrong Sure. You can even name one of them: $$-1\cdot\pmatrix{\frac1{\sqrt2}\\\frac1{\sqrt2}} = \pmatrix{-\frac1{\sqrt2}\\-\frac1{\sqrt2}}$$ $\endgroup$ – AlexR Aug 17 '15 at 19:51
  • $\begingroup$ that just reminded me of a previous example and after fanatically looking for it in my notes I found it, I had the following bases, s(transpose(-0.5 1)) which was the same thing as s(transpose(1 -2)). Thank you! $\endgroup$ – Dick Armstrong Aug 17 '15 at 19:56
  • $\begingroup$ @DickArmstrong Indeed. BTW If you want to typeset vectors, have a look at the \pmatrix code from my comment. $\endgroup$ – AlexR Aug 17 '15 at 19:57
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    $\begingroup$ Allright. In case you don't have them already: here, here, here and here are posts on this site. $\endgroup$ – AlexR Aug 17 '15 at 20:05

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