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If $A$ is a square matrix, and $A^\dagger$ is the conjugate transpose of $A$. Let us assume $A$ has no nontrivial Jordan Block, i.e. $A$ is diagonalizable. Suppose $x_{\lambda, a}$ is the eigenvector of $A$ with eigenvalue $\lambda$, i.e. $$A\cdot x_{\lambda, a} = \lambda x_{\lambda, a}$$ Note that there can be multiple eigenvectors with the same eigenvalue $\lambda$, and we use the index $a$ to label them. Then $\lambda^*$ must be a eigenvalue of $A^\dagger$. I am wondering whether the following property holds:

Any eigenvector of $A$ with eigenvalue $\lambda$, i.e. $x_{\lambda, a}$, can be expressed as a linear combination of the eigenvector of $A^\dagger$ with the eigenvalue $\lambda^*$. Concretely, denote $A^\dagger\cdot y_{\lambda^*, a} = \lambda^* y_{\lambda^*, a}$, then there exists some coefficients $t_{ab}$ such that $$x_{\lambda, a}= \sum_{b} t_{ab} y_{\lambda^*, b}$$

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  • $\begingroup$ I've no idea what "Here a is the index labeling the degenerate eigenvectors of λ sub-eigen space" means and what it has to do with the question about $\;\lambda^*\;$... $\endgroup$
    – DonAntonio
    Oct 25, 2019 at 15:18
  • $\begingroup$ @DonAntonio I've added some extra details to explain your puzzle. $\endgroup$
    – user34104
    Oct 25, 2019 at 15:22

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No. Consider the matrix $$ A = \begin{pmatrix}0&1\\0&0\end{pmatrix}. $$ The only eigenvector of $A$ is $e_1$ (first standard basis vector). Now, $$ A^* = \begin{pmatrix}0&0\\1&0\end{pmatrix}, $$ whose only eigenvector is $e_2$. Even if $A$ is diagonalizable, this is false, as the simple example $$ A = \begin{pmatrix}0&1\\0&1\end{pmatrix} $$ shows.

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  • $\begingroup$ But the example you mentioned $A$ is not diagonalizable. Maybe I need to restrict to the case where $A$ is diagonalizable. But thanks for the answer! $\endgroup$
    – user34104
    Oct 25, 2019 at 16:44
  • $\begingroup$ I edited....... $\endgroup$
    – amsmath
    Oct 25, 2019 at 16:57

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