# Ambiguity of the Eigenvalue

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 $c \neq 1$.

• Nope, because $\mu \frac{1}{c}= \lambda$ Dec 14, 2016 at 13:54
• 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.
– mfl
Dec 14, 2016 at 13:56

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