Results 1 to 2 of 2

Thread: Problem of rank

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    66

    Problem of rank

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

    show that:

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

    then $\displaystyle rank(A+B)=rank(A)+rank(B)$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Xingyuan View Post
    Assume $\displaystyle A,B \in M_n(\mathbb{K})$

    show that:

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

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

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

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

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

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

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

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

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

    $\displaystyle =n -\text{rank}(T)-\text{rank}(S) + \text{rank}(T+S). \ \ \ \Box$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Aug 20th 2010, 05:32 AM
  2. rank sum problem
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Apr 12th 2010, 05:20 PM
  3. Problem regarding rank
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jul 6th 2009, 10:02 PM
  4. Short proof that rows-rank=column-rank?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Jun 26th 2009, 10:02 AM
  5. rank problem.....
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Apr 23rd 2009, 06:36 PM

Search Tags


/mathhelpforum @mathhelpforum