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. 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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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).

But

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

and

\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