Let \(A \in M_n(k)\) be an \(n \times n\) matrix over a field \(k\).
Suppose that \(v \in k^n\) is a column vector such that the set \(\{v, A v, \dots, A^{n-1} v\}\) is linearly independent. Show that every matrix \(B \in M_n(k)\) that commutes with \(A\) (\(A B = B A\)) is a polynomial in \(A\) with coefficients in \(k\).
Show that there exists a vector \(v \in k^n\) such that \(\{v, A v, \dots, A^{n-1} v\}\) is linearly independent if and only if the characteristic polynomial \(\chi_A(x)\) equals the minimal polynomial \(m_A(x)\).
Solution
Goal: Prove that the centralizer of a cyclic operator is the polynomial algebra \(k[A]\) in (a), and characterize cyclicity via \(\chi_A(x) = m_A(x)\) in (b).
1. Part (a): Any commuting matrix \(B\) is a polynomial in \(A\).
1.1. Since \(\{v, A v, \dots, A^{n-1} v\}\) contains \(n\) linearly independent vectors in \(k^n\), it forms a basis \(\mathcal{B} = \{v, A v, \dots, A^{n-1} v\}\) for \(k^n\).
1.2. Since \(\mathcal{B}\) spans \(k^n\), the vector \(B v \in k^n\) can be expressed as a linear combination of basis vectors:\[B v = \sum_{j=0}^{n-1} c_j A^j v = p(A) v,\]where \(p(x) = \sum_{j=0}^{n-1} c_j x^j \in k[x]\).
1.3. Because \(A B = B A\), it follows by induction that \(A^k B = B A^k\) for every integer \(k \ge 0\).
1.4. Evaluate the action of \(B\) on each basis vector \(A^k v\) (\(0 \le k \le n - 1\)):\[B (A^k v) = (B A^k) v = (A^k B) v = A^k (B v) = A^k (p(A) v) = p(A) (A^k v),\]using that \(A^k\) and \(p(A)\) commute.
1.5. Since the linear maps \(B\) and \(p(A)\) agree on all \(n\) basis vectors of \(\mathcal{B}\), they are equal as linear transformations on \(k^n\).
1.6. Thus \(B = p(A)\), so \(B\) is a polynomial in \(A\).
2. Part (b) (\(\implies\)): Linear independence implies \(\chi_A(x) = m_A(x)\).
2.1. Suppose \(\{v, A v, \dots, A^{n-1} v\}\) is linearly independent in \(k^n\).
2.2. If \(\deg m_A(x) = d < n\), write \(m_A(x) = \sum_{j=0}^d a_j x^j\) with \(a_d = 1\).
2.3. Then \(0 = m_A(A) v = \sum_{j=0}^d a_j A^j v\).
2.4. Since \(d < n\), this is a non-trivial linear combination of a subset of \(\{v, A v, \dots, A^{n-1} v\}\) equaling zero, contradicting linear independence.
2.5. Thus \(\deg m_A(x) \ge n\).
2.6. By the Cayley–Hamilton Theorem, \(m_A(x) \mid \chi_A(x)\).
2.7. Since \(\deg \chi_A(x) = n\) and both \(m_A(x)\) and \(\chi_A(x)\) are monic polynomials of degree \(n\), we conclude \(\chi_A(x) = m_A(x)\).
3. Part (b) (\(\impliedby\)): \(\chi_A(x) = m_A(x)\) implies existence of a cyclic vector.
3.1. Regard \(V = k^n\) as a finitely generated module over the principal ideal domain \(k[x]\), where the action is given by \(x \cdot w = A w\) for \(w \in V\).
3.2. By the Structure Theorem for finitely generated modules over a PID (Invariant Factor Decomposition), there exist monic polynomials \(d_1(x) \mid d_2(x) \mid \dots \mid d_r(x)\) in \(k[x]\) such that\[V \cong \bigoplus_{i=1}^r \frac{k[x]}{\langle d_i(x) \rangle}\]as \(k[x]\)-modules.
3.3. Under this decomposition:- The characteristic polynomial is the product of all invariant factors: \(\chi_A(x) = \prod_{i=1}^r d_i(x)\).- The minimal polynomial is the largest invariant factor: \(m_A(x) = d_r(x)\).
3.4. The hypothesis \(\chi_A(x) = m_A(x)\) implies \(\prod_{i=1}^r d_i(x) = d_r(x)\).
3.5. Since each \(d_i(x)\) is monic of degree \(\ge 1\), this equality forces \(r = 1\) and \(d_1(x) = \chi_A(x)\).
3.6. Therefore, \(V \cong k[x]/\langle \chi_A(x) \rangle\) is a cyclic \(k[x]\)-module generated by the element \(\bar{1} = 1 + \langle \chi_A(x) \rangle\).
3.7. Let \(v \in V\) be the image of \(\bar{1}\) under the isomorphism.
3.8. In the quotient ring \(k[x]/\langle \chi_A(x) \rangle\), the elements \(\{\bar{1}, \bar{x}, \bar{x}^2, \dots, \bar{x}^{n-1}\}\) form a \(k\)-vector space basis because \(\deg \chi_A(x) = n\).
3.9. Applying the \(k[x]\)-module isomorphism, the corresponding elements \(\{v, A v, A^2 v, \dots, A^{n-1} v\}\) form a \(k\)-basis for \(V = k^n\), and are in particular linearly independent.
4. Conclusion:
The matrix $B$ is a polynomial in $A$, and a cyclic vector exists if and only if $\chi_A(x) = m_A(x)$.