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I have a matrix:

$$ \left[ \begin{array}{ccc} 2 & 1 & 1 \\ -11 & 4 & 5 \\ -1 & 1 & 0 \\ \end{array}\right] $$

I got the eigenvalues to be $[-1,1,2]$.

For the eigenvalue $-1$, I got an eigenvector $[0,1,-1]$.

On the solutions, it says the correct answer is $[0,-1,1]$.

Is there any actual difference?

ramgorur
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Alex
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  • This looks a bit like a Q&A for an assignment. Expand it and make it more informative for the wider audience of CV. – Giuseppe Biondi-Zoccai Apr 05 '16 at 22:56
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    Yes your answer would be acceptable to anyone who knows that eigenvectors are not unique but are defined only up a to multiple. – JohnK Apr 05 '16 at 23:01
  • This is becoming a FAQ: see http://stats.stackexchange.com/search?q=eigenvector+sign – whuber Apr 05 '16 at 23:08
  • Usually for consistency, people will require that the $i$'th component (for some fixed $i$) of each eigenvector is positive, which sets a convention. This is useful in particular if you're trying to establish a reference direction with respect to that component. – Alex R. Apr 05 '16 at 23:17
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    @whuber This is indeed a frequently asked question. This thread http://stats.stackexchange.com/questions/88880 is probably a better one to use for closing duplicates. – amoeba Aug 17 '16 at 18:27
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    @Amoeba I like the existing duplicate because it explicitly is about eigenvalues and eigenvectors. It seems like a reach to redirect questions purely about that topic to answers that focus on PCA and FA. – whuber Aug 17 '16 at 18:30
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    @whuber Indeed, that is fair enough. Perhaps the link I left here in the comments will also be helpful to somebody. – amoeba Aug 17 '16 at 18:31

1 Answers1

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No, there is no difference. Notice that if $v$ is an eigenvector to $A$ with eigenvalue $\lambda$ and $\alpha$ is a scalar, then

$$ A \alpha v = \alpha A v = \lambda \alpha v $$

and thus $\alpha v$ is also an eigenvector with eigenvalue $\lambda$. Since $\alpha$ is any scalar, if you let $\alpha = -1$ then you see that $v$ being an eigenvector implies $-v$ is an eigenvector. So there is no mathematical difference between which "scaling" of the eigenvector you choose ($\alpha$ just scales the eigenvector and flips it).

Note: Normally one chooses the normalized eigenvalue (norm = 1) but even then that doesn't account for the "flipping".

Chris Rackauckas
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