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For matrix $A$ let $\lambda$ be eigenvalue corresponding to eigenvector $v$.

$$Av = \lambda v = \frac{c}{1} \lambda \frac{1}{c} v = \mu \frac{1}{c} v,$$

where $c$ is some real number. It is clear that if $c \neq 1$, then $\mu \neq \lambda$. This would imply that the eigenvalue can be any real number and it is only the direction of $v$ that matters? Without normalization we could not talk about the greatest eigenvalue?

EDIT: $$\mu \frac{1}{c} v = \mu u = Av = cAu$$ thus $Au \neq \mu u$, if $c \neq 1$.

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  • $\begingroup$ Nope, because $\mu \frac{1}{c}= \lambda$ $\endgroup$
    – Arnaldo
    Dec 14, 2016 at 13:54
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    $\begingroup$ You have $Av=\mu \frac vc$ and not $Av=\mu v$ or $A\frac vc=\mu \frac vc.$ Thus, if $c\ne 1$ you can't say that $\mu$ is an eigenvalue. $\endgroup$
    – mfl
    Dec 14, 2016 at 13:56

2 Answers 2

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That is not true because your "new" eigenvalue is $\mu \frac{1}{c}$ (not $\mu$) which keep beeing $\lambda$.

$\mu \frac{1}{c}=\lambda$

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    $\begingroup$ My idea was that there would be unambiguous eigenvector, so even though this is true it was not the answer I was looking for. I made the answer to my edit. $\endgroup$
    – user3644640
    Dec 14, 2016 at 14:06
  • $\begingroup$ So, are you good now? $\endgroup$
    – Arnaldo
    Dec 14, 2016 at 14:22
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You wrote $Av = \mu \frac 1 c v$, hence by definition of an eigenvector $$A \frac 1c v = \frac 1c\mu\frac 1c v ,$$ so the eigenvalue is $\frac \mu c=\lambda$.

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