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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). 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. – 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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