## 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

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

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

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