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.
Solution
Choose nonzero eigenvectors \(v_i\) with \(Tv_i=c_i v_i\), where \(c_1,\ldots,c_n\) are the distinct characteristic values. Eigenvectors belonging to distinct eigenvalues are linearly independent: if a shortest nontrivial relation \(\sum_{i=1}^r a_i v_i=0\) existed, applying \(T-c_rI\) would give \[ \sum_{i=1}^{r-1}a_i(c_i-c_r)v_i=0, \] a shorter nontrivial relation, since \(c_i-c_r\ne0\). Thus \(v_1,\ldots,v_n\) are linearly independent. Since \(\dim V=n\), they form a basis, and the matrix of \(T\) in this basis is diagonal. Hence \(T\) is diagonalizable.