Let \(T\) be the linear operator on \(\RR^3\) which is represented by the matrix \[ \begin{bmatrix} 3 & 1 & -1 \\ 2 & 2 & -1 \\ 2 & 2 & 0 \end{bmatrix} \] in the standard ordered basis. Show that there is a diagonalizable operator \(D\) on \(\RR^3\) and a nilpotent operator \(N\) on \(\RR^3\) such that \(T = D + N\) and \(DN = ND\). Find the matrices of \(D\) and \(N\) in the standard basis. (Just repeat the proof of Theorem 12 for this special case.)
Solution
Let \[ A=\begin{pmatrix} 3&1&-1\\ 2&2&-1\\ 2&2&0 \end{pmatrix}. \] The required matrices are \[ \boxed{ D= \begin{pmatrix} 1&1&0\\ 0&2&0\\ -2&2&2 \end{pmatrix}, \qquad N= \begin{pmatrix} 2&0&-1\\ 2&0&-1\\ 4&0&-2 \end{pmatrix}.} \]
1. Determine the polynomial relations of \(A\).
A direct determinant computation gives \[ \chi_A(x)=(x-1)(x-2)^2. \] The matrix is not diagonalizable at the eigenvalue \(2\): for example, \[ A-2I= \begin{pmatrix} 1&1&-1\\ 2&0&-1\\ 2&2&-2 \end{pmatrix} \] has rank \(2\), so the \(2\)-eigenspace is one-dimensional although \(2\) has algebraic multiplicity two. Hence the minimal polynomial is \[ m_A(x)=(x-1)(x-2)^2 \] [@HK71].
2. Construct the diagonalizable part as a polynomial in \(A\).
We seek a polynomial \(q\) that equals \(1\) at the eigenvalue \(1\), equals \(2\) at the eigenvalue \(2\), and is constant to first order on the repeated primary factor at \(2\). Thus impose \[ q(1)=1, \qquad q(2)=2, \qquad q'(2)=0. \] The unique polynomial of degree at most two satisfying these conditions is \[ q(x)=-x^2+4x-2. \] Set \[ D=q(A)=-A^2+4A-2I. \] Direct multiplication gives \[ A^2= \begin{pmatrix} 9&3&-4\\ 8&4&-4\\ 10&6&-4 \end{pmatrix}, \] and hence the displayed matrix for \(D\).
On the generalized \(1\)-eigenspace, \(D\) acts as multiplication by \(1\); on the generalized \(2\)-eigenspace, the congruence \[ q(x)\equiv2\pmod{(x-2)^2} \] shows that \(D\) acts as multiplication by \(2\). Therefore \(D\) is diagonalizable, with minimal polynomial dividing \((x-1)(x-2)\) [@HK71].
3. The remainder is nilpotent and commutes with \(D\).
Define \[ N=A-D=A^2-3A+2I=(A-I)(A-2I). \] This gives the displayed matrix for \(N\). Since \(D\) and \(N\) are polynomials in \(A\), they commute. Moreover, \[ N^2=(A-I)^2(A-2I)^2. \] The minimal polynomial \((A-I)(A-2I)^2=0\) implies \[ N^2=(A-I)\bigl((A-I)(A-2I)^2\bigr)=0. \] Thus \(N\) is nilpotent. Finally its definition gives \[ A=D+N, \] so the corresponding operators satisfy \[ \boxed{T=D+N,\qquad DN=ND,} \] with \(D\) diagonalizable and \(N\) nilpotent.