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Solve Tr(a+b)=tr(a)+tr(b)

  1. Aug 22, 2007 #1
    I am not sure if this the correct forum, but

    Problem: if the tr(c) is defined as [tex]tr(c)=\sum^{n}_{i=1}c_{ii}[/tex]

    Than deduce that tr(a+b)= tr(a)+tr(b) and that tr(ka)=ktr(a).

    Attempt at the solution: For tr(a+b)=tr(a)+tr(b) I have somerthing to the effect of ,


    I do not know if this is the simplest approach, but I know that now I have to employ the associative property of addition, but I am not so sure how to bring this in quantitatively..or that just it...am I done?

    The tr(ka) should be easier after I get the first one.

    Any thoughts?

  2. jcsd
  3. Aug 22, 2007 #2
    [tex]\displaystyle tr(a+b)=\sum_{i=1}^n(a_{ii}+b_{ii})=\sum_{i=1}^na_{ii}+\sum_{i=1}^nb_{ii}=tr(a)+tr(b)[/tex]
    [tex]\displaystyle tr(ka)=\sum_{i=1}^nka_{ii}=k\sum_{i=1}^na_{ii}=ktr(a)[/tex]
  4. Aug 22, 2007 #3
    The function is defined as a sum. Sums are linear. Therefore, the function is linear. QED.
  5. Aug 23, 2007 #4


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    Yes, but the problem is, I suspect to demonstrate it in this case.
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