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

1. Aug 22, 2007

I am not sure if this the correct forum, but

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

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 ,

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

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

2. Aug 22, 2007

### red_dog

$$\displaystyle tr(a+b)=\sum_{i=1}^n(a_{ii}+b_{ii})=\sum_{i=1}^na_{ii}+\sum_{i=1}^nb_{ii}=tr(a)+tr(b)$$
$$\displaystyle tr(ka)=\sum_{i=1}^nka_{ii}=k\sum_{i=1}^na_{ii}=ktr(a)$$

3. Aug 22, 2007

### Nesk

The function is defined as a sum. Sums are linear. Therefore, the function is linear. QED.

4. Aug 23, 2007

### HallsofIvy

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