Thread: Diagonalization

Let be a linear operator on the n-dimensional vector space , and suppose that has distinct characteristic values. Prove that is diagonalizable

Originally Posted by Chandru1

Let be a linear operator on the n-dimensional vector space , and suppose that has distinct characteristic values. Prove that is diagonalizable

T is diagonalisable if V has a basis consisting of eigenvectors of T. But eigenvectors associated with distinct eigenvalues are linearly independent. If all n eigenvalues are distinct then there must be n linearly independent eigenvectors, and they form a basis.