# Signs in the orthonormal bases

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?

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 $x$.
• @DickArmstrong Sure. You can even name one of them: $$-1\cdot\pmatrix{\frac1{\sqrt2}\\\frac1{\sqrt2}} = \pmatrix{-\frac1{\sqrt2}\\-\frac1{\sqrt2}}$$ – AlexR Aug 17 '15 at 19:51
• @DickArmstrong Indeed. BTW If you want to typeset vectors, have a look at the \pmatrix code from my comment. – AlexR Aug 17 '15 at 19:57