Theorem: Let $V$ be an inner product finite space with an orthonormal basis $\mathcal B$. Let $L$ be an operator on $V$, and let $A = [L]_\mathcal{B}$, the matrix associate to $L$. Then the matrix elements of $A$ are $$A_{ij} = \langle b_i, Lb_j\rangle.$$
If the basis $\mathcal B$ is only orthogonal, is it true that$$A_{ij}=\frac{\langle b_i, Lb_j\rangle}{\langle b_i, b_i\rangle}?$$
Best Answer
Assume that $\mathcal{B}=\{b_1,b_2,\ldots,b_n\}$, where $n:= \dim(V)$. Since $$L(b_j)=\sum_{k=1}^n\,A_{k,j}\,b_k\text{ for each }j\in\{1,2,\ldots,n\}=:[n]\,,$$ we have $$\big\langle b_i,L(b_j)\big\rangle=\sum_{k=1}^n\,A_{k,j}\,\langle b_i,b_k\rangle\text{ for all }i,j\in[n]\,.$$ If $\mathcal{B}$ is an orthogonal basis, then $$\big\langle b_i,L(b_j)\big\rangle=A_{i,j}\,\langle b_i,b_i\rangle\text{ for all }i,j\in[n]\,,$$ proving your claim.
In general, let $\langle\_,\_\rangle$ be a nondegenerate symmetric bilinear form on $V$ and $\{\beta_1,\beta_2,\ldots,\beta_n\}$ the dual basis of $\{b_1,b_2,\ldots,b_n\}$. Then, $\langle \beta_i,b_j\rangle =\delta_{i,j}$ for all $i,j\in[n]$, where $\delta$ is the Kronecker delta. Then, the matrix $[A_{i,j}]_{i,j\in[n]}$ of $L$ in the basis $\mathcal{B}=\{b_1,b_2,\ldots,b_n\}$ is given by $$A_{i,j}=\big\langle \beta_i,L(b_j)\big\rangle\text{ for all }i,j\in [n]\,.$$ In your particular case, $$\beta_i=\frac{b_i}{\langle b_i,b_i\rangle}\text{ for every }i\in[n]\,.$$
Remark. In the case the base field is $\mathbb{C}$, we can also take $\langle \_,\_\rangle$ to be a nondegenerate sesquilinear form on $V$ that is antilinear in the first entry, and linear in the second entry. The work is the same.