$\boxed{\impliedby}$ : Supposons que
$$\ds\forall(i,j)\in\llbracket1,n\rrbracket^{2},\;\left\langle u(\vv{e_i}),\vv{ej}\right\rangle =\left\langle \vv{e_i},u(\vv{e_j})\right\rangle $$ Soit $\ds\vv{x}=\sum_{i=1}^{n}{x_{i}\vv{e_i}}$ et $\ds\vv{y}=\sum_{i=1}^{n}{y_{i}\vv{e_j}}$. Alors, par linéarité de $u$, on a : $$\begin{array}{rcl}\left\langle u(\vv{x}),\vv{y}\right\rangle & = & \ds\left\langle \sum_{i=1}^{n}{x_{i}u(\vv{e_i})},\sum_{j=1}^{n}{y_{j}\vv{e_j}}\right\rangle \\ & = & \ds\sum_{i=1}^{n}{\sum_{j=1}^{n}{x_{i}y_{j}\left\langle u(\vv{e_i}),\vv{e_j}\right\rangle }} \\ & = & \ds\sum_{i=1}^{n}{\sum_{j=1}^{n}{x_{i}y_{j}\left\langle \vv{e_i},u(\vv{e_j})\right\rangle }} \\ & = & \ds\left\langle \sum_{i=1}^{n}{x_{i}\vv{e_i}},\sum_{j=1}^{n}{y_{j}u(\vv{e_j})}\right\rangle \\ & = & \ds\left\langle \vv{x},u(\vv{y})\right\rangle \end{array}$$ donc $u$ est symétrique.