Index Notation, Identity Matrix

  • Thread starter Thread starter PhDeezNutz
  • Start date Start date
  • Tags Tags
    Index notation
AI Thread Summary
The discussion centers on proving that multiplication with the identity matrix is commutative, specifically showing that IA = AI. The derivation involves using index notation and the Kronecker delta, leading to the conclusion that both expressions yield A_{ij}. Concerns are raised about the use of repeated indices in Einstein notation, questioning whether it complicates the derivation unnecessarily. It is suggested that explicit notation is preferable for clarity in mathematical exercises, while Einstein notation may be more suitable for physics contexts. Ultimately, the conversation emphasizes the importance of notation in accurately conveying mathematical concepts.
PhDeezNutz
Messages
849
Reaction score
557
Homework Statement
Prove that multiplication with the Identity Matrix is commutative
Relevant Equations
My approach is to compute expressions for ##IA## and ##AI## and show that ##IA = AI##

Of course If ##A## is an ##m \times n## matrix it would be ##I_m## on the left and ##I_n## on the right.
Terms only generate when ##k = i ##

##\left( IA \right)_{ij} = \delta_{ik}A_{kj} = \delta_{ii}A_{ij} = A_{ij}##

##\left( AI \right)_{ij} = A_{ik} \delta_{kj} = A_{ii} \delta_{ij} = A_{ij}##

Therefore ##IA = AI##

I’m bothered by three repeated indices so I’m questioning my derivation.
 
Physics news on Phys.org
PhDeezNutz said:
Homework Statement: Prove that multiplication with the Identity Matrix is commutative
Relevant Equations: My approach is to compute expressions for ##IA## and ##AI## and show that ##IA = AI##

Of course If ##A## is an ##m \times n## matrix it would be ##I_m## on the left and ##I_n## on the right.

Terms only generate when ##k = i ##

##\left( IA \right)_{ij} = \delta_{ik}A_{kj} = \delta_{ii}A_{ij} = A_{ij}##

##\left( AI \right)_{ij} = A_{ik} \delta_{kj} = A_{ii} \delta_{ij} = A_{ij}##

Therefore ##IA = AI##

I’m bothered by three repeated indices so I’m questioning my derivation.
If you are using the (Einstein) notation with summation suppressed, then technically you must go directly:
$$\delta_{ik}A_{kj} = A_{ij}$$If you are using standard notation, then there is no limit to how many terms you can have with the same index.
 
So does that mean I’m splitting hairs over what is essentially a definition?
 
PhDeezNutz said:
So does that mean I’m splitting hairs over what is essentially a definition?
Not really. The Einstein notation cannot handle the second of these expressions:
$$\sum_i a_ib_i \equiv a_ib_i$$$$a_ib_i \ \ \text{for some specific i}$$It also can't handle something like:
$$\exists i: a_i = b_i$$Instead:
$$(\forall i: a_i = b_i) \equiv a_i = b_i$$
 
My advice is to use explicit notation for math exercises and keep Einstein for physics later on.

I am unhappy with $$\left( AI \right)_{ij} = \sum_k A_{ik} \delta_{kj} = \sum_? A_{ii} \delta_{ij} = A_{ij}$$and would stay with $$\left( AI \right)_{ij} = \sum_k A_{ik} \delta_{kj} = A_{ij} $$because ## \sum A_{ii} \delta_{ij} \ne A_{ij}##

##\ ##
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top