For arbitrary \(a,b,c\) in a field \(F\), compute the minimal polynomial of \[ \begin{pmatrix} 0&0&a\\ 1&0&b\\ 0&1&c \end{pmatrix}. \]
Solution
Let \[ A \coloneqq \begin{pmatrix} 0&0&a\\ 1&0&b\\ 0&1&c \end{pmatrix}, \] and let \(e_1,e_2,e_3\) be the standard basis of \(F^3\).
1. The vector \(e_1\) is cyclic for \(A\):\[ e_1, \qquad Ae_1=e_2, \qquad A^2e_1=e_3 \]form a basis of \(F^3\).
The first column of \(A\) is \(e_2\), so \(Ae_1=e_2\). The second column of \(A\) is \(e_3\), so \[ A^2e_1=Ae_2=e_3. \] Thus the three displayed vectors are exactly the standard basis.
2. One has\[ A^3e_1 = a e_1+bAe_1+cA^2e_1. \]
By step 1, \[ A^3e_1=Ae_3. \] The third column of \(A\) is \[ a e_1+b e_2+c e_3. \] Substitute \(e_2=Ae_1\) and \(e_3=A^2e_1\).
3. The polynomial\[ q(t)\coloneqq t^3-ct^2-bt-a \]annihilates \(A\).
Step 2 says \[ q(A)e_1=0. \] Since \(q(A)\) commutes with \(A\), \[ q(A)Ae_1=Aq(A)e_1=0 \] and \[ q(A)A^2e_1=A^2q(A)e_1=0. \] By step 1, the vectors \(e_1,Ae_1,A^2e_1\) form a basis. Hence \(q(A)\) vanishes on a basis and therefore \[ q(A)=0. \]
4. No nonzero polynomial of degree at most \(2\) annihilates \(A\).
Suppose \[ p(t)=u+vt+wt^2 \] satisfies \(p(A)=0\). Applying this operator to \(e_1\) gives \[ u e_1+vAe_1+wA^2e_1=0. \] The three vectors are linearly independent by step 1, so \[ u=v=w=0. \] Thus \(p=0\).
5. The minimal polynomial of \(A\) is\[ \boxed{ m_A(t)=t^3-ct^2-bt-a }. \]
Step 3 gives a monic degree-\(3\) annihilating polynomial. Step 4 shows that no nonzero annihilating polynomial has smaller degree. By the definition of the minimal polynomial, it must therefore equal \(q(t)\).
6. Q.E.D.
Step 5 computes the requested minimal polynomial for arbitrary \(a,b,c\in F\).