Let m \[ A \da \begin{bmatrix} 4 & 1 & -1 \\ -6 & -1 & 2 \\ 2 & 1 & 1 \end{bmatrix} \in \Mat(3\times 3, \CC) .\]
Find the Jordan canonical form \(J\) of \(A\).
Find an invertible matrix \(P\) such that \(J = \inverseof{P} A P\).
Write down the minimal polynomial of \(A\).
You should not need to compute \(\inverseof{P}\)
\(\chi_A(t) = t^n - \tr\qty{\Extpower^1 A}t^{n-1} + \tr\qty{\Extpower^2 A}t^{n-2} - \cdots \pm \det(A)\)
Finding generalized eigenvectors: let \(B = A-\lambda I\), get eigenvector \(v\), solve \(Bw_1 = v, Bw_2 = w_1, \cdots\) to get a Jordan block. Repeat with any other usual eigenvectors.
Convention: construct Jordan blocks in decreasing order of magnitude of eigenvalues.
Polynomial exponent data:
Minimal polynomial exponents: sizes of largest Jordan blocks.
Characteristic polynomial exponents: sum of sizes of Jordan blocks, i.e. how many times \(\lambda\) is on the diagonal of \(\JCF(A)\).
Solution
(parts a and b)
Write \(\chi_A(t) = t^3 - T_1 t^2 + T_2 t - T_3\) where \(T_i \da \tr\qty{\Extpower^i A}\):
\(T_1 = \tr(A) = 4-1+1=4\).
\(T_2 = (-1-2) + (4+2) + (-4-6) = 5\).
\(T_3 = \det(A) = 4(-1-2) -1(-10) + (-1)(-6+2) = 2\).
So \(\chi_A(t) = t^3 - 4t^2 + 5t-2\).
Try rational roots test: \(r \in \ts{\pm 2/1}\), and check that 2 is root.
By polynomial long division, \(\chi_A(t) / (t-2) = t^2-2t+1 = (t-1)^2\).
So the eigenvalues are \(\lambda = 2, 1\).
\(\lambda = 2\):
- Set \(U\da A-\lambda I\), then find \(\RREF(U)\) to compute its kernel: \[ U \da \begin{bmatrix} 2 & 1 & -1 \\ -6 & -3 & 2 \\ 2 & 1 & -1 \end{bmatrix} \leadsto \begin{bmatrix} 2 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} ,\] which yields \(v_1 = [1,-2,0]\).
\(\lambda = 2\):
Similarly, \[ U \da \begin{bmatrix} 3 & 1 & -1 \\ -6 & -2 & 2 \\ 2 & 1 & 0 \end{bmatrix} \leadsto \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} ,\] which yields \(v_2 = [1,-2,1]\).
Solve \(Uw = v_3\): \[ \begin{bmatrix} 3 & 1 & -1 & 1 \\ -6 & -2 & 2 & -2 \\ 2 & 1 & 0 & 1 \end{bmatrix} \leadsto \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} ,\] so take \(v_3 = [0,1,0]\).
Putting things together: \[ A &= \inverseof{P} J P \text{ where } \\ J = J_1(\lambda = 2) \oplus J_2(\lambda = 1) &= \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \\ P = [v_1, v_2, v_3] &= \begin{bmatrix} 1 & 1 & 0 \\ -2 & -2 & 1 \\ 0 & 1 & 0 \end{bmatrix} .\]
(part c)
Write \(\min_A(t) = (t-2)(t-1)^{\ell_1}\), then since \(\min_A(t)\) divides \(\chi_A(t)\) either \(\ell_1 = 1, 2\).
\(\ell_1\) is the size of the largest block corresponding to \(\lambda = 1\), which is size 2, so \(\lambda_1=2\).
Thus \[ \min_A(t) = (t-2)(t-1)^2 .\]