= Solution
Let $m_T$ be the <minimal polynomial>. If $Tv=\lambda v$ with $v\ne0$, then $0=m_T(T)v=m_T(\lambda)v$, so $m_T(\lambda)=0$. Conversely, if $m_T(\lambda)=0$, write $m_T(X)=(X-\lambda)q(X)$. If $T-\lambda I$ were invertible, the identity $(T-\lambda I)q(T)=0$ would imply $q(T)=0$, contradicting minimality of the degree. Thus $T-\lambda I$ has a nonzero <kernel of a linear map>, and $\lambda$ is an <eigenvalue>. This proves the <eigenvalue criterion using the minimal polynomial> for roots in the field under consideration.
For the given matrix, direct expansion gives
$$
\boxed{\chi_T(X)=\det(XI-A)=X^3+3X^2+3X+1=(X+1)^3.}
$$
Put $N=A+I$. Direct multiplication gives
$$
N^2=\begin{pmatrix}1&2&1\\-1&-2&-1\\1&2&1\end{pmatrix}\ne0,\qquad N^3=0.
$$
By the <Cayley-Hamilton theorem>, $m_T$ divides $(X+1)^3$; the nonzero square rules out every smaller power. Therefore $\boxed{m_T(X)=(X+1)^3}$. Its sole <eigenvalue> is $-1$, but the size of the nilpotent part makes the minimum polynomial cubic rather than linear.
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