Calculating the operator (or operator matrix) of a minimal polynomial. Here is the question I am trying to answer:
Let $V$ be a real vector space and $T: V \to V$ a linear transformation with minimal polynomial $T^3 - T = 0.$ For real numbers $\lambda,$ let $V_{\lambda} = \operatorname{ ker} (T - \lambda).$
$(a)$ Show that $V_{\lambda} = 0$ unless $\lambda \in \{0, 1, -1\}.$
$(b)$ Define a linear transformation $V \to V_0 \oplus V_1 \oplus V_{-1}$ and show that it is an isomorphism.
My questions are:
1- Is part $(a)$ asking me to calculate the eigenvalues of the operator $T$? If so, how can I know the operator (or its matrix) if I know its minimal polynomial?If not, what actually is the question asking?
2- For part $(b),$ I do not know how to tackle it, could someone give me the intuition behind answering this part?
3- Should it be written $V_{\lambda} = \operatorname{ ker} (T - \lambda I)$ instead?
 A: $\newcommand\A{\mathscr A}\newcommand\B{\mathscr B}$
(a)
You do not have to know the operator or its matrix in order to find its eigenvalues. It happens that we can recover all eigenvalues from the minimal polynomial.
Here is an easier example. Suppose $\A:V\to V$ is a linear transformation.
If $v$ is an eigenvector of $\A$ with eigenvalue $\alpha\in\Bbb R$, then $\A v=\alpha v$ and $$\A^2v=\A(\A v)=\A(\alpha v)=\alpha(\A v)=\alpha(\alpha v)=\alpha^2 v.$$ Hence $(\A^2-\A)v=\alpha^2v-\alpha v = (\alpha^2-\alpha)v$.
If we also know $\A$ has minimal polynomial $\A^2-\A$, then by definition of minimal polynomial, $(\A^2-\A)v=0$, i.e., $(\alpha^2-\alpha)v=0.$  Since as an eigenvector, $v\not=0$, we must have $\alpha^2-\alpha=0$. Solving this quadratic equation for $\alpha\in\Bbb R$, we find that $\alpha =0,1$. 
Back to $T$ in the question.
$T^3-T=0$ means $(T^3-T)(u)=0$, i.e. $T^3u=Tu$ for all $u\in V$.
Suppose $0\not=v\in V_\lambda$, i.e. $\ Tv=\lambda v$.
$T^3v=T(T(T(v))=T(T(\lambda v))=T(\lambda(T(v))=T(\lambda\lambda v)=T(\lambda^2v)=\lambda^2(Tv)=\lambda\lambda^2v=\lambda^3 v$
So $\lambda^3v=\lambda v$, i.e., $(\lambda^3-\lambda)v=0$. Since $0\not=v$,
we must have $\lambda^3-\lambda=0$, which means $\lambda \in \{0, 1, -1\}$.

(b)
The idea is to express the identity map $I$ as a linear combination of three factors of $T^3-T=T(T-I)(T+I)$. 
$$I =-(T-I)(T+I)+\frac{1}{2}T(T+I)+\frac{1}{2}T(T-I) \tag{*}\label{*}$$
Applying both sides to an arbitrary $v\in V$, we have
$$ v = v_0 + v_1 + v_{-1}\tag{**}\label{**}$$
where $v_0=(-(T-I)(T+I))v$, $\ v_1=(\frac{1}{2}T(T+I))v$, $\ v_{-1}=(\frac{1}{2}T(T-I))v$
Note that $f(T)(g(T)v)=(f(T)g(T))v$ for any polynomial $f$ and $g$. Also note $(T^3-T)v=0$ by the definition of minimal polynomial.

*

*$Tv_0=-(T(T-I)(T+I))v=-(T^3-T)v=0$. So $v_0\in V_0$.

*$(T-I)v_1=(\frac12T(T-I)(T+I))v=\frac12(T^3-T)v=0$. So $v_1\in V_1$.

*$(T+I)v_{-1}=(\frac12T(T-I)(T+I))v=\frac12(T^3-T)v=0$. So $v_{-1}\in V_{-1}$.

Hence we can define a linear transformation $\phi: V\to V_0\oplus V_1 \oplus V_{-1}$, $\phi(v)=(v_0, v_1, v_{-1})$.

*

*$\phi(v)=0$ means $v_0=0, v_1=0, v_{-1}=0$, which means $v=0$ thanks to $\eqref{**}$.
So $\phi$ is injective.


*Suppose $(u_0, u_1, u_{-1})\in V_0\oplus V_1 \oplus V_{-1}$. Let $u=u_0+u_1+u_{-1}\in V$.

*

*$(-(T-I)(T+I))u = (-(T-I)(T+I))u_0+(-(T-I)(T+I))u_1+(-(T-I)(T+I))u_{-1}=(-(0-1)(0+1))u_0+(-(1-1)(1+1))u_1+(-(-1-1)(-1+1))u_{-1}=u_0$.

*Similarly, $(\frac12T(T+I))u =u_1$.

*Similarly, $(\frac12T(T-I))u =u_{-1}$.

Hence $\phi(u)=(u_0, u_1, u_{-1})$.
So $\phi$ is surjective.
So $\phi$ is an isomorphism.

I was struck by awe the first time I had been taught formulas like $\eqref{*}$ and $\eqref{**}$ and the ensuing arguments.  It was something that I had zero intuition. Later on I have done several similar exercises and I have seen various variations and generalizations. Those kind of formulas have become part of my intuition by now. The minimal polynomial of a linear transformation is indeed an interesting and powerful concept.
