Index Notation, Identity Matrix

[PhDeezNutz]

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.

 

[PeroK]

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.

 

[PhDeezNutz]

So does that mean I’m splitting hairs over what is essentially a definition?

 

[PeroK]

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:
$$(\forall i: a_i = b_i) \equiv a_i = b_i$$

 

[BvU]

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 $\sum A_{ii} \delta_{ij} \ne A_{ij}$

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