For each $1 \leq i,j \leq n$ let $E_{ij} \in \mathrm{End}(\mathbf{V})$ be the linear operator defined by
\begin{equation*} E_{ij} \mathbf{e}_k = \langle \mathbf{e}_j,\mathbf{e}_k \rangle \mathbf{e}_i, \quad 1 \leq k \leq n. \end{equation*}
What is the matrix of $E_{ij}$ relative to $\mathcal{B}$ (the standard basis)?
I believe $E_{ij}$ is a matrix of zeroes with 1 in the $i,j$ position. How would you obtain the matrix relative to $\mathcal{B}$?
The $(k,l)$-matrix element of $[E_{ij}]_\mathcal{B}$ is
\begin{equation*} \langle \mathbf{e}_k, E_{ij} \mathbf{e}_l \rangle = \langle \mathbf{e}_k, \langle \mathbf{e}_j,\mathbf{e}_l \rangle\mathbf{e}_i \rangle = \langle \mathbf{e}_i,\mathbf{e}_k\rangle \langle \mathbf{e}_j,\mathbf{e}_l\rangle. \end{equation*}
Thus, the matrix $[E_{ij}]_\mathcal{B}$ has a $1$ in the $(i,j)$-position and all other matrix elements are $0$.