# 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.

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 $$(M_{\mathcal A,\mathcal A}(T))_{12}$$.

• Thanks for the elaborate answer! I can finally grasp why this formula works. The key thing that I overlooked was the change of basis matrix being orthogonal and thus its inverse being equal to the transpose. Commented Jan 16, 2023 at 21:46
• Thank you :D. Yeah, That was a delicate point and it was important to spend some words more Commented Jan 16, 2023 at 21:50

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.