Taking vectors from one basis to another [Byron and Fuller]

  • #1
207
0
B&F have the following:
[itex]\delta_{i j} = e'_i \cdot e'_j = a_{i k} \left( e_k \cdot e'_j \right) = a_{i k} a_{j k} [/itex]

and they ask the reader to show that
[itex] a_{k i} a_{k j} = \delta_{i j} [/itex]

Does it suffice to show the following? :

[itex] \delta_{i j} = a_{i k} a_{j k} \to \delta_{j i} = \left( a_{i k} a_{j k} \right)^T = a_{k i} a_{k j} [/itex]
and note that

[itex] \delta_{j i} = \delta_{i j} [/itex]
?

I believe this is sufficient, but I have a feeling I short-cut this.
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
132
B&F have the following:
[itex]\delta_{i j} = e'_i \cdot e'_j = a_{i k} \left( e_k \cdot e'_j \right) = a_{i k} a_{j k} [/itex]

and they ask the reader to show that
[itex] a_{k i} a_{k j} = \delta_{i j} [/itex]

Does it suffice to show the following? :

[itex] \delta_{i j} = a_{i k} a_{j k} \to \delta_{j i} = \left( a_{i k} a_{j k} \right)^T = a_{k i} a_{k j} [/itex]
and note that

[itex] \delta_{j i} = \delta_{i j} [/itex]
?

I believe this is sufficient, but I have a feeling I short-cut this.
Hey Elwin.Martin.

One suggestion for you is to use the fact that the inner product is commutative. So swap the e^k and the e'^j around, to show the result.

One note is that you can't just take things like this for granted: it's a pain in the arse but its just a necessary habit in mathematics. However in saying this, things will become more intuitive as you get more experience with them.
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
I'm not a fan of notation like ##\left( a_{i k} a_{j k} \right)^T##. The transpose operation acts on matrices, not numbers. I'd write ##a_{ki}a_{kj}=(A^T)_{ik}A_{kj}=(A^TA)_{ij}##. If the matrix A is known to be orthogonal, then ##A^TA=I##, and the right-hand side of the first equality is ##\delta_{ij}##.
 

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