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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?
## 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?