Finding $\det(I+A^{100})$ where $A\in M_3(R)$ and eigenvalues of $A$ are $\{-1,0,1\}$ I have a matrix $A \in M_3(R)$ and it is known that $\sigma (A)=\{-1, 0, 1\}$, where $\sigma (A)$ is a set of eigenvalues of matrix $A$. I am now supposed to calculate $\det(I + A^{100})$.
I know that $A^{100}$ could be calculated using a diagonal matrix which has the eigenvalues of $A$ on it's diagonal and using matrices which are formed using the eigenvectors of $A$, but I am not sure how to get there. Or it might not even be the right approach.
I know there is a similar question, but I don't really understand the answer given there. It's not fully explained. So if anyone could help, that would be great. Thanks
 A: Since $A$ has three different eigenvalues, we know it must be diagonalizable, so there is an invertible $P$ such that
$$ PAP^{-1} = \begin{pmatrix}-1\\ & 0 \\ && 1 \end{pmatrix} $$
Now the determinant of any matrix such as for example $I+A^{100}$ is the same as any matrix that is is similar to, so
$$ \det(I+A^{100}) = \det(P(I+A^{100})P^{-1}) =
\det(PIP^{-1} + PA^{100}P^{-1}) =
\det(I+ (PAP^{-1})^{100}) $$
which you can easily calculate directly.
Note that you don't actually need to know exactly what $P$ is.
A: The eigenvalues of $A$ are $\{-1, 0, 1 \}$, 
Therefore, the eigenvalues of $A^{100}$ are $\{(-1)^{100}, 0^{100}, 1^{100} \} = \{1, 0, 1 \}$ (counting multiplicities).
Therefore, the eigenvalues of $I + A^{100}$ are $\{2,1,2 \}$ (counting multiplicities).
The determinant is the product of the eigenvalues, so:
$$ \det(I + A^{100}) = 1 \cdot 2^2 = 4 $$
A: wehave the next fact:
$$det(I+A^{100})=det(P\cdot I\cdot P^{-1}+PD^{100}P^{-1})=det(P)det(I+diag(1,0,1))det(P^{-1})$$
hence $D^{100}=diag(-1^{100},0,1^{100})=diag(1,0,1)$
$$=det(P)det(P^{-1})det(diag(2,1,2))=4det(PP^{-1})=4$$
