 Solution

All calculations here take place in the finite field $F={\mathbb{F}}_3$. For the given matrix $S$, its characteristic polynomial isThe values $q(0),q(1),q(2)$ are $2,2,1$, so this quadratic polynomial has no polynomial root in $F$ and is an irreducible polynomial. Direct calculation givesNeither $S$ nor $q(S)$ is zero. The Cayley-Hamilton theorem says that the minimal polynomial divides $xq(x)$; its proper monic polynomial divisors cannot annihilate $S$, so $m_S=xq$.

The two kernels give the required primary decomposition theorem for linear maps explicitly:The determinant of the matrix with these three basis vectors as columns is $1$ in $F$, proving $W=W_1\oplus W_2$. To check these are invariant subspaces, writing $u=(1,2,0{)}^T$ and $z=(0,0,1{)}^T$ gives $Su=u+2z$ and $Sz=2u$. Hence the matrix of the restriction to $W_2$ in the basis $(u,z)$ iswith minimal polynomial $q$; the nonzero vector space $W_1$ has zero restriction and minimal polynomial $x$.