# Double summation

1. Aug 18, 2012

### IniquiTrance

If I know that $\sum_{k=1}^n a_{ik} = 1$ and $\sum_{j=1}^n b_{kj} = 1$, why is the following permitted?

$$\sum_{j=1}^n \sum_{k=1}^n a_{ik}b_{kj} = \left(\sum_{j=1}^n b_{kj}\right) \left(\sum_{k=1}^n a_{ik}\right) = 1\cdot 1 = 1$$

Thanks!

2. Aug 18, 2012

### micromass

Staff Emeritus
It's not permitted. What you wrote makes no sense since you sum over k and one of the $b_{kj}$ is outside that sum. That is not allowed.

What you could do is:

$$\sum_{j=1}^n\sum_{k=1}^n a_{ik}b{kj}= \sum_{k=1}^n \sum_{j=1}^n a_{ik}b_{kj} = \sum_{k=1}^n \left(a_{ik} \sum_{j=1}^n b_{kj}\right)= \sum_{k=1}^n a_{ik}=1$$

3. Aug 18, 2012

### IniquiTrance

Thank you very much. I knew I wasn't understanding something.