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.
Best Answer
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.$