----- Let $R$ be an [[integral domain]]. > [!proposition] Proposition. ([[change of basis formula]]) > Let $\alpha:F \to G$ be a [[linear map|homomorphism]] of [[submodule generated by a subset|finitely generated]] [[free module|free]] $R$-[[module|modules]]. Let $P$ be a [[matrix]] representation of $\alpha$ with respect to [[basis|bases]] $A \subset F$ and $C \subset G$. > > **1.** If $Q$ is [[equivalent matrices|equivalent]] to $P$ — say, $Q$ is a [[matrix]] representation of $\alpha$ with respect to [[basis|bases]] $B \subset F$ and $D \subset G$ — then we have $Q=M_{C}^{D} \cdot P \cdot N_{B}^{A}$ > where $N_{B}^{A}$ and $M_{C}^{D}$ are the ([[matrix of the identity w.r.t. two bases|change of basis]]) [[matrix|matrices]] corresponding to $\nu_{A}^{B}=\psi ^{-1} \circ \varphi$ and $\mu_{C}^{D}=\sigma ^{-1} \circ \rho$ [^2] per the diagram below. > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZARgBpiBdUkANwEMAbAVxiRADEQBfU9TXfIRQAmclVqMWbAOLdeIDNjwEiABlKrx9Zq0QgASgD1gAHRMQ0zOAAIAglzl8lgtaWFbJug8bMWr1gCEHHicBFRQAZg0PHTYjU3NLJhsAYWD5RTChZCj3am0pPXjfJJsAEWDxGCgAc3giUAAzACcIAFskdRAcCCQyCVi9M0Y0AAs6RxAW9qRRbt7ELoKvM3pmsaxJ6Y7EKPnO-M82MzRsEGoGOgAjGAYABX5lIRBmrBrRnC3WnbmepD3lscTGAmAB9WyGALnECXG73R4uPSvd6fEJTb5IAAs1D+iAArIdBiAzG0wSlDGUvjNENj9oh+oChiZmqNehdrrcHs5wi83h8qTsCXSGUcmdgam0JlwKFwgA > \begin{tikzcd} > R^{\oplus A} \arrow[rd, "\varphi"] \arrow[dd, "\nu_A^B"'] & & & R^{\oplus C} \arrow[dd, "\mu_C^D"] \arrow[ld, "\rho"'] \\ > & F \arrow[r, "\alpha"] & G & \\ > R^{\oplus B} \arrow[ru, "\psi"'] & & & R^{\oplus D} \arrow[lu, "\sigma"] > \end{tikzcd} > \end{document} > ``` > > **2.** Conversely, if we have $Q=MPN$ for [[inverse matrix|invertible]] $M,N$, then $Q$ is the [[matrix]] of $\alpha$ with respect to the [[basis|bases]] of $F,G$ which get identified with $B=\{ N_{:, i} \}_{i} \subset R^{\oplus C}$ and $D=\{M^{-1}_{:, j}\}_{j} \subset R^{\oplus A}$ under the [[coordinate isomorphism|coordinate isomorphisms]] $\rho$ and $\varphi$ and is therefore [[equivalent matrices|equivalent]] to $P$. > > > [!proof]+ Proof > **1.** There is not really much to show... definitions and the diagram immediately yield $\begin{align} Q = \mathcal{M}(\alpha; B, D)= & \mathcal{M}(\sigma ^{-1} \circ \alpha \circ \psi) \ \ (*) \\ = & \mathcal{M}(\mu_{C}^{D} \circ \rho ^{-1} \circ \alpha \circ \varphi \circ (\nu_{A}^{B})^{-1}) \\ = & \mathcal{M} (\mu_{C}^{D}) \mathcal{M}(\rho ^{-1} \circ \alpha \circ \varphi) \mathcal{M}(\nu_{B}^{A}) \\ = & M_{C}^{D} \mathcal{M}(\alpha; A, C) N_{B}^{A} \\ = & M_{C}^{D} P N_{B}^{A}. \end{align}$ > (See [^3] to discuss $(*)$.) > **2.** As bases (and they *are* bases, by invertibility), $B$ and $D$ are identified with standard bases of $R^{\oplus B}$ and $R^{\oplus D}$ under appropriate [[coordinate isomorphism|coordinate isomorphisms]] $\psi$ and $\sigma$. What are the matrices of $\varphi ^{-1}\circ \psi: R^{\oplus B} \to R^{\oplus A}$ and $\rho ^{-1} \circ \sigma:R^{\oplus D} \to R^{\oplus C}$? The $i^{th}$ columns are precisely $\varphi ^{-1} \circ \psi(\mathbb{1}_{i})=\varphi ^{-1}(\boldsymbol b_{i})=N_{:, i}$ and $\rho ^{-1} \circ \sigma(\mathbb{1}_{i})=\rho ^{-1}(\boldsymbol d _{i})=M^{-1}_{:, i}$. Thus these matrices are just $N$ and $M^{-1}$. It follows that > > $\begin{align} > \mathcal{M}(\alpha; \{ M_{:, i} \}, \{ N_{:, i} \}) = & \mathcal{M}( \sigma ^{-1} \circ \alpha \circ \psi) \\ > = & \mathcal{M}(\sigma ^{-1} \circ \rho \circ \textcolor{Thistle}{\rho ^{-1} \circ \alpha \circ \varphi } \circ \varphi ^{-1} \circ \psi ) \\ > = & \mathcal{M} (\sigma ^{-1} \circ \rho) \mathcal{M} (\rho ^{-1} \circ \alpha \circ \varphi) \mathcal{M}(\varphi ^{-1} \circ \psi) \\ > = & \mathcal{M}^{-1}(\rho ^{-1} \circ \sigma) \mathcal{M}(\alpha; A, C) N \\ > = & M P N. > \end{align}$ > > (This seems too laborious?) $\begin{align} \end{align}$ [^3]: If viewed in light of the main definition in [[matrix]], $(*)$ reminds that if [[basis|bases]] are not offered as arguments to $\mathcal{M}$ then the [[free module|standard basis]] of the relevant [[free module|coordinate space]] is used. Or we can view $(*)$ as indicating that we are now working with 'step 1' of the equivalent definition (which only knows of standard bases of coordinate spaces...) > [!specialization] Specialization to linear operators. Let $V$ be a [[vector space]] and suppose $\textcolor{Skyblue}{T} \in \text{End}(V)$. Let $\textcolor{LimeGreen}{A=\{ u_{j} \}_{j=1}^{n}}$ and $\textcolor{Apricot}{B=\{ v_{j} \}_{j=1}^{n}}$ be [[basis|bases]] of $V$. Let $N_{A}^{B}=\MM\big(\textcolor{Thistle}{\id}; \textcolor{LimeGreen}{A}, \textcolor{Apricot}{B}\big)=\MM\big(\textcolor{Thistle}{\id}; \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}\big).$ Then $\MM\big(\textcolor{Skyblue}{T},\textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}} \big)=N_{B}^{A}\MM\big(\textcolor{Skyblue}{T}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}\big)N_{A}^{B},$ where $N_{A}^{B}$ is the [[matrix of the identity w.r.t. two bases|matrix of the change of basis]] from $A$ to $B$ and $N_{B}^{A}$ is its [[inverse matrix|inverse]] (so that $\mathcal{M}(\textcolor{Skyblue}{T}; \textcolor{Apricot}{B})$ arises as a [[conjugate|conjugation]] of $\mathcal{M}(\textcolor{Skyblue}{T}; \textcolor{LimeGreen}{A})$). > >[!proof]- > > 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 $S=I$, getting $(*)$ $\begin{align} > > \MM\big(\textcolor{Skyblue}{T\textcolor{Thistle}{}}, \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}} \big) = & \MM\big(\textcolor{Skyblue}{\textcolor{Thistle}{I}T\textcolor{Thistle}{}}, \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}},\textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}},\textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}} \big) \\ > > = & \MM\big(\textcolor{Thistle}{I}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}},\textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}} \big)\MM\big(\textcolor{Skyblue}{T}, \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}\big) \\ > > = & A^{-1}\MM\big(\textcolor{Skyblue}{T}, \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}\big), > > \end{align}$ > > where we have used [[matrix of the identity w.r.t. two bases]]. > > > > > >Now in the **for [[linear operator|operators]]** definition replace $S$ with $T$ and $T$ with $I$, getting $\begin{align}&\MM\big( \textcolor{Skyblue}{T\textcolor{Thistle}{}}, \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}},\textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}} \big) \\ > &=\MM\big( \textcolor{Skyblue}{T\textcolor{Thistle}{I}}, \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}},\textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}} \big) \\ > &= \MM\big(\textcolor{Skyblue}{T}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}\big)\MM\big(\textcolor{Thistle}{I}, \textcolor{LimeGreen}{\{ u_{j} \}_{j=1}^{n}}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}\big) \\ > &= \MM\big(\textcolor{Skyblue}{T}, \textcolor{Apricot}{\{ v_{j} \}_{j=1}^{n}}\big)A. > \end{align}$ > Substitution of the above into $(*)$ yields the result. $\qedin$ > ^specialization [^1]: This is [[matrix]] in the sense of its [[matrix#^equivalence|equivalent definition]] ('step 1'), defined in terms of the 'standard basis' of [[free module|coordinate space]]. [2]: And the greek letters are the relevant [[coordinate isomorphism|coordinate isomorphisms]]. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```