A query on the eigenvalues of anti-involution matrix

Question

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.

Answer 1

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 $-1$.

Answer 2

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.$