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

B&F have the following:
$\delta_{i j} = e'_i \cdot e'_j = a_{i k} \left( e_k \cdot e'_j \right) = a_{i k} a_{j k}$

$a_{k i} a_{k j} = \delta_{i j}$

Does it suffice to show the following? :

$\delta_{i j} = a_{i k} a_{j k} \to \delta_{j i} = \left( a_{i k} a_{j k} \right)^T = a_{k i} a_{k j}$
and note that

$\delta_{j i} = \delta_{i j}$
?

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

chiro
B&F have the following:
$\delta_{i j} = e'_i \cdot e'_j = a_{i k} \left( e_k \cdot e'_j \right) = a_{i k} a_{j k}$

$a_{k i} a_{k j} = \delta_{i j}$

Does it suffice to show the following? :

$\delta_{i j} = a_{i k} a_{j k} \to \delta_{j i} = \left( a_{i k} a_{j k} \right)^T = a_{k i} a_{k j}$
and note that

$\delta_{j i} = \delta_{i j}$
?

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.

Fredrik
Staff Emeritus