Posts Tagged ‘trace of a 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. 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 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