Proving AI_n = A in Linear Algebra Class: Understanding the Final Step

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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?

 

[D H]

##\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}##

 

[phospho]

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]

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.

 

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Thank you I understand now,

is this proof ok for BI_n: http://gyazo.com/40ea1f81c22d6e315963a8ceb0b530bc

 

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anyone?