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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 ,

[tex]tr(a+b)=\sum^{n}_{i=1}a_{ii}+b_{ii}=a_{11}+a_{22}+...a_{nn}+b_{11}+b_{22}+...b_{nn}[/tex]

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?

Casey

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# Homework Help: Solve Tr(a+b)=tr(a)+tr(b)

**Physics Forums | Science Articles, Homework Help, Discussion**