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

1. Mar 28, 2012

### Elwin.Martin

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.

2. Mar 29, 2012

### chiro

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. Mar 29, 2012

### Fredrik

Staff Emeritus
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 $\delta_{ij}$.