Let \(T\) be the linear operator on \(\RR^8\) which is represented in the standard basis by the matrix \[ A = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 \\ 0 & -1 & -1 & -1 & -1 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}. \]
Find the characteristic polynomial and the invariant factors.
Find the primary decomposition of \(\RR^8\) under \(T\) and the projections on the primary components. Find cyclic decompositions of each primary component as in Theorem 3.
Find the Jordan form of \(A\).
Find a direct-sum decomposition of \(\RR^8\) into \(T\)-cyclic subspaces as in Theorem 3. (Hint: One way to do this is to use the results in (b) and an appropriate generalization of the ideas discussed in Example 4.)
Solution
Put \(N_0=T\) and \(N_1=T-I\). Direct calculation gives \[ \boxed{\chi_T(x)=x^4(x-1)^4, \qquad m_T(x)=x^2(x-1)^2.} \] Thus the invariant factors are \[ \boxed{x^2(x-1)^2,\quad x^2(x-1)^2.} \]
1. Characteristic polynomial and invariant factors.
For the displayed matrix \(A\), expansion or polynomial row reduction gives \[ \det(xI-A)=x^4(x-1)^4. \] Moreover direct multiplication gives \[ A^2(A-I)^2=0, \] while neither \(A(A-I)^2\) nor \(A^2(A-I)\) is zero. Hence \(m_T=x^2(x-1)^2\). Equivalently, both eigenvalues \(0\) and \(1\) have largest Jordan block of size two.
There are four dimensions in each primary component and the largest block in each has size two. The explicit cyclic decompositions in step 3 below show that each primary component is the sum of two size-two cyclic blocks. Pairing the two \(x^2\) blocks with the two \((x-1)^2\) blocks yields precisely two invariant factors, both \(x^2(x-1)^2\) [@HK71].
2. Primary decomposition and projections.
The primary components are \[ W_0=\ker A^2, \qquad W_1=\ker(A-I)^2. \] One computes \[ W_0=\operatorname{span}\{e_2-e_8,e_3-e_8,e_4-e_8,e_5-e_8\}, \] and \[ W_1=\operatorname{span}\{e_1,e_5,e_6,e_7\}. \] Both have dimension four, their intersection is zero because their annihilating polynomials are relatively prime, and hence \[ \mathbb R^8=W_0\oplus W_1. \]
As in Exercise 7.4.4, let \[ e_0(x)=1-3x^2+2x^3, \qquad e_1(x)=3x^2-2x^3. \] These satisfy the required CRT congruences modulo \(x^2\) and \((x-1)^2\), so the primary projections are \[ E_0=e_0(A),\qquad E_1=e_1(A). \] Explicitly, \[ E_0= \begin{pmatrix} 0&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&-1&-1&-1&0&0&0&-1\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&1 \end{pmatrix}, \] and \(E_1=I-E_0\).
3. Cyclic decompositions of the primary components.
For \(W_0\), set \[ u_1=e_2-e_8, \qquad u_2=e_4-e_8. \] Then \[ Au_1=-e_2+e_3, \] and \[ Au_2=-e_2+e_3-e_4+e_5. \] Both images are nonzero and lie in \(\ker A\), and they are linearly independent. Therefore \[ W_0=Z(u_1;T)\oplus Z(u_2;T), \] where each cyclic summand has basis \((u_i,Au_i)\) and annihilator \(x^2\).
For \(W_1\), set \[ v_1=e_5, \qquad v_2=e_6. \] Then \[ (A-I)v_1=e_1+e_6-e_7, \qquad (A-I)v_2=e_1. \] These two images are nonzero and linearly independent in \(\ker(A-I)\). Hence \[ W_1=Z(v_1;T)\oplus Z(v_2;T), \] with bases \((v_i,(A-I)v_i)\) and annihilator \((x-1)^2\) for each summand.
4. Jordan form.
The four cyclic chains from step 3 give two Jordan blocks of size two at eigenvalue zero and two of size two at eigenvalue one. Thus \[ \boxed{J(A)=J_2(0)\oplus J_2(0)\oplus J_2(1)\oplus J_2(1).} \] This agrees with both the characteristic and minimal polynomials found in step 1.
5. A direct sum into \(T\)-cyclic subspaces.
Because \(x^2\) and \((x-1)^2\) are coprime, the sum of a cyclic vector from a zero-primary block and one from a one-primary block is cyclic for the direct sum of those two blocks [@HK71]. Put \[ \alpha_1=u_1+v_1=e_2+e_5-e_8, \qquad \alpha_2=u_2+v_2=e_4+e_6-e_8. \] Then \[ Z(\alpha_1;T)=Z(u_1;T)\oplus Z(v_1;T), \] and similarly for \(\alpha_2\). Each has annihilator \(x^2(x-1)^2\), hence dimension four. Since the four primary cyclic blocks in step 3 were mutually independent and exhaust \(V\), \[ \boxed{\mathbb R^8=Z(\alpha_1;T)\oplus Z(\alpha_2;T).} \] These two annihilators are exactly the two invariant factors stated in step 1.