Finding a single matrix-element of a linear transformation with a change of basis The question:
Let $A = \begin{pmatrix} a & b & c\\ d & e & f \\ g & h & i\end{pmatrix}$ be the standard matrix of linear transformation $T : \mathbb{R}^3 \rightarrow \mathbb{R}^3$. Additionally let there be an orthonormal basis $\mathcal{A} = \{a_1, a_2, a_3 \}$ with $a_1 = \frac{e_1+2e_2+e_3}{\sqrt{6}}, a_2=\frac{e_1-e_3}{\sqrt{2}}, a_3=\frac{e_1-e_2+e_3}{\sqrt{3}}$ and $\{ e_1,e_2,e_3\}$ the standard basis of $\mathbb{R}^3$.
determine matrix element $(T^{\mathcal{A}}_{\mathcal{A}})_{12}$
I thought of this approach: $T^{\mathcal{A}}_{\mathcal{A}} = C^{\mathbb{R}^3}_{\mathcal{A}}T^{\mathbb{R}^3}_{\mathbb{R}^3}C^{\mathcal{A}}_{\mathbb{R}^3}$ but it was a mess to compute and gives all of the matrix elements. The provided answer to the question was as follows:
$(T^{\mathcal{A}}_{\mathcal{A}})_{12} = \langle a_1, T(a_2)\rangle$ and is apparently contingent on the basis $\mathcal{A}$ being orthonormal.
It's a neat formula but I have no idea why it works and haven't been able to find it on the internet. I have tried to derive this myself but haven't made any progress.
 A: Generally, if you have a $n\times n$-matrix $M$ with columns $m_1,..,m_n$ then for any other Matrix $T$
\begin{align*}
\left(M^TTM\right)_{ij} = \langle m_i,Tm_j \rangle\,.
\end{align*}
This follows directly from the definition of matrix multiplication.
You can derive their formular from yours using this.
A: Interesting question. I think the idea is the following:
Let $e=\lbrace e_1, e_2, e_3\rbrace$ be the canonical basis of $\mathbb R^3$ and denote by $M_{\alpha, \beta}(T)$ the matrix associated to an operator $T\colon \mathbb R^3\to \mathbb R^3$ where in the domain is fixed the basis $\alpha$ and in the codomain is fixed the basis $\beta$.
To get more clean what I mean, let give me an example. In my notation, your matrix $A$ is simply the matrix $M_{e,e}(T)$ and $T_{\mathcal{A}}^{\mathcal A}$ is $M_{\mathcal A, \mathcal A}(T)$. Moreover, for example, in general $M_{\alpha,\beta}(Id_{\mathbb R^3})$ is the change base matrix from the base $\beta$ to the base $\alpha$.
Let us come back to our problem. We have just to use the formula which gives how the associated matrix of an operator changes if you change the basis. Therefore
$$M_{\mathcal A,\mathcal A}(T)=M_{\mathcal A, e}(Id)\cdot A\cdot M_{e,\mathcal A}(Id)$$
At this point we observe that $\mathcal A$ is orthonormal, and so the change base matrix $M_{e,\mathcal A}(Id)$ is orthogonal, that means $M_{\mathcal A, e}(Id)=M_{e,\mathcal A}(Id)^{-1}= M_{e,\mathcal A}(Id)^t$. The previous formula becomes
$$M_{\mathcal A,\mathcal A}(T)=M_{e,\mathcal A}(Id)^t\cdot A\cdot M_{e,\mathcal A}(Id).$$
At this point we have to talk about the columns of the change base matrix $M_{e,\mathcal A}(Id)$. Since $e$ is the canonical basis, then its columns are simply the vectors of $\mathcal A$, just by definition of change base matrix.
Now we are interested in computing $(M_{\mathcal A,\mathcal A}(T))_{ij}$. Observe that this is general equivalent to $e_i^tM_{\mathcal A,\mathcal A}(T)e_j$, and so we can conclude
$$(M_{\mathcal A,\mathcal A}(T))_{ij}=e_i^tM_{\mathcal A,\mathcal A}(T)e_j=e_i^tM_{e,\mathcal A}(Id)^t\cdot A\cdot M_{e,\mathcal A}(Id)e_j=(M_{e,\mathcal A}(Id)e_i)^t\cdot A \cdot (M_{e,\mathcal A}(Id)e_j)=a_i^t\cdot A\cdot a_j=\langle a_i, Aa_j\rangle =\langle a_i, T(a_j)\rangle,$$
and the thesis follows. If you have questions, please ask me because there are some hidden steps that I decided to skip.
You can observe that what I wrote is completely general. We don't need to stay on $\mathbb R^3$ and we can compute any entry and not only $(M_{\mathcal A,\mathcal A}(T))_{12}$.
