Diagonalisable matrices over different fields

Question

I believe this fits in with my knowledge of Jordan Normal form, however I am not sure how to approach the question itself? I am especially lost with $F_7$

Answer 1

Calculate its eigenvalues, characteristic and minimal polynomials and etc. iver the different fields:

$$p_A(x)=\det(xI-A)=\begin{vmatrix}x&-1&0\\ 0&x&-1\\ -1&0&x\end{vmatrix}=x^3-1=(x-1)(x^2+x+1)$$

Now, hints:

== Over $\;\Bbb C\;$ , the above polynomial has three different roots, so...

== Over $\;\Bbb R\;$ , the above polymomial has only one root and then its minimal polymomial cannot decompose as a product of different linear factors, thus...

== Over $\;\Bbb F_7\;$, the polynomial has another root besides $\;x=1\;$ if $\;\Delta=1-4=-3=4\;$ is a square, which it obviously is, thus...

Answer 2

The matrix obviously satisfies $A^3=I$. Over the complex numbers, being of finite order implies being diagonalisable. Over the real numbers however the only possible eigenvalue is $1$ (real root of $X^3-1$), so if it were diagonalisable it would have to be $I$, which it is not. Over $\Bbb F_7$ there are $3$ distinct roots of $X^3-1$, namely (the classes of) $1,2,4$. So $X^3-1=(X-1)(X-2)(X-4)$, and $A$ is annihilated by this split polynomial with simple roots; therefore it is diagonalisable (in the end the same argument that works for$~\Bbb C$).

If you are unfamiliar with the theorem that says being annihilated by a split polynomial with simple roots implies being diagonalisable, then you can also easily see that $X^3-1$ is the minimal and also characteristic polynomial of $A$, so that its roots must be eigenvalues; then having $3$ distinct eigenvalues in dimension$~3$ implies $A$ is diagonalisable