Let $A,H \in \mathbb{C}^{n \times n}$ be such that $H$ is Hermitian and invertible and $A = H^{-1} A^* H$. In this case, $A$ is said to be $H$-self-adjoint. This is due to the fact that if $\langle \cdot, \cdot \rangle$ is the usual complex inner product, and we define the sesquilinear product $$ [x, y] := \langle Hx, y\rangle, $$ then $$ [Ax, y] = [x, Ay]. $$ Note when $H$ is positive definite, the product $[\cdot, \cdot]$ is a standard inner product studied in elementary linear algebra. However, when $H$ is not definite, there exist vectors $x$ such that $[x, x] = 0$. In this case, it is called an indefinite inner product (see e.g. [1, 2, 3]).

Similarly, one can define the analogous concepts of $H$-unitary and $H$-normal matrices. The case of $H$-unitary matrices is interesting because it generalizes some properties of standard unitary matrices. In particular, whereas standard unitary matrices have all their eigenvalues on the unit circle, the spectrum of $H$-unitary matrices is merely symmetric with respect to the unit circle, that is, if $A$ is $H$-unitary for some $H$ Hermitian and invertible, and $\lambda \in \sigma(A)$, then $\bar\lambda^{-1} \in \sigma(A)$. Similarly, the spectrum of $H$-self-adjoint matrices (not necessarily $H$-unitary) is symmetric with respect to the real line, that is if $\lambda \in \sigma(A)$ then $\bar\lambda \in \sigma(A)$.

My question is for a matrix $A$ that is invertible, $H$-self-adjoint, not $H$-unitary, but is known to have an eigenvalue $\lambda_0$ that is unimodular (i.e. $|\lambda_0| = 1$). My conjecture is that if $Av = \lambda_0 v$, then $A^*v = \bar\lambda_0 v$. I have not been able to ascertain this in general, only through observation of (several hundred) of examples. I know the conjecture is false when $\lambda$ is not unimodular.

The general results on $H$-unitary matrices do not apply to the projection of $A$ onto the eigenspace of $\lambda_0$ because the latter is singular. And I have not been able to find results on unimodular eigenvalues of $H$-self-adjoint matrices. Any help is appreciated!

[1] Gohberg, Israel, Peter Lancaster, and Leiba Rodman. Indefinite linear algebra and applications. Vol. 13. Basel: Birkhäuser, 2005.

[2] Bognár, János. Indefinite inner product spaces. Vol. 78. Springer Science & Business Media, 2012.

[3] Mehl, Christian. "On classification of normal matrices in indefinite inner product spaces." The Electronic Journal of Linear Algebra 15 (2006): 50-83.

Let $A,H \in \mathbb{C}^{n \times n}$ be such that $H$ is Hermitian and invertible and $A = H^{-1} A^* H$. In this case, $A$ is said to be $H$-self-adjoint. This is due to the fact that if $\langle \cdot, \cdot \rangle$ is the usual complex inner product, and we define the sesquilinear product $$ [x, y] := \langle Hx, y\rangle, $$ then $$ [Ax, y] = [x, Ay]. $$ Note when $H$ is positive definite, the product $[\cdot, \cdot]$ is a standard inner product studied in elementary linear algebra. However, when $H$ is not definite, there exist vectors $x$ such that $[x, x] = 0$. In this case, it is called an indefinite inner product (see e.g. [1, 2, 3]).

Similarly, one can define the analogous concepts of $H$-unitary and $H$-normal matrices. The case of $H$-unitary matrices is interesting because it generalizes some properties of standard unitary matrices. In particular, whereas standard unitary matrices have all their eigenvalues on the unit circle, the spectrum of $H$-unitary matrices is merely symmetric with respect to the unit circle, that is, if $A$ is $H$-unitary for some $H$ Hermitian and invertible, and $\lambda \in \sigma(A)$, then $\bar\lambda^{-1} \in \sigma(A)$. Similarly, the spectrum of $H$-self-adjoint matrices (not necessarily $H$-unitary) is symmetric with respect to the real line, that is if $\lambda \in \sigma(A)$ then $\bar\lambda \in \sigma(A)$.

My question is for a matrix $A$ that is invertible, $H$-self-adjoint, not $H$-unitary, but is known to have an eigenvalue $\lambda_0$ that is unimodular (i.e. $|\lambda_0| = 1$). My conjecture is that if $Av = \lambda_0 v$, then $A^*v = \bar\lambda_0 v$. I have not been able to ascertain this in general, only through observation of (several hundred) of examples. I know the conjecture is false when $\lambda$ is not unimodular. If necessary, one may assume $H$ has real entries.

The general results on $H$-unitary matrices do not apply to the projection of $A$ onto the eigenspace of $\lambda_0$ because the latter is singular. And I have not been able to find results on unimodular eigenvalues of $H$-self-adjoint matrices. Any help is appreciated!

[1] Gohberg, Israel, Peter Lancaster, and Leiba Rodman. Indefinite linear algebra and applications. Vol. 13. Basel: Birkhäuser, 2005.

[2] Bognár, János. Indefinite inner product spaces. Vol. 78. Springer Science & Business Media, 2012.

[3] Mehl, Christian. "On classification of normal matrices in indefinite inner product spaces." The Electronic Journal of Linear Algebra 15 (2006): 50-83.