Eigenvalues of Normal Matrix

Question

Suppose $A^*A = AA^*$ (* is the conjugate transpose) and $A$ has eigenvalues $\lambda_1, \cdots, \lambda_n$. How to show that $A^*$ has eigenvalues $\bar \lambda_i$ ?

I could find a proof for the special case where the eigenvalues are distinct, but not for the general case.

Answer to @user and KaiWang: no, "this" does not "suffice". You are forgetting the transposition in the "conjugate transpose". — Anne Bauval, Feb 12 '23 at 20:08
Sorry for the mistake. $(Av)^{\ast} = \overline{(Av)^T} = \overline{v^T A^T} =\overline{v}^T \overline{A}^T$. We therefore see that $\overline{v}^T A^{\ast} = \overline{v}^T \overline{\lambda}$. Is this right? — Kai Wang, Feb 12 '23 at 20:12
Why does this not prove that $\overline{\lambda}$ is an eigenvalue of $A^{\ast}$? — Kai Wang, Feb 12 '23 at 20:17
@KaiWang well, yes it does, but for the same reason as in your answer, i.e. quite indirectly, not by exhibiting some eigenvector. — Anne Bauval, Feb 12 '23 at 22:29
Since we don't need here your hypothesis $A^A = AA^,$ I found a better duplicate: https://math.stackexchange.com/questions/139694 — Anne Bauval, Feb 12 '23 at 22:52

score 0 · Accepted Answer · answered Feb 12 '23 at 19:59

Let $A\in \mathbb{C}^{m\times m}$. Suppose $A-\lambda I$ is not invertible. Then its rows are linearly dependent; i.e. if $r_1,\cdots,r_m$ are the row vectors of $A-\lambda I$ then there exists complex numbers $c_1,\cdots,c_m$, not all zero, such that $c_1r_1 + \cdots + c_mr_m = 0$ (is the zero vector in $\mathbb{C}^m$).

Conjugating everything gives $\overline{c_1} \text{ } \overline{r_1} + \cdots + \overline{c_m} \text{ }\overline{r_m} =0$. This implies that the column vectors of $A^{\ast} - \overline{\lambda}I$ are linearly dependent because $\overline{r_j}$ is the $j$th column vector of $A^{\ast}-\overline{\lambda}I$.

Thus we can see that

$$A-\lambda I \text{ not invertible } \to A^{\ast}-\overline{\lambda} I \text{ not invertible } \to (A^{\ast})^{\ast} - \overline{\overline{\lambda}} I = A-\lambda I \text{ not invertible }$$

Or simply: $\det(A^-\bar\lambda I)=\det((A-\lambda I)^)=\overline{\det(A-\lambda I)}.$ — Anne Bauval, Feb 12 '23 at 20:12

Eigenvalues of Normal Matrix

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