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I solved this homework question and when I compared my solution to the textbook solution, the eigenvector is slightly different.

I got $[0,\frac{1}{2},1]$

The book says: $[0,1,2]$

It's basically the same vector multiplied by two. Can you do that? Thanks.

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    Yes, an eigenvector is only unique upto multiplication by a constant. Think about how it's defined. It solves an equatuon $Av = \lambda v$. What happens if you change $v$ to say, $2v$? Would it still solve it?2012-12-10
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    True, now I realize it was a very stupid question. Thanks :)2012-12-10
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    @Ahsan: why not add that as an answer?2012-12-10
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    @Ahsan, up to multiplication by a *nonzero* constant.2012-12-10

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This is a community wiki answer intended to remove this question from the unanswered queue.


As Ahsan noted in the comments, an eigenvector (in a 1-dimensional eigenspace) is only unique up to multiplication by a non-zero constant.

If the eigenspace is more than 1-dimensional, there are even more possible answers. In general every basis of the eigenspace is a possible solution to such a problem.