Determine the Jordan canonical form of \[ A=\begin{pmatrix} 1&2&3\\ 0&4&5\\ 0&0&4 \end{pmatrix}. \]
Solution
1. The characteristic polynomial of \(A\) is\[ \chi_A(t) = (t-1)(t-4)^2. \]
The matrix \(A\) is upper triangular. Hence its characteristic polynomial is the product of \[ t-a_{ii} \] over its diagonal entries: \[ \chi_A(t) = (t-1)(t-4)(t-4). \]
2. The eigenspace for the eigenvalue \(4\) is one-dimensional.
One has \[ A-4I = \begin{pmatrix} -3&2&3\\ 0&0&5\\ 0&0&0 \end{pmatrix}. \] If \[ \begin{pmatrix} x\\ y\\ z \end{pmatrix} \in \ker(A-4I), \] then the second row gives \[ 5z=0, \] so \(z=0\). The first row then gives \[ -3x+2y=0. \] Thus \[ \ker(A-4I) = \operatorname{span} \left\{ \begin{pmatrix} 2\\ 3\\ 0 \end{pmatrix} \right\}, \] which has dimension \(1\).
3. The eigenvalue \(4\) contributes exactly one Jordan block, and thatblock has size \(2\).
2. Thus there is one block whose total size is\(2\), namely\[ J_2(4) = \begin{pmatrix} 4&1\\ 0&4 \end{pmatrix}. \]
4. The eigenvalue \(1\) contributes the one-dimensional block\[ (1). \]
By step 1, the algebraic multiplicity of \(1\) is \(1\). Hence its entire Jordan contribution is the unique block of size \(1\).
5. The Jordan canonical form is\[ \boxed{ \begin{pmatrix} 1&0&0\\ 0&4&1\\ 0&0&4 \end{pmatrix}. } \]
Combine the blocks from steps <1>3–<1>4. Jordan blocks may be reordered, so any permutation of these diagonal blocks is the same Jordan canonical form up to block ordering.
6. Q.E.D.
Step 5 is the requested Jordan canonical form.