- Show that if \(\theset{\vector v_i}\) is a basis for \(V\) where \(\dim(V) = n\) and \(T(\vector v_i) = \vector v_{i+1 \mod n}\) then \(T\) is diagonalizable with minimal polynomial \(x^n-1\).
Solution
1. Show that \(x^n - 1\) annihilates \(T\):
1.1. For each basis vector \(\mathbf{v}_i\) (\(0 \le i \le n - 1\)), the operator \(T\) satisfies:\[ T^n(\mathbf{v}_i) = \mathbf{v}_{i + n \pmod n} = \mathbf{v}_i = I(\mathbf{v}_i). \]
applying the permutation \(T\) \(n\) times.
1.2. Since \(T^n\) agrees with the identity operator \(I\) on a basis, \(T^n = I\), so \((T^n - I) = 0\).Thus the minimal polynomial \(m_T(x)\) divides \(x^n - 1\).
definition of minimal polynomial.
2. Show that \(\deg(m_T) = n\), so \(m_T(x) = x^n - 1\):
2.1. Consider the vector \(\mathbf{v}_0\). Its images under powers of \(T\) are:\[ \mathbf{v}_0, \, T(\mathbf{v}_0) = \mathbf{v}_1, \, T^2(\mathbf{v}_0) = \mathbf{v}_2, \, \dots, \, T^{n-1}(\mathbf{v}_0) = \mathbf{v}_{n-1}. \]These \(n\) vectors form the given basis of \(V\), hence are linearly independent.
hypothesis on basis.
2.2. If \(g(x) = \sum_{j=0}^d c_j x^j\) is a non-zero polynomial of degree \(d < n\) with \(g(T) = 0\), then:\[ 0 = g(T)(\mathbf{v}_0) = \sum_{j=0}^d c_j T^j(\mathbf{v}_0) = \sum_{j=0}^d c_j \mathbf{v}_j. \]By linear independence of \(\{\mathbf{v}_0, \dots, \mathbf{v}_{n-1}\}\), \(c_0 = c_1 = \cdots = c_d = 0\), a contradiction.
linear independence of basis vectors.
2.3. Thus no non-zero polynomial of degree \(< n\) annihilates \(T\), so \(\deg(m_T) \ge n\).Since \(m_T(x) \mid (x^n - 1)\) and \(\deg(x^n - 1) = n\), the monic minimal polynomial is:\[ m_T(x) = x^n - 1. \]
monic divisibility of equal degree polynomials.
3. Prove diagonalizability:
3.1. Over \(\mathbb{C}\), the polynomial \(x^n - 1\) factors as:\[ x^n - 1 = \prod_{k=0}^{n-1} (x - \omega^k), \quad \text{where } \omega = e^{2\pi i/n}. \]The \(n\) roots \(\omega^0, \omega^1, \dots, \omega^{n-1}\) are distinct.
distinct \(n\)-th roots of unity.
3.2. A linear operator is diagonalizable if and only if its minimal polynomial splits into distinct linear factors.Since \(m_T(x) = x^n - 1\) has \(n\) distinct linear factors, \(T\) is diagonalizable.
diagonalizability criterion via minimal polynomials.
3.3. Explicitly, for each \(k \in \{0, \dots, n-1\}\), the discrete Fourier vector:\[ \mathbf{w}_k = \sum_{j=0}^{n-1} \omega^{-jk} \mathbf{v}_j \]is a non-zero eigenvector satisfying \(T(\mathbf{w}_k) = \omega^k \mathbf{w}_k\), providing an eigenbasis \(\{\mathbf{w}_0, \dots, \mathbf{w}_{n-1}\}\) for \(V\).
\(T(\mathbf{w}_k) = \sum \omega^{-jk} \mathbf{v}_{j+1} = \omega^k \sum \omega^{-(j+1)k} \mathbf{v}_{j+1} = \omega^k \mathbf{w}_k\).
4. Conclusion:\(T\) is diagonalizable with minimal polynomial \(m_T(x) = x^n - 1\). Q.E.D.
2 and 3.