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Math Help - Problem of rank

  1. #1
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    Problem of rank

    Assume A,B \in M_n(\mathbb{K})

    show that:

    If AB=BA=0 and rank(A^2)=rank(A)

    then rank(A+B)=rank(A)+rank(B)
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  2. #2
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    Quote Originally Posted by Xingyuan View Post
    Assume A,B \in M_n(\mathbb{K})

    show that:

    If AB=BA=0 and rank(A^2)=rank(A)

    then rank(A+B)=rank(A)+rank(B)
    let V=\mathbb{K}^n and T,S : V \longrightarrow V be the corresponding linear transformations defined by Tv=Av, \ Sv = Bv. let's put K_1=\ker T, \ K_2=\ker S, and W_1=\text{im}(T).

    since ST=0, we have W_1 \subseteq K_2. thus, using the rank nullity theore, we have K_1 + K_2 \supseteq K_1 \oplus W_1=V and hence \dim (K_1+K_2)=n. \ \ \ \ \ \ (1)

    on the other hand, since \text{rank}(T^2)=\text{rank}(T) and \ker T \subseteq \ker T^2, we have, again by the rank-nullity theorem, \ker T^2=\ker T = K_1. now suppose v \in \ker (T+S).

    then Tv=-Sv and thus T^2v=-TSv=0, i.e. v \in \ker T^2=K_1. hence Tv=0 and therefore Sv=-Tv=0. so \ker(T+S)=K_1 \cap K_2. \ \ \ \ \ (2)

    finally, using (1) and (2) and the rank-nullity theorem, completing the proof is easy:

    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) - \dim(\ker(T+S))

    =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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