Determine \(\operatorname{JCF}(B)\) for \[ B := \begin{pmatrix} 5 & -1 & 0 & 0 \\ 9 & -1 & 0 & 0 \\ 0 & 0 & 7 & -2 \\ 0 & 0 & 12 & -3 \end{pmatrix} .\]
Solution
Write \[ B=B_1\oplus B_2, \qquad B_1=\begin{pmatrix}5&-1\\9&-1\end{pmatrix}, \quad B_2=\begin{pmatrix}7&-2\\12&-3\end{pmatrix}. \]
For \(B_1\), \[ \chi_{B_1}(t)=(t-5)(t+1)+9=(t-2)^2. \] Moreover \[ B_1-2I=\begin{pmatrix}3&-1\\9&-3\end{pmatrix} \] has rank \(1\), so the eigenspace for \(2\) is one-dimensional. Hence \(B_1\) contributes one Jordan block \(J_2(2)\).
For \(B_2\), \[ \chi_{B_2}(t)=(t-7)(t+3)+24=t^2-4t+3=(t-1)(t-3), \] so \(B_2\) has two distinct eigenvalues and is diagonalizable.
Therefore, up to ordering of Jordan blocks, \[ \boxed{ \operatorname{JCF}(B)= \begin{pmatrix} 2&1&0&0\\ 0&2&0&0\\ 0&0&1&0\\ 0&0&0&3 \end{pmatrix}.} \]