Let \(T\) be the linear operator on \(F^4\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} c & 0 & 0 & 0 \\ 1 & c & 0 & 0 \\ 0 & 1 & c & 0 \\ 0 & 0 & 1 & c \end{bmatrix}. \] Let \(W\) be the null space of \(T - cI\).
Prove that \(W\) is the subspace spanned by \(\epsilon_4\).
Find the monic generators of the ideals \(S(\epsilon_4; W)\), \(S(\epsilon_3; W)\), \(S(\epsilon_2; W)\), \(S(\epsilon_1; W)\).
Solution
Put \[ N=T-cI. \] Then the given matrix shows \[ N\epsilon_1=\epsilon_2, \qquad N\epsilon_2=\epsilon_3, \qquad N\epsilon_3=\epsilon_4, \qquad N\epsilon_4=0. \]
1. The null space \(W\) is spanned by \(\epsilon_4\).
For \[ v=a_1\epsilon_1+a_2\epsilon_2+a_3\epsilon_3+a_4\epsilon_4, \] one has \[ Nv=a_1\epsilon_2+a_2\epsilon_3+a_3\epsilon_4. \] Thus \(Nv=0\) if and only if \(a_1=a_2=a_3=0\). Hence \[ \boxed{W=\ker(T-cI)=\operatorname{span}\{\epsilon_4\}.} \]
2. Membership in \(S(\epsilon_j;W)\) is controlled by the vanishing order at \(c\).
By definition, \[ S(\alpha;W)=\{f\in F[x]:f(T)\alpha\in W\} \] [@HK71]. Write any polynomial uniquely as \[ f(x)=a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+(x-c)^4q(x). \] Since \(N^4=0\), this gives \[ f(T)=a_0I+a_1N+a_2N^2+a_3N^3. \] Applying this to the four chain vectors yields \[ \begin{aligned} f(T)\epsilon_4&=a_0\epsilon_4,\\ f(T)\epsilon_3&=a_0\epsilon_3+a_1\epsilon_4,\\ f(T)\epsilon_2&=a_0\epsilon_2+a_1\epsilon_3+a_2\epsilon_4,\\ f(T)\epsilon_1&=a_0\epsilon_1+a_1\epsilon_2+a_2\epsilon_3+a_3\epsilon_4. \end{aligned} \] Because \(W=\operatorname{span}\{\epsilon_4\}\), these vectors lie in \(W\) respectively with no restriction, with \(a_0=0\), with \(a_0=a_1=0\), and with \(a_0=a_1=a_2=0\).
3. Read off the four monic generators.
The conditions in step 2 are exactly divisibility by successive powers of \(x-c\). Therefore \[ \boxed{ S(\epsilon_4;W)=(1),\qquad S(\epsilon_3;W)=(x-c),\qquad S(\epsilon_2;W)=((x-c)^2),\qquad S(\epsilon_1;W)=((x-c)^3).} \] Thus the requested monic generators are \[ \boxed{1,\ x-c,\ (x-c)^2,\ (x-c)^3.} \]