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Hello,I have a question. If A and B are NND matrices, how to prove

  1. Sep 1, 2011 #1
    Hello,

    I have a question. If A and B are NND matrices, how to prove C(A) belongs to C(A+B)?

    I can prove that C(A)<C(A,B) by using A=(A,B)transpose[(I,0)], and I also can prove C(A+B)<C(A,B) using the similar approach.

    But I cannot move further because my thoughs maybe not related to the answer at all.

    Can someone help? Many thanks.
     
  2. jcsd
  3. Sep 1, 2011 #2
    Re: C(a)<c(a+b)

    You need to explain your notation. What's NND? what's "C" ?
     
  4. Sep 1, 2011 #3
    Re: C(a)<c(a+b)

    NND is nonnegative definite, C is column space
     
  5. Sep 1, 2011 #4

    Mark44

    Staff: Mentor

    Re: C(a)<c(a+b)

    Also, I get that C(A + B) is the column space of the matrix sum, A + B, but what does C(A, B) mean? And this - (A,B)transpose[(I,0)].
     
  6. Sep 1, 2011 #5
    Re: C(a)<c(a+b)

    Please no worry about those stuff, I am just confused about my original question, why C(A) belongs to C(A+B), is there any tricks?

     
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