Posts Tagged ‘Hermitian matrix’

For a \in \mathbb{C} let \overline{a} denote the complex conjugate of a. Recall that a matrix [a_{ij}] \in M_n(\mathbb{C}) is called Hermitian if a_{ij}=\overline{a_{ji}}, for all 1 \leq i,j \leq n. It is known that if A is Hermitian, then A is diagonalizable and every eigenvalue of A is a real number. In this post, we will 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 a_1, \ldots , a_m \in \mathbb{R}, then (a_1 + \ldots + a_m)^2 \leq m(a_1^2 + \ldots + a_m^2).

Solution.  We have a^2+b^2 \geq 2ab for all a,b \in \mathbb{R} and so

(m-1)\sum_{i=1}^m a_i^2=\sum_{1 \leq i < j \leq m}(a_i^2+a_j^2) \geq \sum_{1 \leq i < j \leq m}2a_ia_j.

Adding the term \sum_{i=1}^m a_i^2 to both sides of the above inequality will finish the job. \Box

Problem 2. Prove that if 0 \neq A \in M_n(\mathbb{C}) is Hermitian, then {\rm{rank}}(A) \geq ({\rm{tr}}(A))^2/{\rm{tr}}(A^2).

Solution. Let \lambda_1, \ldots , \lambda_m be the nonzero eigenvalues of A. Since A is diagonalizable, we have {\rm{rank}}(A)=m. We also have {\rm{tr}}(A)=\lambda_1 + \ldots + \lambda_m and {\rm{tr}}(A^2)=\lambda_1^2 + \ldots + \lambda_m^2. Thus, by Problem 1,

({\rm{tr}}(A))^2 \leq {\rm{rank}}(A) {\rm{tr}}(A^2)

and the result follows. \Box

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