I am stuck on this problem I have found and was wondering if any one had any ideas. I want to prove, that if $V$ is a finite dimensional inner product space with inner product $\langle,\rangle$, and $R:V \rightarrow V, S:V \rightarrow V$ and $\{e_1,...e_n\}$ is an orthonormal basis for $V$, then $$\langle Su,v\rangle =\langle u, Rv\rangle$$ if and only if $$Re_j=\sum_{i=1}^n\langle e_j,Se_i\rangle e_i$$
I think you can prove one direction (the forward complication) by working out the matrix for $S$ with respect to a basis, and, nothing that $R$ is the adjoint of $S$, you can follow the usual proof for the form of the adjoint's matrix, and you get the required sum. However, I'm assuming I now need to show a converse direction, and that this is not strong enough to show both sides? If so, how do I do this, starting with sum, I just can't see where to go from there.
Thanks.
You don't need no representation of the matrices. The forward case is just one line. Recall that for any $v\in V$ you know that $$v=\sum_{i=1}^n\langle v,e_i\rangle e_i,$$ therefore $$Re_j=\sum_{i=1}^n\langle Re_j,e_i\rangle e_i =\sum_{i=1}^n\langle e_j,Se_i\rangle e_i.$$ For the converse you only need to show $\langle Su,v\rangle =\langle u, Rv\rangle$ for $u,v\in\{e_1,\cdots,e_n\}$: \begin{align}\langle e_k,Re_j\rangle&=\bigl\langle e_k,\sum_{i=1}^n \langle e_j,Se_i\rangle e_i\bigr\rangle\\ &=\sum_{i=1}^n\langle e_j,Se_i\rangle\langle e_k,e_i\rangle\\ &=\sum_{i=1}^n\langle e_j,Se_i\rangle\delta_{ki}\\ &=\langle e_j,Se_k\rangle\\ &=\langle Se_k,e_j\rangle. \end{align}