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

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Comments
  1. Mary says:

    I was looking for this, thanks! 🙂

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