= Solution
Using $\chi_M(t)=\det(tI-M)$, expansion or triangular reordering gives
$$
\boxed{\chi_M(t)=(t-1)^2(t-2)^2.}
$$
The <eigenspace> for $1$ is spanned by $e_1$, so its algebraic multiplicity two requires a size-two <Jordan block>. A convenient <Jordan chain> is
$$
v_1=e_1,\qquad v_2=e_2+e_3,\qquad (M-I)v_2=v_1.
$$
For <eigenvalue> $2$, the <eigenspace> has <basis> $w_1=e_1+e_3$, $w_2=e_4$, hence both corresponding blocks have size one. Therefore a <Jordan normal form> and the <minimal polynomial> are
$$
\boxed{J=\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&2&0\\0&0&0&2\end{pmatrix},
\qquad m_M(t)=(t-1)^2(t-2).}
$$
The ordered <basis> $(v_1,v_2,w_1,w_2)$ supplies the <change-of-basis matrix>
$$
\boxed{P=\begin{pmatrix}1&0&1&0\\0&1&0&0\\0&1&1&0\\0&0&0&1\end{pmatrix}.}
$$
Its <determinant> is one, and the chain equations give $MP=PJ$, proving $P^{-1}MP=J$. The square at $t=1$ in the <minimal polynomial> is essential: $M$ is not <diagonalizable>.
Back to article page
