I am searching for orthogonal matrices $A$ with the property

$$|A|_F^2 = \deg( \chi_A(t) ) = 2 \deg( m_A(t)), tr(A) = 1$$

where $\chi_A(t)$ is the characteristic polynomial and $m_A(t)$ is the minimal polynomial and $|.|_F$ is the Frobenius norm. For some $n \in \mathbb{N}$, I have already found a matrix $A_n$ with this property, but now I am asking myself how to construct or find other matrices with this property.

If it helps: The matrices I am considering are defined as follows: The matrix $A_n$ for a natural number $n$ is defined as $A_n = \oplus_{d|n} Z_d$ where $Z_d$ is a circulant matrix with $0/1$ which is defined on Wikipedia as $Z$. If $n$ is a perfect number then one has the additional property: $2\deg(m_{A_n}(t)) = \deg(\chi_{A_n}(t))$

The other properties are fullfilled for all $n$.

Any help is highly appreciated.

  • $\begingroup$ Notice: orthogonal matrices are normal, and are diagonalizable through unitary matrices, but all the quantities you have (degrees, norm, trace) does not change with an unitary base change $\endgroup$ – Exodd May 3 '18 at 16:37
  • $\begingroup$ What does that mean? I do not really understand what you mean. Thanks for your comment by the way. $\endgroup$ – orgesleka May 3 '18 at 16:39
  • $\begingroup$ btw, there's an easy way to build those kind of matrix from rotation and circulant matrices $\endgroup$ – Exodd May 3 '18 at 17:03
  • $\begingroup$ Could you please provide an answer with more details? $\endgroup$ – orgesleka May 3 '18 at 17:04

Remember that a rotation matrix $$ R = \begin{pmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{pmatrix} $$ has $\exp(i\theta)$ and $\exp(-i\theta)$ as eigenvalues. Suppose now that $tr(R) = 2\cos\theta=1/2$. Such a $\theta$ is not a rational multiple of $\pi$, so $\exp(i\theta)$ is different from any eigenvalue of $Z_n$.

For any $n\ne 1$, the matrix $A = R\oplus R\oplus Z_n \oplus Z_n$ satisfies all your conditions, since $tr(A)=2 tr(R) = 1$.

  • $\begingroup$ Thank you for your answer. What is $\deg(m_A(t))$? $\endgroup$ – orgesleka May 3 '18 at 17:14
  • $\begingroup$ $n+2$, that is half of the dimension $\endgroup$ – Exodd May 3 '18 at 17:21

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