Let \(a\), \(b\), and \(c\) be elements of a field \(F\), and let \(A\) be the following \(3 \times 3\) matrix over \(F\): \[ A = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix}. \] Prove that the characteristic polynomial for \(A\) is \(x^3 - ax^2 - bx - c\) and that this is also the minimal polynomial for \(A\).
Solution
1. Computation of the characteristic polynomial:
1.1. The characteristic matrix \(xI - A\) is:\[ xI - A = \begin{pmatrix} x & 0 & -c \\ -1 & x & -b \\ 0 & -1 & x - a \end{pmatrix}. \]
matrix subtraction \(xI_3 - A\).
1.2. Compute the determinant \(\det(xI - A)\) by cofactor expansion along the first row:\[ \det(xI - A) = x \cdot \det\begin{pmatrix} x & -b \\ -1 & x - a \end{pmatrix} - c \cdot \det\begin{pmatrix} -1 & x \\ 0 & -1 \end{pmatrix}. \]
Laplace expansion along row 1.
1.3. Evaluating the \(2 \times 2\) determinants:\[ \det(xI - A) = x\big(x(x - a) - b\big) - c(1 - 0) = x(x^2 - ax - b) - c = x^3 - ax^2 - bx - c. \]Thus the characteristic polynomial is \(p(x) = x^3 - ax^2 - bx - c\).
polynomial expansion.
2. Proof that the minimal polynomial is \(p(x)\):
2.1. By the Cayley–Hamilton Theorem, the minimal polynomial \(m(x)\) divides the characteristic polynomial \(p(x)\), so \(\deg(m) \le 3\).
Cayley–Hamilton Theorem.
2.2. Consider the first standard basis vector \(e_1 = (1, 0, 0)^T \in F^3\).Compute the successive applications of \(A\):\[ A e_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = e_2, \qquad A^2 e_1 = A e_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = e_3. \]
matrix-vector multiplication.
2.3. The set \(\{e_1, A e_1, A^2 e_1\} = \{e_1, e_2, e_3\}\) is the standard basis of \(F^3\), hence linearly independent over \(F\).
standard unit vectors are linearly independent.
2.4. For any non-zero polynomial \(g(x) = d_0 + d_1 x + d_2 x^2 \in F[x]\) of degree at most 2:\[ g(A) e_1 = d_0 e_1 + d_1 A e_1 + d_2 A^2 e_1 = (d_0, d_1, d_2)^T \neq (0, 0, 0)^T. \]Thus \(g(A) \neq 0\) for all non-zero polynomials of degree \(< 3\).
linear independence of \(\{e_1, Ae_1, A^2e_1\}\).
2.5. Therefore \(\deg(m) \ge 3\).Since \(m(x)\) is monic, divides \(p(x)\), and has degree 3, we must have \(m(x) = p(x) = x^3 - ax^2 - bx - c\).
uniqueness of monic divisor of equal degree.
3. Conclusion:Both the characteristic polynomial and minimal polynomial of \(A\) equal \(x^3 - ax^2 - bx - c\). Q.E.D.
1 and 2.