91Ó°ÊÓ

Since A ∈ M_{n × n}(F) has n distinct eigenvalues, let them be λ1, λ2, ..., λn. Now, for each eigenvalue λi, find the eigenvectors vi corresponding to this eigenvalue by solving (A - λiI)vi = 0, where I is the identity matrix.

As we have n eigenvectors (v1, v2, ..., vn) corresponding to the n distinct eigenvalues, let's check if the eigenvectors are linearly independent. To do this, consider the linear combination: c1v1 + c2v2 + ... + cnvn = 0. (1) Our goal is to show that the only solution for equation (1) is when all the coefficients are zero (c1 = c2 = ... = cn = 0).

Multiply both sides of equation (1) by \(A - λ_nI\): \(c1(A - λ_nI)v1 + c2(A - λ_nI)v2 + ... + cn(A - λ_nI)vn = 0\) Since \(vi\) is an eigenvector for the eigenvalue \(λ_i\), we have \((A - λ_iI)vi = 0\). Therefore, we can simplify the equation as: \(c1(λ_1 - λ_n)v1 + c2(λ_2 - λ_n)v2 + ... + cn(λ_n - λ_n)vn = 0\) Rearranging the terms, we get: \(c1(λ_1 - λ_n)v1 + c2(λ_2 - λ_n)v2 + ... + cn(0)vn = 0\) Now, multiply both sides of equation (1) by \(A - λ_{n - 1}I\): \(c1(λ_1 - λ_{n - 1})v1 + c2(λ_2 - λ_{n - 1})v2 + ... + cn(λ_n - λ_{n - 1})vn = 0\) By subtracting the last two equations, we get: \(c1((λ_1 - λ_n) - (λ_1 - λ_{n - 1}))v1 + c2((λ_2 - λ_n) - (λ_2 - λ_{n - 1}))v2 + ... = 0\) Simplifying, we get: \(c1v1 + c2v2 + ... = 0\) Since the eigenvalues are distinct, we can continue subtracting these equations (up to n - 1 of them) until we deduce c1 = c2 = ... = cn = 0. Therefore, the eigenvectors are linearly independent.

Since we have n linearly independent eigenvectors, we can form a matrix P using these eigenvectors, such that each column of P represents an eigenvector [v1, v2, ..., vn]. Now, let's define a diagonal matrix D, where D_{ii} = λi. Then, using our analysis, we can write A as: \(A = PDP^{-1}\) Therefore, A is diagonalizable since P is an invertible matrix and D is a diagonal matrix.