## 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 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. $\Box$

## Rank of the product of two matrices

Posted: September 9, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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

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

## Rank of sum of two matrices

Posted: May 13, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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

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

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

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