# Does the sign of eigenvectors matter? [duplicate]

I had an matrix ((2,1,1),(-11,4,5),(-1,1-0)) I got the eigen values to be -1,1,2 for the eigenvalue -1 I got an eigenvector (0,1,-1) on the answers it says the answer is (0,-1,1). Is there an actual difference?

## marked as duplicate by whuber♦Apr 5 '16 at 23:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• This looks a bit like a Q&A for an assignment. Expand it and make it more informative for the wider audience of CV. – Joe_74 Apr 5 '16 at 22:56
• 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 5 '16 at 23:01
• @whuber This is indeed a frequently asked question. This thread stats.stackexchange.com/questions/88880 is probably a better one to use for closing duplicates. – amoeba Aug 17 '16 at 18:27
• @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
• @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 Answer

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".