# $Av=\lambda v \Rightarrow A^*v=\bar \lambda v$ (general case)

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 $$(A^*-\bar\lambda_i)v_i=0$$ ?

• I have the same confusion. Do you find the answer now? Jun 16, 2020 at 5:03
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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}
• Yes. $\phantom{}$+1. Nov 16, 2018 at 17:23
Let $$M=[A]_\beta$$, where $$\beta=\{v_i\}_{1\le i\le n}$$. Then $$\det\,(M-\lambda I)=0\implies\det\,(M-\lambda I)^*=0$$.