# Proof that AI_n = A

1. Oct 3, 2013

### phospho

I'm trying to prove AI_n = A for my linear algebra class:

$A = (a_{st})$
where $1 \leq s \leq m$ and $1 \leq t \leq n$

then $(AI_n)_{sk} = \displaystyle\sum_{t=1}^n a_{st}(I_n)_{tk} = \displaystyle\sum_{t=1}^n a_{st}\delta_{tk}$

we know that most of $\delta_{tk} = 0$ from the definition,

therefore

$\displaystyle\sum_{t=1}^n a_{st}\delta_{tk} = a_{sk}\delta_{kk} + \displaystyle\sum_{t\not=k} a_{st}\delta_{tk} = a_{sk}(1) + 0 = a_{sk}$
I don't understand the final bit of the proof, we start with defining A = a_{st}, and we end up with a_{sk}, how is this equivalent?

2. Oct 3, 2013

### D H

Staff Emeritus
Which part don't you understand?
1. $\sum_t a_{st}\delta_{tk} = a_{sk}\delta_{kk} + \sum_{t\ne k} a_{st}\delta_{tk}$
2. $a_{sk}\delta_{kk} + \sum_{t\ne k} a_{st}\delta_{tk} = a_{sk}(1) + 0$
3. $a_{sk}(1) + 0 = a_{sk}$

3. Oct 3, 2013

### phospho

the second one, i.e. how you get from 1. to $a_{sk}\delta_{kk}$ how does the t change to a k? for a and delta, am I righ tin thinking it's because of the deifnition of matrix multiplication?

4. Oct 3, 2013

### D H

Staff Emeritus
The first step is just taking one specific term out of the sum, the term for which t=k. This results in $\sum_t a_{st}\delta_{tk} = a_{sk}\delta_{kk} + \sum_{t\ne k} a_{st}\delta_{tk}$.

The second step merely replaces the value of the Kronecker delta with its value. The value of $\delta_{kk}$ in first term, $a_{sk}\delta_{kk}$, is just one. The second term is the sum $\sum_{t\ne k} a_{st}\delta_{tk}$, and here $\delta_{tk}$ is identically zero since $\delta_{tk}=0$ for all t ≠ k.

5. Oct 3, 2013

6. Oct 4, 2013

anyone?