Posts Tagged ‘rank’

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

Problem. Let A, B be m \times n and n \times k matrices, respectively, with entries in some field. Prove that

\text{rank}(AB) \geq \text{rank}(A)+\text{rank}(B) - n.

Solution. Let T_1,T_2 be the corresponding linear transformations.

Claim. \text{nul}(T_1T_2) \leq \text{nul}(T_1) + \text{nul}(T_2).

Proof of the claim. Define f: \ker (T_1T_2) \to \ker T_1 by f(x)=T_2(x) for all x \in \ker(T_1T_2). See that f is well-defined, linear  and \ker f = \ker T_2. So, by the rank-nullity theorem

\text{rank}(f)+ \text{nul}(f)=\dim \ker(T_1T_2)=\text{nul}(T_1T_2).


\text{rank}(f)=\dim \text{im}(f) \leq \dim \ker T_1=\text{nul}(T_1)


\text{nul}(f)=\dim \ker f=\dim \ker T_2 = \text{nul}(T_2). \Box

Now applying the claim and the rank-nullity theorem gives

\text{rank}(T_1T_2)+n=k- \text{nul}(T_1T_2)+n \geq n- \text{nul}(T_1)+ k- \text{nul}(T_2)= \text{rank}(T_1) + \text{rank}(T_2). \ \Box

Problem. Let F be a field. Let A,B be two n \times n matrices with entries in F. Suppose that AB=BA=0 and \text{rank}(A)=\text{rank}(A^2). Prove that \text{rank}(A+B)=\text{rank}(A) + \text{rank}(B).

Solution. Let V=F^n and supppose that T,S : V \longrightarrow V are the corresponding linear transformations defined by A and B respectively. Let K_1=\ker T, K_2=\ker S and W_1=\text{im}(T). Note that, since \text{rank}(T)=\text{rank}(T^2) and \ker T \subseteq \ker T^2, we have \ker T = \ker T^2 by the rank-nullity theorem and thus K_1 \cap W_1={0}. As a result K_1 \oplus W_1=V. Also, since ST=0, we have W_1 \subseteq K_2. Therefore, by the rank-nullity theorem, K_1 + K_2 \supseteq K_1 \oplus W_1=V and hence

 \dim (K_1 + K_2)=n. \ \ \ \ \ \ \ \ \ (1)

Now let v \in \ker (T+S). Then Tv=-Sv and therefore T^2v=-TSv=0, i.e. v \in \ker T^2=K_1. Hence Tv=0 and so Sv=-Tv=0. Thus

 \ker (T+S) = K_1 \cap K_2. \ \ \ \ \ \ (2)

Finally, (1), (2) and the rank-nullity theorem give us

n=\dim(K_1+K_2)=\dim K_1 + \dim K_2 - \dim(K_1 \cap K_2) 

=2n-\text{rank}(T)-\text{rank}(S) - (n - \text{rank}(T+S))

= n - \text{rank}(T) - \text{rank}(S) + \text{rank}(T+S). \ \Box