= Solution
A direct determinant calculation gives
$$
\chi_A(t)=\det(tI-A)=(t-1)^3.
$$
Moreover,
$$
A-I=
\begin{pmatrix}
-3&-6&-9\\
3&6&9\\
-1&-2&-3
\end{pmatrix}
$$
has rank one, so its kernel has dimension two, while $(A-I)^2=0$ and $A\ne I$. Hence the <minimal polynomial> is
$$
\boxed{m_A(t)=(t-1)^2}.
$$
There are two <Jordan block>[Jordan blocks], because the eigenspace has dimension two, and the largest has size two, because the minimal polynomial has exponent two. Thus
$$
\boxed{\operatorname{JNF}(A)=J_2(1)\oplus J_1(1)}.
$$
Back to article page
