# Eigenvalue and eigenvectors of a matrix and its Hermitian conjugate

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 $$x_{\lambda, a}= \sum_{b} t_{ab} y_{\lambda^*, b}$$

• 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 $\;\lambda^*\;$... Oct 25, 2019 at 15:18
• @DonAntonio I've added some extra details to explain your puzzle. Oct 25, 2019 at 15:22

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.
• 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! Oct 25, 2019 at 16:44