Rank of Hermitian matrices

Posted: October 2, 2012 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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 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. $\Box$