Let \(T\) be a linear operator on \(\RR^3\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix}. \] Express the minimal polynomial \(p\) for \(T\) in the form \(p = p_1p_2\), where \(p_1\) and \(p_2\) are monic and irreducible over the field of real numbers. Let \(W_i\) be the null space of \(p_i(T)\). Find bases \(B_i\) for the spaces \(W_1\) and \(W_2\). If \(T_i\) is the operator induced on \(W_i\) by \(T\), find the matrix of \(T_i\) in the basis \(B_i\) (above).
Solution
Let \[ A=\begin{pmatrix} 6&-3&-2\\ 4&-1&-2\\ 10&-5&-3 \end{pmatrix}. \] The answer is \[ \boxed{p(x)=(x-2)(x^2+1),} \] with \[ \boxed{B_1=((1,0,2)), \qquad B_2=((1,1,0),(0,0,1)),} \] and \[ \boxed{[T_1]_{B_1}=(2), \qquad [T_2]_{B_2}=\begin{pmatrix}3&-2\\5&-3\end{pmatrix}.} \]
1. Determine the minimal polynomial.
A direct determinant computation gives \[ \chi_T(x)=\det(xI-A)=(x-2)(x^2+1). \] The two factors are monic and irreducible over \(\mathbb R\). The characteristic polynomial is squarefree, so the minimal polynomial, which has the same irreducible factors as the characteristic polynomial and divides it, must be \[ p(x)=(x-2)(x^2+1) \] [@HK71]. Thus take \[ p_1(x)=x-2, \qquad p_2(x)=x^2+1. \]
2. Compute the two primary subspaces.
Row reduction of \[ A-2I= \begin{pmatrix} 4&-3&-2\\ 4&-3&-2\\ 10&-5&-5 \end{pmatrix} \] gives a one-dimensional kernel spanned by \((1,0,2)\). Hence \[ W_1=\ker(A-2I)=\operatorname{span}\{(1,0,2)\}. \]
Also \[ A^2+I= \begin{pmatrix} 5&-5&0\\ 4&-4&0\\ 10&-10&0 \end{pmatrix}, \] whose kernel is \[ W_2=\{(x,x,z):x,z\in\mathbb R\} =\operatorname{span}\{(1,1,0),(0,0,1)\}. \] Thus the displayed \(B_1,B_2\) are bases.
3. Compute the induced matrices.
For \(v_1=(1,0,2)\), \[ Av_1=(2,0,4)=2v_1, \] so \([T_1]_{B_1}=(2)\).
For \(u_1=(1,1,0)\) and \(u_2=(0,0,1)\), \[ Au_1=(3,3,5)=3u_1+5u_2, \] and \[ Au_2=(-2,-2,-3)=-2u_1-3u_2. \] Therefore \[ [T_2]_{B_2} =\begin{pmatrix}3&-2\\5&-3\end{pmatrix}. \] As a check, \[ \begin{pmatrix}3&-2\\5&-3\end{pmatrix}^2=-I_2, \] which is exactly the relation \(p_2(T_2)=0\) on the second primary component.