Prove or give a counterexample if T V is an operator on a fi

Prove or give a counterexample if T (V) is an operator on a finite-dimensional complex inner product space, then there is an orthonormal basis of V consisting of generalized eigenvectors of T.

Solution

By definition a generalized eigenvector of a complex matrix is a non-zero vector satisfying

                                         (A-cI)^k v=0 for some k.

The generalized eigenvectors of A form a basis for V.

Given any set S of linearly independent vectors (any number of them), the Gram -Schmidt process produces the same number of orthonormal vectors spanning the same subspace as S.

Let S be the set of generalized eigenvectors of A forming a basis of V. Applying the above remark, we obtain an orthonormal basis for V.