Show that the determinant is $\pm 1$ Suppose that $A \in M_n(\mathbb{C})$ satisfies the relations $(A^3+I_n)(A-I_n)=0$ and $A^4=I_n$. I want to show that $\det(A)=\pm 1$.
($M_n(k)$=the $n \times n$ matrices with elements over the field $k=\mathbb{R}$ or $k=\mathbb{C}$, $I_n$=the $n \times n$ identity matrix )
I have thought the following:
$$ (A^3+I_n)(A-I_n)=0 \\ \Rightarrow A^4-A^3+A-I_n^2=0 \\ \Rightarrow I_n-A^3+A-I_n=0 \\ \Rightarrow A-A^3=0 \\ \Rightarrow A(I-A^2)=0 \\ \overset{A \neq 0}{\Rightarrow} A^2=I_n \\ \Rightarrow A=\pm I_n , \text{ and thus } \det(A)=\pm 1. $$
Is my idea right? Or have I done something wrong?
 A: $$A^3=A$$
Since $A$ is non-singular,
$$A^2=I$$
$$\det(A)^2 = 1.$$
Note that from the first step to my second step, the reason is $A$ is non-singular (and not $A$ is non-zero). Since it is non-singular, we can multiply its inverse on both sides.
A: We know that $A$ is invertible (as $A^4 = A(A^3) = I$, so $A^{-1} = A^3$). So we can go from $A^3 - A = 0$ to $A^3 =A$ to $A^2 = A^{-1}A^3 = A^{-1}A = I$.
If then $\lambda$ is an eigenvalue with non-zero eigenvector $x$ for $A$ we know that $x= A^2 x = A(Ax) = A(\lambda x) = \lambda Ax = \lambda^2 x$ and as $x \neq 0$ we can conclude that $\lambda^2 = 1$, so $\lambda = \pm1$ as $\lambda \in \mathbb{C}$ which is a field. Or directly $\det(A) \cdot \det(A) = \det(A^2) = \det(I) =1$ and so $\det(A)^2 = 1$ hence $\det(A) = \pm 1$.
A matrix can have $A^2 = I$ without being equal to $\pm I$: any reflection in a line through the origin is an example.  
A: We have $$p(x) = x^4-1 = (x-1)(x+1)(x-i)(x+i)$$ and $$q(x) = (x^3+1)(x-1) = (x+1)(x-1)\left(x-\frac{1+i\sqrt3}2\right)\left(x-\frac{1-i\sqrt3}2\right)$$
The polynomials $p$ and $q$ annihilate $A$ so the minimal polynomial $m_A$ of $A$ divides both $p$ and $q$. Hence, $m_A \mid \gcd(p,q) = (x-1)(x+1)$.
We conclude that $\sigma(A) \subseteq \{\text{zeroes of }m_A\} \subseteq \{-1,1\}$.
Hence $\det A = \pm 1$ because it is the product of eigenvalues.
