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After an eigenvector computation for PCA, I get the eigenvector $(a, b)$ where $a=b$. This means that if I choose $a=1$, then also $b$ would be equal to $1$, and if I choose $a=2$, $b$ would also be equal to $2$. So the vectors $(1, 1)$, $(2, 2)$, $(3, 3)$, etc. would all be valid as they are pointing in the same direction. I can use any of those vectors to compute my principal component in that direction.

My issue is now as follows: What if I had chosen $a=-1$, then I would also choose $b=-1$. So the vector $(-1, -1)$ would also be considered valid. But, because of the negative, I would instead generate a principal component facing the opposite direction. This would completely change my PCA result.

My question then is: How do we handle these negatives in PCA? They seem to produce a completely different result relative to the positive vectors.

amoeba
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Minaj
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  • First of all, it's true that $(1,1)$, $(2,2)$ and any $(a,a)$ is a valid eigenvector, in PCA we usually use eigenvectors normalized to the unit length, i.e. $(1/\sqrt{2}, 1/\sqrt{2})$, because only then the projection on this direction is given by simple multiplication with this eigenvector ([see here](http://stats.stackexchange.com/questions/177099)). Regarding the sign issue, yes, it can be positive or negative, it does not matter. The sign of all PCs is arbitrary. This has been discussed before, [please see here](http://stats.stackexchange.com/questions/88880). – amoeba Oct 21 '15 at 15:14
  • This issue crops up in interesting places. At http://stats.stackexchange.com/questions/34396, for instance, this flexibility in representing principal components was exploited to track an entire time series of PCAs. – whuber Oct 21 '15 at 15:31

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