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

  1. Mar 28, 2012 #1
    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.
  2. jcsd
  3. Mar 29, 2012 #2


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    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.
  4. Mar 29, 2012 #3


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    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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