Hi I have this question here.
Suppose that $A \in M_4(\mathbb{C})$ has the characteristic polynomial $$p_A(x)=(x^2+3)(x^2+\sqrt{3}x+3)$$ Find $A^{12}$.
I get my eigenvalues are $\frac{-\sqrt{3}+3i}{2},\frac{-\sqrt{3}-3i}{2},\sqrt{3}i,-\sqrt{3}i$. Since the eigenvalues are distinct I know that $A$ is diagonalizable over the complex numbers if $A$ is of size $4 \times 4$ (Which it is).
I also know that $A^{12}=PD^{12}P^{-1}$. How do I figure out $P$ (And subsequently $P^{-1}$ ) though? I don't even know if that's possible...
If you expand $D^{12}$, you get a constant multiple of $I_{4}$, so $A^{12}=D^{12}$.