Linear operator and inner product 
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$$A_{ij}=\frac{\langle b_i, Lb_j\rangle}{\langle b_i, b_i\rangle}?$$
 A: Yes, it is, because $[L]_{\mathcal B}=[id]^{\mathcal B'}_{\mathcal B}[L]_{\mathcal B'}[id]_{\mathcal B'}^{\mathcal B}$, where $\mathcal B'=\left\{\frac{b_i}{\sqrt{\langle b_i,b_i\rangle}}\right\}_{i=1}^n$.
By the previous lemma, $\left([L]_{\mathcal B'}\right)_{ij}=\frac{\langle b_i,Lb_j\rangle}{\sqrt{\langle b_j,b_j\rangle\langle b_i,b_i\rangle}}$. Moreover, $\left([id]^{\mathcal B'}_{\mathcal B}\right)_{ij}=\delta_{ij}\frac1{\sqrt{\langle b_i,b_i\rangle}}$ and $\left([id]_{\mathcal B'}^{\mathcal B}\right)_{ij}=\delta_{ij}\sqrt{\langle b_i,b_i\rangle}$, so $$\left([L]_{\mathcal B}\right)_{ij}=\sum_{k,h}\left([id]^{\mathcal B'}_{\mathcal B}\right)_{ik}\left([L]_{\mathcal B'}\right)_{kh}\left([id]_{\mathcal B'}^{\mathcal B}\right)_{hj}=\\=\sum_{k,h}\delta_{ik}\frac1{\sqrt{\langle b_i,b_i\rangle}}\frac{\langle b_k,Lb_h\rangle}{\sqrt{\langle b_k,b_k\rangle\langle b_h,b_h\rangle}}\delta_{hj}\sqrt{\langle b_h,b_h\rangle}=\frac{\langle b_i,Lb_j\rangle}{\langle b_i,b_i\rangle}$$
A: 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.
