# Math Help - Diagonalization

1. ## Diagonalization

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

2. Originally Posted by Chandru1
Let $T$ be a linear operator on the n-dimensional vector space $V$, and suppose that $T$ has $n$ distinct characteristic values. Prove that $T$ 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.