- #1
Saladsamurai
- 3,020
- 7
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 ,
[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
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