# 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.

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 $$\lambda^m=-1.$$

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 $$-1$$.