---- $R$ is an [[integral domain]]. > [!definition] Definition. ([[coordinate isomorphism]]) > Let $F$ be a [[free module|free]] $R$-[[module]] (e.g., a [[vector space]] if $R=k$ is a [[field]]). [[the rank theorem for free modules|Recall]] that a choice of [[basis]] $B$ for $F$ amounts to a choice of [[module isomorphism|isomorphism]] $F \cong^{\varphi} R^{\oplus B}$. > Moreover, we know from [[free module|universal property of free modules]] that this [[isomorphism]] is uniquely determined: it identifies an element $v \in F$ with the element of $R^{\oplus B}$ consisting of the scalars needed to write $v$ as a [[linear combination]] of elements in $B$. > $\varphi$ is called a [[coordinate isomorphism]]. ^definition > [!specialization] Specialization to finite-dimensional vector spaces. > Let $V$ be a [[vector space]]. > Occasionally we want to think of the elements of $V$ as relabled to be $n$-by-$1$ [[matrix|matrices]]. Once a [[basis]] $v_1, \dots, v_n$ is chosen (that is, one a choice of identification of $V$ with a [[direct sum of modules|direct sum]] [[free module|coordinate space]] $R^{\oplus \{ v_{j} \}_{j=1}^{n}}$ is chosen), the function $\MM$ that takes $v \in V$ to [[matrix of a vector]] is an [[linear isomorphism]] of $V$ onto $\ff^{n \times 1}$ that implements this relabeling. We call it a **coordinate isomorphism**. > Justification: Each $v \in V$ corresponds to a [[basis#Equivalence|unique linear combination of]] $v_1, \dots, v_n$, hence $\MM(v)$ is [[injection|injective]]. Each $M \in \ff^{n \times 1}$ is comprised of of $n$ arbitrary scalars $a_1, \dots, a_n \in \ff$. Since $v_1, \dots, v_n$ [[spans]] $V$, there exists $v\in V$ such that $v = a_1v_1 + \dots + a_nv_n$ and thus $\MM(v)=M$, hence $\MM(v)$ is [[surjection|surjective]]. So the notion of a **coordinate isomorphism** is well-justified by [[linear invertibility is equivalent to injectivity and surjectivity]]. > In general, we also know (without constructing an explicit [[linear isomorphism]]) that $V$ and $\mathbb{F}^{n \times 1}$ are [[linear isomorphism|linearly isomorphic]] because $\dim V = \len(\text{basis})=n=n\cdot 1 = \dim \ff^{n \times 1}$ (see [[vector space of m-by-n matrices#On Dimension|here]] and [[the (finite) dimension theorem|here]]). ^specialization ---- #### ---- #### 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 > ```