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Consider the vector space $V =left{a_0+a_1x+cdots+a_{11}x^{11},;a_iinmathbb{R}right}$. 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 $lambda^{12}-4lambda^9+6lambda^6-4lambda^3+1$.$:$(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 $A(x)=x^5,:A^2(x)=x^9,:A^3(x)=ximplies (A^3-I)=0$ 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 $displaystyleprod_ip_i(lambda)$. If $existslambdaneqlambda_i forall: i$ such that $p(lambda)=0$ then $V$ is not $oplus U_i$ $Big)$

$rule{17cm}{1pt}$

Minimal polynomial : $X^3-1qquad$ Characteristic polynomial : $(X^3-1)^4$.

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 $iin{1,2,3,4}$ let $U_i:=operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_icap U_j=0$ whenever $ineq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
$$V=U_1oplus U_2oplus U_3oplus U_4,$$
of $A$-invariant subspaces. The minimal polynomial of $Avert_{U_i}$ 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 $Avert_{U_i}$, so it is $(X^3-1)^4$.