Results 1 to 4 of 4

Math Help - Rank f+g

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Rank f+g

    Let f,g:X\to Y be a linear transformation. Prove that \text{rank}(f+g)\leq \text{rank} \ f+\text{rank} \ g

    Is this done by using that fact that \text{dim}(X+Y)=\text{dim}X+\text{dim}Y-\text{dim}(X\cap Y) since \text{rank}f\leq\text{dim}X\mbox{?}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Rank f+g

    Quote Originally Posted by dwsmith View Post
    Let f,g:X\to Y be a linear transformation. Prove that \text{rank}(f+g)\leq \text{rank} \ f+\text{rank} \ g

    Is this done by using that fact that \text{dim}(X+Y)=\text{dim}X+\text{dim}Y-\text{dim}(X\cap Y) since \text{rank}f\leq\text{dim}X\mbox{?}
    Note that if y\in \text{im }f+g then y=(f+g)(x)=f(x)+g(x)\in\text{im }f+\text{im }g and so \text{im }(f+g)\subseteq\text{im }f+\text{im }g. Thus, \text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Re: Rank f+g

    Quote Originally Posted by Drexel28 View Post
    Note that if y\in \text{im }f+g then y=(f+g)(x)=f(x)+g(x)\in\text{im }f+\text{im }g and so \text{im }(f+g)\subseteq\text{im }f+\text{im }g. Thus, \text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g).
    Can you explain the top line in more detail?

    Why can say \text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g)?

    Go from looking at rank and then say it is less than the dim and back to rank?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Rank f+g

    Quote Originally Posted by dwsmith View Post
    Can you explain the top line in more detail?

    Why can say \text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g)?

    Go from looking at rank and then say it is less than the dim and back to rank?
    Does it make more sense to recall that \text{rank}(f)=\text{dim }\text{im }f?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Row Rank = Column Rank?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 9th 2011, 01:10 AM
  2. Proof: rank(AB)+n >= rank(A)+rank(B)
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: September 9th 2010, 06:28 PM
  3. Replies: 3
    Last Post: August 20th 2010, 06:32 AM
  4. Row rank and column rank
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 13th 2010, 08:40 AM
  5. Short proof that rows-rank=column-rank?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: June 26th 2009, 11:02 AM

Search Tags


/mathhelpforum @mathhelpforum