Let \(T\) be the linear operator on \(\RR^3\) which is represented in the standard ordered basis by \[ \begin{bmatrix} 3 & -4 & -4 \\ -1 & 3 & 2 \\ 2 & -4 & -3 \end{bmatrix}. \] Find non-zero vectors \(\alpha_1, \ldots, \alpha_r\) satisfying the conditions of Theorem 3.
Solution
Goal. Find cyclic decomposition vectors for the operator \(T\) on \(\RR^3\).
1. Compute the characteristic polynomial of \(T\).
1.1. \(T = \begin{pmatrix} 3 & -4 & -4 \\ -1 & 3 & 2 \\ 2 & -4 & -3 \end{pmatrix}\).
given.
1.2. \(\chi_T(x) = \det(xI - T) = (x-1)^2(x+1)\).
compute the determinant of \(xI - T = \begin{pmatrix} x-3 & 4 & 4 \\ 1 & x-3 & -2 \\ -2 & 4 & x+3 \end{pmatrix}\), which factors as \((x-1)^2(x+1)\).
2. The minimal polynomial is \((x-1)(x+1)\) or \((x-1)^2(x+1)\).
2.1. Check if \((T - I)(T + I) = 0\).
compute \((T-I)(T+I) = T^2 - I\).
2.2. \(T^2 = \begin{pmatrix} 3 & -4 & -4 \\ -1 & 3 & 2 \\ 2 & -4 & -3 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I\).
direct computation.
2.3. Hence \(T^2 = I\), so the minimal polynomial is \(x^2 - 1 = (x-1)(x+1)\).
\(T^2 - I = 0\) and \(T \neq \pm I\).
3. \(T\) is diagonalizable with eigenvalues \(1\) (multiplicity \(2\)) and \(-1\) (multiplicity \(1\)).
the minimal polynomial \((x-1)(x+1)\) has distinct roots, so \(T\) is diagonalizable.
4. Find the cyclic decomposition.
4.1. The eigenspace for \(\lambda = 1\) is \(2\)-dimensional, and for \(\lambda = -1\) is \(1\)-dimensional.
multiplicities of the eigenvalues.
4.2. The cyclic decomposition is \(\RR^3 = Z(\alpha_1; T) \oplus Z(\alpha_2; T) \oplus Z(\alpha_3; T)\) where \(\alpha_1, \alpha_2\) are eigenvectors for \(\lambda = 1\) and \(\alpha_3\) is an eigenvector for \(\lambda = -1\).
since \(T\) is diagonalizable, the cyclic decomposition is the eigenspace decomposition, with each cyclic subspace spanned by an eigenvector.
4.3. Eigenvectors: for \(\lambda = 1\), solve \((T - I)\alpha = 0\); for \(\lambda = -1\), solve \((T + I)\alpha = 0\).
standard eigenvector computation.
5. Q.E.D.
the cyclic decomposition vectors are a basis of eigenvectors (two for \(\lambda = 1\), one for \(\lambda = -1\)), each generating a cyclic subspace.