# Proof that AI_n = A

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?

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D H
Staff Emeritus
##\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?
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}##

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}##
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?

D H
Staff Emeritus