I'm having a problem with this theorem: Matrix $A_{n\times n}$ has no more than $n$ different eigenvalues.
Now I have the matrix $A = \begin{pmatrix} 0&2&0\\ 0&0&1\\ 4&0&0 \end{pmatrix}$ and the elements are from $\mathbb{Z}_7$.
I found its characteristic polynomial to be $\lambda^3 - 8$ which is $\lambda^3 + 6$ above $\mathbb{Z}_7$, now using the polynomial, I found the eigenvalues are $1,2,4$ because $1^3 = 1$, $2^3 = 8 = 1$, $4^3 = 64 = 1$ above $\mathbb{Z}_7$ which means I have 3 different eigenvalues.
But my problem is that from $\lambda^3 - 1$ , I also get the eigenvalues $1\operatorname{cis}(0)$, $1\operatorname{cis}(120)$ and $1\operatorname{cis}(240)$, for a total of 6 eigenvalues.
Are those not eigenvalues because I'm not in the complex number field?, if yes that conflicts the question asked in the following link: Conflicting theorems regarding eigenvalues & matrix diagonalisibility, (the matrix is above $\mathbb{R}$, but has complex eigenvalues).
Your field of $\mathbb{Z}_7$ does not contain complex numbers.
In $\mathbb{Z}_7$ the class of $-1$ does not have a square root because there is no square congruent to $6$ mod (7).
That is your field does not have a complex extension.