Compute \(\JCF(A)\) for \[ A \da \mattt{1}{-1}{0}{-1}{4}{-1}{-4}{13}{-3}. \]
Solution
1. The characteristic polynomial of \(A\) is\[ \chi_A(t)=t(t-1)^2. \]
Direct expansion gives \[ \det(tI-A) = \det\begin{pmatrix} t-1&1&0\\ 1&t-4&1\\ 4&-13&t+3 \end{pmatrix} =(t-1)(t^2-t+1)-(t-1) =t(t-1)^2. \]
2. The eigenspace for the eigenvalue \(0\) is one-dimensional.
The eigenvalue \(0\) has algebraic multiplicity \(1\) by 1, so its geometric multiplicity is also \(1\).
3. The eigenspace for the eigenvalue \(1\) is one-dimensional.
One has \[ A-I= \begin{pmatrix} 0&-1&0\\ -1&3&-1\\ -4&13&-4 \end{pmatrix}. \] The first two rows are linearly independent, while the third row equals \(-13\) times the first row plus \(4\) times the second row. Hence \(\operatorname{rank}(A-I)=2\), so by rank-nullity \[ \dim\ker(A-I)=1. \]
4. The Jordan blocks for the eigenvalue \(1\) consist of one block of size \(2\).
By 1, the eigenvalue \(1\) has algebraic multiplicity \(2\). By 3, its eigenspace has dimension \(1\), which equals the number of Jordan blocks for that eigenvalue. Therefore there is exactly one such block, and its total size must be \(2\).
5. Therefore\[ \JCF(A)=J_2(1)\oplus J_1(0), \]up to permutation of Jordan blocks; explicitly,\[ \JCF(A)\sim \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&0 \end{pmatrix}. \]
Combine 2 and 4.