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 give a lower bound for the rank of a Hermitian matrix. 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.

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