Diagonalizable matrix proof using minimal polynomial

[rideabike]

Homework Statement

A matrix A$\in$M_n(ℂ) is diagonalizable if and only if m_A(x) has no repeated roots.

Homework Equations

If A$\in$Hom(V,V) = {A:V→V | A is a linear map}, the minimal polynomial of A, m_A(x), is the smallest degree monic polynomial f(x) such that f(A)=0.

The Attempt at a Solution

In one direction, want to prove if m_A(x) has repeated roots, then there are not n linearly independent eigenvectors in A (with A n X n)

 

[lippyka]

so A is not diagonalizable.Let A\inMn(ℂ). Assume mA(x) has repeated roots, so there exists a polynomial g(x) such that g(A)=0 where g(x) is a factor of mA(x). Since g(A)=0, then it follows that ker(g(A))≠{0}, which implies that there are at least n-1 linearly independent eigenvectors in A (with A n X n). So A is not diagonalizable. In the other direction, want to prove if there are n linearly independent eigenvectors in A, then mA(x) has no repeated roots.Let A\inMn(ℂ). Assume there are n linearly independent eigenvectors in A, so ker(mA(A))={0}. This implies that mA(x) has no repeated roots. So A is diagonalizable.