Independence of eigenvectors for distinct eigenvalues (${\prod }_{j\ne k}(M-{\lambda }_{j}I){\sum }_i{c}_i{e}_i=0$)

Applying the product of the other eigenvalue factors to a linear relation isolates each coefficient. Distinct eigenvalues therefore give linearly independent eigenvectors. A complete eigenvector basis diagonalizes the associated linear system of ordinary differential equations.