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In this question and answers to such question people say that $v$ cannot be $\vec{0}$, but this not correct in my opinion. We assume it is different from $\vec{0}$ because otherwise $Av = \lambda v$ would be trivially true, right?

So, instead of saying "but this is a contradiction because $v$ cannot be $\vec{0}$" we should say "but this is a contradiction because, by assumption, $v \neq \vec{0}$", right?

I just wanted to clarify this point.

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  • $\begingroup$ If you assume $v$ is an eigenvector and conclude that $v=0$ then you have derived a contradiction and can conclude that $v$ is not an eigenvector. $\endgroup$
    – Ian
    Commented Jan 16, 2016 at 23:48
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    $\begingroup$ You could separately assume $v$ is nonzero, or you could assume $v$ is an eigenvector, it is the same. $\endgroup$
    – Ian
    Commented Jan 16, 2016 at 23:51
  • $\begingroup$ To add to Ian’s point: by definition, eigenvectors are non-zero. $\endgroup$
    – amd
    Commented Jan 17, 2016 at 0:37

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As it states already in the question, thats the contradiction...

(sorry cant comment as of yet)

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