I was hoping someone could help with the following:
[Working in index notation]
I understand that we can write a new basis e'_i = (a_ij)(e_j) , where e'_i is the new basis, a_ij is the rotation (change of basis) , and e_j is the old basis.
I found something in my lectures notes I do not understand....
"The dot product of e'_i and e_k = a_ik , clearly, a_ik is the component of e'_i in the e_k direction. *****Also, a_ik can be considered as the component of e_k in the e'_i direction*****"
I understand why l_ik is the component of e'_i in the e_k direction, purely since e'_i = a_ik (e_k) ... which makes sense, but I don't understand why it's the other way around?
At this stage we haven't yet shown A is orthogonal , so please do not use the fact that we can just use the transpose.
Help would be much appreciated.
Expand ${e_i}'$ in the basis $\{ e_1, \dots, e_n\}$.
$${e_i}' = a_{i1}e_1 + a_{i2}e_2 + \cdots + a_{ij}e_j + \cdots + a_{in}e_n$$
Taking the dot product with $e_j$ yields
$$\begin{align}e_j\cdot {e_i}' &= e_j \cdot(a_{i1}e_1 + a_{i2}e_2 + \cdots + a_{ij}e_j + \cdot + a_{in}e_n) \\ &= a_{i1}(e_j\cdot e_1) + \cdots + a_{in}(e_j\cdot e_n) \\ &= a_{i1}(0) + \cdots a_{i(j-1)}(0) + a_{ij}(1) + a_{i(j+1)}(0) + \cdots a_{in}(0) \\ &= a_{ij}\end{align}$$
Thus $a_{ij}$ is the component ${e_i}'$ in the $e_j$ direction.
Now because $\{{e_1}', \dots, {e_n}'\}$ is a basis for the space also, we can expand $e_j$ in them: $$e_j = b_{1j}{e_1}' + b_{2j}{e_2}' + \cdots + b_{ij}{e_i}' + \cdots + b_{nj}{e_n}'$$ and take the dot product with ${e_i}'$. After the same steps as above you'll get ${e_i}'\cdot e_j = b_{ij}$. But because ${e_i}'\cdot e_j = e_j \cdot {e_i}'$, and we already found $e_j \cdot {e_i}' = a_{ij}$, that means that $b_{ij}=a_{ij}$. What can you conclude from here?
BTW, this only works because $\{e_1, \dots, e_n\}$ and $\{{e_1}', \dots, {e_n}'\}$ are orthonormal bases. If they were general bases you'd have to take the dot product with an element of the dual basis ($e^j$ or ${e^i}'$). Pretty fun, right? ;)