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We have the linear operator $\mathcal{O}$ acting on an arbitrary vector $\vec{a}$ which can be written in terms of the set of orthonormal basis vectors {$\vec{e_i}$}: $\vec{a} = \sum\limits_i \vec{e_i} a_i$. It is also given that

\begin{equation} \mathcal{O} \vec{e_i} = \sum\limits_j \vec{e_j} O_{ji} \end{equation}

Now, show that $O_{ji} = \vec{e_i} \cdot \mathcal{O}\vec{e_j} $

I can "show" this just substituting into the centered equation above and using the completeness relation., but that is not the same as deriving it from basic relations. How to proceed?

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    This does not hold in general; it does if the basis is orthonormal. – 2017-02-03
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    Right, the basis is orthonormal! – 2017-02-03
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    Well, in that case, just taking the inner product of both sides of displayed equation with $e_j$ you get $$e_j \cdot \mathcal O e_i = \sum_k e_j \cdot e_k O_{ki}=O_{ji}$$ since $e_j \cdot e_k=0$ when $k \neq j$ and $e_j \cdot e_j=1$. – 2017-02-03

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