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- Suppose $A=\{ u_{j} \}_{j=1}^{n}$ and $B=\{ v_{j} \}_{j=1}^{n}$ are [[basis|bases]] of a [[submodule generated by a subset|finitely generated]] [[free module]] (e.g., a finite-dimensional [[vector space]]) $F$.
> [!proposition] Proposition. ([[matrix of the identity w.r.t. two bases]])
> The two [[basis|bases]] correspond to '[[coordinate isomorphism|coordinate]] [[module isomorphism|isomorphisms]]' [^3] $R^{\oplus A} \xrightarrow{\varphi} F \text{ and } R^{\oplus B} \xrightarrow{\psi} F;$
> now $R^{\oplus A} \xrightarrow{ \psi ^{-1} \circ \varphi}R ^{\oplus B}$
> is an [[linear map|isomorphism]] whose [[matrix]] [^2] $N_{A}^{B}$ is the matrix of the [[identity map]] wrt two bases: $ \textcolor{Skyblue}{N_{A}^{B}= \MM\big(\id_{F}; A, B\big)}.$
> $N_{A}^{B}$ is called the **matrix of the change of basis**. Its [[inverse matrix|inverse]] is $\textcolor{Thistle}{\MM\big(\id_{F}, B,A)=: N_{B}^{A}}$. [[dual vector space|Its]] [[matrix transpose|transpose]] [[dual basis|is]] $\mathcal{M}(\id_{F^{\vee}}, B^{\vee}, A^{\vee})$. The transpose of its inverse is $\mathcal{M}(\id_{F ^{\vee}}, A ^{\vee}, B^{\vee})$.
> [!basicexample]
> Consider the [[basis|bases]] $\textcolor{Skyblue}{(4,2),(5,3)}$ and $\textcolor{Thistle}{(1,0),(0,1)}$ of $\rr^{2}$. We have $\begin{align}
> \MM\big(\id, \textcolor{Skyblue}{(4,2),(5,3)}, \textcolor{Thistle}{(1,0),(0,1)}\big) \\
> = \begin{pmatrix} | & | \\
> [\id \big(\textcolor{Skyblue}{(4,2)}\big)]_{\textcolor{Thistle}{(1,0),(0,1)}} & [\id \big(\textcolor{Skyblue}{(5,3)}\big)]_{\textcolor{Thistle}{(1,0),(0,1)}} \\
> | & |
> \end{pmatrix} \\
> = \begin{pmatrix}
> 4 & 5 \\
> 2 & 3
> \end{pmatrix}.
> \end{align}$
> The [[inverse matrix|inverse]] of the matrix above is[^1] $\begin{pmatrix}
> \frac{3}{2} & -\frac{5}{2} \\
> -1 & 2
> \end{pmatrix}.$
> Then the **statement** implies that $\begin{align}
> \MM\big(I, \textcolor{Thistle}{(1,0),(0,1)},\textcolor{Skyblue}{(4,2),(5,3)}\big) = \begin{pmatrix}
> \frac{3}{2} & -\frac{5}{2} \\
> -1 & 2
> \end{pmatrix}.
> \end{align}$
>
[^1]: We used the quick trick $A^{-1}=\frac{1}{\det A}\begin{pmatrix}a & -b \\-c & d\end{pmatrix}$ here.
[^2]: When we say 'whose matrix', we are referring to [[matrix]] in the sense of its [[matrix#^equivalence|equivalent definition]] ('step 1'), defined (only) in terms of the 'standard basis' of [[free module|coordinate space]]. But then this obviously is the original definition given $A,B$, hence the notation $\mathcal{M}(\id; A, B)$.
[^3]: Review [[free module|universal property of free modules]] together with the [[basis]] definition if confused.
> [!proof]- Proof. ([[matrix of the identity w.r.t. two bases]])
> Consider the [[product of matrices is the matrix of the product|matrix of the product of linear maps]]. In the **for [[linear operator|operators]]** definition, let $S=T=I$. Recalling that the [[identity matrix|matrix of the identity operator w.r.t. any basis is the identity matrix]], we have $\begin{align}I = \textcolor{Skyblue}{I} \textcolor{Thistle}{I} = \textcolor{Skyblue}{\MM\big(I,(\{ u_{j} \}_{j=1}^{n}), \{ v_{j} \}_{j=1}^{n}\big)} \textcolor{Thistle}{\MM\big(I,(\{ v_{j} \}_{j=1}^{n}), \{ u_{j} \}_{j=1}^{n}\big)}.
\end{align}$
Now interchange the roles of the $u