If $L: \mathbb{V} \to \mathbb{V}$ is any map and we have two inner products defined on $\mathbb{V}$, $[ , ]$ and $⟨ , ⟩$, and we pick two orthonormal bases w.r.t. each of these inner products, how can one prove that if $L(B) = A$ (where $A$ and $B$ are the two bases), then $L$ preserves the inner product, and thus $[x, y] = ⟨L(x), L(y)⟩$?
I know it must be easy, but I got confused and lost.
In the following, I am assuming $V$ is finite-dimensional, of dimension $n$. Let $\{v_1,\ldots, v_n\}$ be an orthonormal basis w.r.t $[\cdot,\cdot]$ and $\{w_1,\ldots,w_n\}$ be an orthonormal basis w.r.t. $\langle \cdot,\cdot\rangle$.
Then we are given that there exists a permutation $\pi$ on $\{1,\ldots,n\}$ such that $L(v_i) = w_{\pi(i)}$, $i = 1,\ldots, n$. (This is just expressing the fact that $L$ maps one basis to the other.)
By relabelling the $w_j$'s if necessary, we can assume $L(v_i) = w_i$.
If, $x,y \in V$, we can write $x = \sum_{i = 1}^{n}a_iv_i$ and $y = \sum_{j = 1}^{n}b_jv_j$, so as $[v_i,v_j] = 0$ unless $i = j$,
\begin{equation} [x,y] = \sum_{i = 1}^{n}a_ib_i. \end{equation}
But also, $L(x) = \sum_{i = 1}^{n}a_iw_{i}$, and $L(y) = \sum_{j = 1}^{n}b_jw_{j}$.
As $\langle w_{i}, w_{j}\rangle = 0$ unless $i = j$, we see that \begin{equation} \langle L(x), L(y)\rangle = \sum_{i = 1}^{n}a_ib_i = [x,y]. \end{equation}