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.

  • $\begingroup$ 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. $\endgroup$ Jun 16, 2020 at 14:38
  • $\begingroup$ 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 $\endgroup$
    – fractal
    Jun 16, 2020 at 14:55
  • $\begingroup$ 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 $A^* v_2 = \bar \lambda_1 v_2.$ $\endgroup$ Jun 16, 2020 at 15:57
  • $\begingroup$ You can use your method to prove the equivalence if you assume a finite dimension and that T is normal. $\endgroup$ Jun 19, 2020 at 14:57

1 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.

  • $\begingroup$ Why is it true that one of the eigenvectors of $A^*$ is not orthogonal to $v_1?$ $\endgroup$ Jun 17, 2020 at 16:22
  • $\begingroup$ You are right, this is not obvious, only if we assume that all eigevalues are distinct. I will reformulate the answer, I think there is a better way. $\endgroup$ Jun 17, 2020 at 16:35
  • $\begingroup$ Fixed the answer. I needed to add the normality of the matrix for it to work. $\endgroup$ Jun 17, 2020 at 17:26

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