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Suppose $V$ is a finite-dimensional complex inner product space and $v_1,v_2,...,v_3$ is an orthonormal basis of $V$. Define $A:V \to V$ by $Av_i=\lambda_i v_i$ for some $\lambda_i \in \mathbb{C}.$ Show that $A^*v_i=\bar \lambda v_i.$

My attempt:

$(Av_i,v_i)=(\lambda_iv_i,v_i)=(v_i,\bar\lambda_i v_i)=(v_i,A^*v_i)$

$\Rightarrow(v_i,(A^*-\bar \lambda_i)v_i)=0 \Rightarrow (A^*-\bar \lambda_i)v_i \perp v_i $

How to show that $(A^*-\bar\lambda_i)v_i=0$ ?

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Just extend what you have a bit: \begin{align} \langle \sum_j a_j v_j,A^*\sum_k b_k v_k \rangle = &\langle A \sum_{j}a_j v_j, \sum_{k}b_k v_k\rangle \\ = & \langle \sum_{j}\lambda_j a_j v_j,\sum_k b_k v_k\rangle \\ = &\sum_j\lambda_j a_j\overline{b_j} \\ = &\langle\sum_j a_j v_j,\sum_k \overline{\lambda_k}b_kv_k\rangle \\ \end{align}

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  • $\begingroup$ @user1561 : Like the way I have redone it? $\endgroup$ – DisintegratingByParts Nov 16 '18 at 17:01
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    $\begingroup$ Yes. $\phantom{}$+1. $\endgroup$ – user1551 Nov 16 '18 at 17:23
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Here is another method:

Let $M=[A]_\beta$, where $\beta=\{v_i\}_{1\le i\le n}$. Then $\det\,(M-\lambda I)=0\implies\det\,(M-\lambda I)^*=0$.

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