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Suppose $\lambda\in K$ and $T\in L(V)$ where $V$ is finite dimensional inner-product space over $K$ .Prove that prove that $\lambda$ is eigenvalue of $T$ if and only if $\bar{\lambda}$ is eigenvalue for $T^*$

$<Tv,u> = <\lambda v,u> = \bar\lambda<v,u>=<v,\bar\lambda u>$

how to solve i am not geeing any idea

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2 Answers 2

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The proof comes down to showing that if $T$ has an eigenvector, then so does $T^*$:

Suppose that $v$ is an eigenvector of a linear operator $T$ on a finite dimensional inner-product space $K$ with corresponding eigenvalue $\lambda$. Then, for arbitrary vector $x \in K$,

$0=<0,x>=<(T-\lambda I)(v),x)>=<v,(T-\lambda I)^*, (x))> = <v,(T^* -\bar{\lambda} I),(x)>$

From here, notice that $v$ is orthogonal to the range of $(T^* -\bar{\lambda} I)$.

What can you say from here about the kernel?

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  • $\begingroup$ ...not understand $\endgroup$ Commented May 3, 2018 at 3:47
  • $\begingroup$ Which part are you having trouble with? $\endgroup$ Commented May 3, 2018 at 3:50
  • $\begingroup$ ..other part if $\bar{\lambda}$ is eigenvalue for $\lambda$ is eigenvaluefor t $\endgroup$ Commented May 3, 2018 at 3:57
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$$\langle\,T^*u, v\,\rangle = \overline{\langle\,Tv, u\,\rangle} = \overline{\langle\,\lambda v, u\,\rangle} = \overline{\bar{\lambda}\langle\,v, u\,\rangle} = \lambda \langle\,u,v\,\rangle = \langle\, \bar{\lambda}u,v\,\rangle $$ Going the other way: $$\langle\,Tv, u\,\rangle = \overline{\langle\,T^*u, v\,\rangle} = \overline{\langle\,\bar{\lambda} u, v\,\rangle} = \overline{\lambda\langle\,u, v\,\rangle} = \bar{\lambda} \langle\,v,u\,\rangle = \langle\, \lambda v,u \,\rangle $$

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  • $\begingroup$ @adfriedman...i have faced problem with other part $\endgroup$ Commented May 3, 2018 at 3:54
  • $\begingroup$ You just have to reverse all the steps to get the other direction, the philosophy is the same. $\endgroup$
    – user284331
    Commented May 3, 2018 at 3:55
  • $\begingroup$ Hopefully my addition makes it clear. $\endgroup$
    – adfriedman
    Commented May 3, 2018 at 4:08
  • $\begingroup$ Are you sure this answer is correct? I think it is incomplete, because we need $v$ to be an eigenvector. $\endgroup$ Commented Jun 17, 2020 at 17:28
  • $\begingroup$ @DaviBarreira sure, as to match the OPs notation, I am assuming $v$ is an eigenvector of $T$ corresponding to $\lambda$ in the first direction. In the other direction we are assuming $\bar{\lambda}$ is an eigenvalue corresponding to eigenvector $u$ of $T^*$. $\endgroup$
    – adfriedman
    Commented Jun 17, 2020 at 17:34

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