Suppose $\lambda$ is a eigenvalue of linear map $T$ if and only if $\bar{\lambda}$ is eigenvalue of the $T^{*}$ with no additional special condition

Question

Suppose $T$ is a linear operator on the $V$($V$ is a real or complex inner product space), $\lambda$ is a eigenvalue of linear map $T$ if and only if $\bar{\lambda}$ is eigenvalue of the $T^{*}$.

I searched at the MSE, there are proof in the form of matrix and I understand it. But when use the following method: $<Av_1,v_2>=\lambda_{1}<v_1,v_2>=<v_1,\bar{\lambda_{1}}v_2>$ And $<Av_1,v_2>=<v_1,A^{*}v_2>$ And use it maybe $A^{*}v_2=\bar{\lambda_{1}}v_2$ can be proved. But I don't understand why $A^{*}v_2=\bar{\lambda_{1}}v_2$ in the last step.

Answer 1

You can use your method for the case of $V$ with finite dimension and if A is a normal matrix. So, you showed that $$ \langle v_1,\bar\lambda v_2 \rangle = \langle \lambda v_1, v_2 \rangle = \langle Av_1, v_2 \rangle = \langle v_1,A^*v_2 \rangle.$$ Now, take $v_2$ as an eigenvector of $A^*$ so that $$ \langle v_1,\bar\lambda v_2 \rangle = \langle v_1,A^*v_2 \rangle = \langle v_1,\lambda' v_2 \rangle, \text{ hence} $$

$$ \langle v_1,v_2(\lambda' - \bar\lambda) \rangle = \overline{(\lambda' - \bar\lambda)} \langle v_1,v_2 \rangle = 0. $$ Finally, since we are assuming that $V$ has finite dimension, we know that one of the eigenvectors of $A^*$ is not orthogonal to $v_1$, from which it follows that $\lambda'=\bar\lambda$. Note that here we use the fact that $A$ is normal, therefore, the eigenvectors form a basis, so we use this to conclude that one of the eigenvectos is not orthogonal.