How to calculate $\det(I +A^{50})$ with eigenvalues given

Question

Let $A$ be a $3 *3$ matrix with the eigenvalues $1,-1,0$. How to calculate $\det(I +A^{50})$ ? I know that the answer is 4, but i have no idea how to approach such a problem

Answer 1

Hint: Since $A$ has three different eigenvalues, it is diagonalizable.

Answer 2

Hint

If we denote the eigenvalues of $A$ by $\lambda$ we have

(1) The eigenvalues of $A^n$ are $\lambda^n$

(2) The eigenvalues of $A+I$ are $1+\lambda$

(3) $\det(A)=\lambda_1\lambda_2\cdots \lambda_n$

Answer 3

The characteristic polynomial of $A$ is $\lambda (\lambda ^2-1)$ According to the Cayley Hamilton theorem $A$ satisfies its characteristic polynomial so $$A^3=A$$

Thus

$$A^{50}=A^{48}A^2=A^{16}A^2=A^{18}=A^2$$

$$\det(I+A^{50})=\det(I+A^2)$$

Let $D$ be the diagonal matrix of eigenvalues of $A$

$$A=P^{-1}DP\implies A^2=P^{-1}D^2P$$

$$\det(I+A^2)=\det(I+D^2)=1(2)(2)=4$$