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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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  • 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. – Ian Jan 16 '16 at 23:48
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    You could separately assume $v$ is nonzero, or you could assume $v$ is an eigenvector, it is the same. – Ian Jan 16 '16 at 23:51
  • To add to Ian’s point: by definition, eigenvectors are non-zero. – amd Jan 17 '16 at 00:37

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

(sorry cant comment as of yet)

NeinDochOah
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