Question

An n x n matrix with fewer than n distinct eigen values is not diagonalizable

Answer 1

The specified matrix will be non-diagonalized if it does not have n linearly independent eigenvectors. The characteristic polynomial will mark multiple roots - repeated eigenvalues of the matrix. Thus, the eigenvectors of such a matrix will have linearly combined scalar components indicating the "stretch" of the eigenvector during the linear transformation. Such a matrix will be defective.