# 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

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: $$=\lambda_{1}=$$ And $$=$$ 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.

• Can you provide a link to the answer that uses this process? Like you mentioned, this question has been asked (and solved before) in this site; in fact, here is a very good solution by Sheldon Axler. Jun 16, 2020 at 14:38
• Thank you for your link. Maybe it can be proved by matrix, I have seen a proof just by the method I mentioned above, I try to find the link but I can not find it again :(, it just prove it as I mentioned above.@Carlo Jun 16, 2020 at 14:55
• By definition, the adjoint is the unique linear operator (or matrix) satisfying $\langle Tv, w \rangle = \langle v, T^* w \rangle$ for all vectors $v, w \in V.$ But in the equation you mention, it seems that $v_1$ is fixed, so I don’t see how to conclude that $A^* v_2 = \bar \lambda_1 v_2.$ Jun 16, 2020 at 15:57
• You can use your method to prove the equivalence if you assume a finite dimension and that T is normal. Jun 19, 2020 at 14:57

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.
• Why is it true that one of the eigenvectors of $A^*$ is not orthogonal to $v_1?$ Jun 17, 2020 at 16:22