If u and v are orthogonal, v t u = 0, so a 2 = (uv t) (uv t) = u (v t u) v t = 0. Let k be the eigenvalue of a, then k 2 = 0, so k=0, and all n eigenvalues of a are 0.

The rank of A is 1, so the basic solution system of the equation group Ax=0 has n- 1 vectors, that is, A has only n- 1 linearly independent eigenvectors belonging to n-fold eigenvalue 0, so A cannot be diagonalized.