For let denote the complex conjugate of Recall that a matrix is called Hermitian if for all It is known that if is Hermitian, then is diagonalizable and every eigenvalue of is a real number. In this post, we will give a lower bound for the rank of a Hermitian matrix. In fact, the lower bound holds for any diagonalizable complex matrix whose eigenvalues are real numbers. To find the lower bound, we first need an easy inequality.
Problem 1. Prove that if then
Solution. We have for all and so
Adding the term to both sides of the above inequality will finish the job.
Problem 2. Prove that if is Hermitian, then
Solution. Let be the nonzero eigenvalues of Since is diagonalizable, we have We also have and Thus, by Problem 1,
and the result follows.