A query on the eigenvalues of anti-involution matrix An involution matrix $A$ is defined by the condition
$$
A^2 = I \tag{1}
$$
The eigenvalues of an involution matrix $A$ are the roots of unity.
Generalizing, an $m$-involution matrix $A$ is defined by the condition
$$
A^m = I \tag{2}
$$
The eigenvalues of an $m$-involution matrix $A$ are the $m$th roots of unity.
In a similar manner, an anti-involution matrix $A$ can be defined as
$$
A^2 = - I
$$
(I hope that this is the standard definition)
An example of an anti-involution matrix is
$$
A = \left[ \begin{array}{cc}
       0 & -1 \\
       1 & 0 \\
\end{array} \right]
$$
The eigenvalues of the above matrix $A$ are: $\pm j$.
The characteristic equation of this $2 \times 2$ matrix $A$ is
$$
\lambda^2 + 1 = 0
$$
Is it possible to show that the eigenvalues of an anti-involution matrix are the
square roots of $-1$ ?
In a similar way, if we define anti-m-involution matrix $A$ as
$$
A^m = - I,
$$
can we show that its eigenvalues are the $m$th roots of $-1$?
I don't ask about the eigenvectors of anti-involution matrices as I am not aware of any results for the eigenvectors of involution matrices.. Kindly help me on their eigenvalues. Thank you.
 A: Let $A^m = - I$ and let $ \lambda$ be an eigenvalue of $A$. By the spectral mapping theorem, $\lambda^m$ is an eigenvalue of $A^m$, hence $\lambda^m$ is an eigenvalue of $-I$. This gives $ \lambda^m=-1.$
A: Let $(\lambda, \mathbf{x})$ be an eigenpair for an anti-involution matrix $A$.
By definition, $A$ satisfies $A^2 = -I$.
By definition, the eigenpair $(\lambda, \mathbf{x})$  satisfies
$$
A \mathbf{x} = \lambda \mathbf{x}, \ \ \mbox{where} \ \mathbf{x} \neq 0.
$$
Since $A^2 = - I$, it follows that $A^{-1} = - A$.
Note that
$$
A \mathbf{x} = \lambda \mathbf{x} \iff A^{-1} [A \mathbf{x}] = A^{-1} [\lambda \mathbf{x}] \iff I \mathbf{x} = \lambda (- A \mathbf{x}) \tag{1}
$$
Simplifying (1), we conclude that
$$
A \mathbf{x} = \lambda \mathbf{x} \iff \mathbf{x} = - \lambda (A \mathbf{x}) =
- \lambda (\lambda \mathbf{x})  \iff (1 + \lambda^2) \mathbf{x} = 0 \iff  \lambda^2 + 1 = 0
$$
Thus, if $\lambda$ is any eigenvalue of $A$, then it satisfies $\lambda^2 + 1 = 0$.
Hence, the eigenvalues of an anti-involution matrix $A$ are the square roots of $-1$.
The above proof can be easily extended to the general case for an anti-$m$-involution matrix.
We can show that if $\lambda$ is an eigenvalue of an anti$-m$-involution matrix,
then it satisfies $\lambda^m + 1 = 0$.
I give an outline of proof for the anti$-m$-involution matrix.
Suppose that $(\lambda, \mathbf{x})$ is an eigenpair of $A$.
Since $A^m = - I$, it follows that $A^{-1} = - A^{m - 1}$.
Thus, we have
$$
A \mathbf{x} = \lambda \mathbf{x} \iff A^{-1} [A \mathbf{x}] = A^{-1} [\lambda \mathbf{x}] \iff I \mathbf{x} = \lambda (- A^{m - 1} \mathbf{x}) \tag{2}
$$
Simplifying (2), we conclude that
$$
A \mathbf{x} = \lambda \mathbf{x} \iff \mathbf{x} = - \lambda (\lambda^{m - 1} \mathbf{x}) =
- \lambda^m \mathbf{x}  \iff (1 + \lambda^m) \mathbf{x} = 0 \iff  \lambda^m+ 1 = 0
$$
Thus, if $\lambda$ is any eigenvalue of an anti-$m$-involution matrix $A$, then it satisfies $\lambda^m + 1 = 0$.
Hence, the eigenvalues of an anti-involution matrix $A$ are the $m$th order roots of $-1$.
