How to show that any rank one matrix is diagonalizable?
I've no idea how to start with.
An n by n matrix is "diagonalizable" if and only if there exist a basis for consisting entirely of eigenvectors of the matrix. If the matrix has "rank 1" then it has one non-zero eigenvalue. An eigenvector corresponding to that eigenvalue will span a one dimensional subspace. The matrix, applied to any vector in the n-1 dimensional orthogonal subspace, will give 0 so any basis for that subspace will be a basis consisting of eigenvectors corresponding to eigenvalue 0. Adding the first eigenvector to that will give a basis for the entire space consisting of eigenvectors.