91Ó°ÊÓ

Assume that V has a basis of eigenvectors of T. Let B = {v1, v2, ..., vn} be such a basis. Since B is a basis, we can write any vector v ∈ V as a linear combination of the basis vectors: \[v = c_1v_1 + c_2v_2 + ... + c_nv_n\] Now, let p(x) be the minimal polynomial of T. We want to show that p(x) has no repeated roots. Let λ be a root of p(x). Then, p(T)(v) = 0 for all v ∈ V. Since T is a linear operator, we have: \[T(v) = T(c_1v_1 + c_2v_2 + ... + c_nv_n) = c_1Tv_1 + c_2Tv_2 + ... + c_nTv_n\] As B is a basis of eigenvectors, then \[T(v_i) = λ_iv_i\] where λ_i is the eigenvalue corresponding to eigenvector v_i. Now, \(p(T)v = 0\), so, \[p(T)(c_1v_1 + c_2v_2 + ... + c_nv_n) = 0\] Since p(λ) is a root, this equation must hold for all linear combinations. Therefore, \(p(T)v_i = 0\) for all i. Hence, p(λ_i) = 0. Since p(x) is the minimal polynomial, all its roots must be distinct. Thus, p(x) has no repeated roots.

Assume that the minimal polynomial p(x) of T has no repeated roots. Let B be a basis for V whose vectors are eigenvectors of T, and let λ be an eigenvalue of T with corresponding eigenvector v. Then, \(T(v) = λv\). Since p(x) has no repeated roots, there exists a polynomial q(x) such that \[p(x) = (x-λ_1)(x-λ_2)...(x-λ_n)\] where - \(\{λ_1, λ_2, ..., λ_n\}\) are the distinct eigenvalues of T. We claim that there exists a vector w ∈ V such that \(w = c_1v_1 + c_2v_2 + ... + c_nv_n\) where c_i ≠ 0 for at least one i. Apply T to w: \[T(w) = T(c_1v_1 + c_2v_2 + ... + c_nv_n) = c_1Tv_1 + c_2Tv_2 + ... + c_nTv_n\] As B is a basis of eigenvectors, we must have: \[T(w) = c_1λ_1v_1 + c_2λ_2v_2 + ... + c_nλ_nv_n\] Now, consider the polynomial r(x) = p(x)/q(x). r(x) is the minimal polynomial of T. Therefore, r(T)w = 0: \[r(T)(c_1v_1 + c_2v_2 + ... + c_nv_n) = 0\] Hence, \(r(T)v_i = 0\) for all eigenvalues λ_i of T. Thus, with the assumption that the minimal polynomial of T has no repeated roots, V has a basis of eigenvectors of T. This completes our proof.