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I want to show that if $A$ is normal then

$$ A v = \lambda v \implies A^* v = \bar{\lambda} v $$

I can show that $A^*v$ is also an eigenvector of $A$, using the fact that $A$ and $A^*$ commute, but I know that this doesn't imply $A^* v \propto v$. So I'm not sure where to go from here. Thanks.

EDIT: The top answer solves the problem, but I would like to know how to prove this result using the polarisation identity as suggested below. This is not the approach taken in the other questions on this site, linked to in the comments below.

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marked as duplicate by Jonas Meyer, user147263, muaddib, Ivo Terek, A.P. Jul 24 '15 at 21:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $(\operatorname{span} \{v\})^\perp$ is $A$-invariant, i.e. $w\perp v \implies Aw \perp v$. $\endgroup$ – Daniel Fischer Jul 24 '15 at 14:11
  • $\begingroup$ Actually that's what I'm trying to show using the above! How do you prove that? $\endgroup$ – gj255 Jul 24 '15 at 14:29
  • $\begingroup$ On second thoughts, it is in fact easier the other way round. $\endgroup$ – Daniel Fischer Jul 24 '15 at 14:39
  • $\begingroup$ math.stackexchange.com/q/879787, math.stackexchange.com/q/436318 $\endgroup$ – Jonas Meyer Jul 24 '15 at 15:26
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For a normal operator $A$, we have $\lVert Aw\rVert = \lVert A^\ast w\rVert$ for all $w$. Now, assuming that $v$ is an eigenvector to the eigenvalue $\lambda$ of $A$, we find

$$\langle A^\ast v, v\rangle = \langle v, Av\rangle = \langle v, \lambda v\rangle = \overline{\lambda}\langle v,v\rangle.\tag{1}$$

Further, we have

$$\lvert \langle A^\ast v, v\rangle\rvert \leqslant \lVert A^\ast v\rVert\cdot \lVert v\rVert = \lvert\lambda\rvert\cdot \lVert v\rVert^2\tag{2}$$

by the Cauchy-Bun'akovskij-Schwarz inequality. By $(1)$, we have the equality case, hence $A^\ast v$ and $v$ are linearly dependent, i.e. $A^\ast v = \mu\cdot v$ for some $\mu \in \mathbb{C}$. But then $(1)$ immediately yields $\mu = \overline{\lambda}$.

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    $\begingroup$ If we are using your first line, then we can note that $A-\lambda I$ is also normal, hence if $Av=\lambda v$ then $\|(A-\lambda I)^*v\|=\|(A-\lambda I) v\|=0$. $\endgroup$ – Jonas Meyer Jul 24 '15 at 15:08
  • $\begingroup$ So true. That's better. $\endgroup$ – Daniel Fischer Jul 24 '15 at 15:12
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Or you do what Daniel did $\langle A^\ast v, v\rangle = \langle v, Av\rangle = \langle v, \lambda v\rangle = \overline{\lambda}\langle v,v\rangle.\tag{1}$ and by polarization you have what you want.

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  • $\begingroup$ What do you mean 'by polarisation'? $\endgroup$ – gj255 Jul 24 '15 at 14:50
  • $\begingroup$ @gj255 en.wikipedia.org/wiki/Polarization_identity this identity is very very useful! You just show it for $<Ax,x>=...$ and you have it for $<Ax,y>=...$ $\endgroup$ – Haha Jul 24 '15 at 15:09

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