Characteristic polynomial modulo 12 
Consider the vector space $V =\left\{a_0+a_1x+\cdots+a_{11}x^{11},\;a_i\in\mathbb{R}\right\}$. Define a linear operator $A$ on $V$ by $A(x^i) = x^{i+4}$ where $i + 4$ is taken modulo $12$.
Find $(a)$ the minimal polynomial of $A$ and $(b)$ the characteristic polynomial of $A$.

My try:
I coulnot find another way so I tried the brute force method. $\:$I found the matrix representation of the operator is $$A=\begin{bmatrix} 0&0&0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&0&0&1&0&0\\0&0&0&0&0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&0&0&0&0&1\\1&0&0&0&0&0&0&0&0&0&0&0\\0&1&0&0&0&0&0&0&0&0&0&0\\0&0&1&0&0&0&0&0&0&0&0&0\\0&0&0&1&0&0&0&0&0&0&0&0\\0&0&0&0&1&0&0&0&0&0&0&0\\0&0&0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&1&0&0&0&0&0\\0&0&0&0&0&0&0&1&0&0&0&0\\\end{bmatrix}.$$
The characteristic polynomial I found to be $\lambda^{12}-4\lambda^9+6\lambda^6-4\lambda^3+1$.$\:$(It took me almost 40 minutes. $\:$Is there another way to do this problem? Provide hints or suggestions please.
$\rule{17cm}{1pt}$
Taking forward the answer provided by $\textbf{Servaes}$ "The minimal polynomial of $A|_{U_i}$ is still $X^3−1$." Taking $U_1=span\{x_1,x_5,x_9\}$ we see $A(x)=x^5,\:A^2(x)=x^9,\:A^3(x)=x\implies (A^3-I)=0$  and since it factors into linear irredeucible factors, we have the minimal polynomial of $A|_{U_i}$ is $X^3−1$.
Completing the proof: We show that characteristic polynomial of $A$ is the product of characteristic polynomials of $A|_{U_i}$ where $V=\oplus U_i$. We have seen that minimal polynomial of $A|_{U_i}$ is $X^3−1$ which is precisely the characteristic polynomial. So let $p_i(\lambda)$ is characteristic polynomial corresponding to eigenvalue $\lambda_i$ and invariant subspace $U_i$ and $p(\lambda)$ is the characteristic polynomial of A. Then $\displaystyle p(\lambda_i)=0 \:\forall\;i  \implies p_i(\lambda)|p(\lambda) \:\forall\;i \implies p(\lambda)=\prod_ip_i(\lambda)$.
$\Big($$p(\lambda)$ is atleast $\displaystyle\prod_ip_i(\lambda)$. If $\exists\lambda\neq\lambda_i \forall\: i$ such that $p(\lambda)=0$ then $V$ is not $\oplus U_i$ $\Big)$
$\rule{17cm}{1pt}$
Minimal polynomial : $X^3-1\qquad$ Characteristic polynomial : $(X^3-1)^4$.
 A: It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
$$A^3(x^i)=x^{i+12}=x^i,$$
for all $i$, so $A$ is a zero of $X^3-I$. This factors as
$$X^3-I=(X-I)(X^2+X+I),$$
which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same irreducible factors and has degree $12$, so the minimal polynomial equals
$$(X-I)^a(X^2+X+I)^b,$$
for some positive integers $a$ and $b$ satisfying $a+2b=12$. 
Note that $a$ is the algebraic multiplicity of the eigenvalue $1$, which is at least $4$ because
$$1+x^4+x^8,\qquad x+x^5+x^9,\qquad x^2+x^6+x^{10},\qquad x^3+x^7+x^{11},$$
are four linearly independent eigenvectors with eigenvalue $1$. So $(a,b)$ is either $(4,4)$, $(6,3)$, $(8,2)$ or $(10,1)$.

The following is a bit contrived and implicitly assumes some slightly advanced ideas, but it is an (almost) computation-free way of determining the characteristic polynomial:
For $i\in\{1,2,3,4\}$ let $U_i:=\operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_i\cap U_j=0$ whenever $i\neq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
$$V=U_1\oplus U_2\oplus U_3\oplus U_4,$$
of $A$-invariant subspaces. The minimal polynomial of $A\vert_{U_i}$ is still $X^3-1$ (verify this!), hence it is the characteristic polynomial of the restriction. The characteristic polynomial of $A$ is the product of the characteristic polynomials of the $A\vert_{U_i}$, so it is $(X^3-1)^4$.
