Why the Coordinate Map Is an Isomorphism

From Basis Expansions to Coordinates

A vector in a finite-dimensional space may look different from a column in \(\mathbb{R}^n\): it could be a polynomial, a matrix, or an element of an abstract vector space. An ordered basis provides a consistent way to encode each such vector by a list of real numbers. The key point is that the encoding loses no information. Every vector has a coordinate vector, and no two different vectors share one.

Work over \(\mathbb{R}\), and let \(V\) be a finite-dimensional vector space with an ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\). A basis is a linearly independent list that spans \(V\). Thus every \(\mathbf{v}\in V\) can be expressed as a linear combination of the basis vectors, and that expression is unique. The real numbers in the expression, in the same order as the basis vectors, form the coordinate vector.

There are two separate reasons the coordinate vector is well-defined. Since the basis spans \(V\), an expansion exists for every \(\mathbf{v}\). Since the basis is linearly independent, two different coefficient lists cannot produce the same vector. Existence means the coordinate map has an output for every input; uniqueness means each input has just one output. The isomorphism theorem below shows how these facts also give surjectivity and injectivity.

Linearity and the Isomorphism Theorem

A map between vector spaces is linear if it preserves addition and scalar multiplication. It is one-to-one (injective) if equal outputs imply equal inputs, and onto (surjective) if every element of its codomain is an output. A linear map that is both one-to-one and onto is called an isomorphism. The coordinate map preserves linear operations because adding vectors or multiplying by a scalar acts on their basis coefficients in the same way.

Proof. First we verify linearity. Let \(\mathbf{u},\mathbf{v}\in V\), and write their unique basis expansions as \(\mathbf{u}=\sum_{i=1}^{n}a_i\mathbf{b}_i\) and \(\mathbf{v}=\sum_{i=1}^{n}c_i\mathbf{b}_i\). For any \(r,s\in\mathbb{R}\), the vector \(r\mathbf{u}+s\mathbf{v}\) has the expansion \(\sum_{i=1}^{n}(ra_i+sc_i)\mathbf{b}_i\). By uniqueness of basis coefficients, its coordinate vector is \(\begin{bmatrix}ra_1+sc_1\\ \vdots\\ ra_n+sc_n\end{bmatrix}=r[\mathbf{u}]_{\mathcal{B}}+s[\mathbf{v}]_{\mathcal{B}}\). Hence \(\Phi_{\mathcal{B}}(r\mathbf{u}+s\mathbf{v})=r\Phi_{\mathcal{B}}(\mathbf{u})+s\Phi_{\mathcal{B}}(\mathbf{v})\).

To prove injectivity, suppose \(\Phi_{\mathcal{B}}(\mathbf{u})=\Phi_{\mathcal{B}}(\mathbf{v})\). The two vectors therefore have the same coefficients in their basis expansions. Substituting those same coefficients into both expansions gives \(\mathbf{u}=\mathbf{v}\). Thus the map is one-to-one.

To prove surjectivity, take any column \(\mathbf{c}=\begin{bmatrix}c_1\\ \vdots\\ c_n\end{bmatrix}\in\mathbb{R}^n\). Form the vector \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\in V\). This is a basis expansion with coefficients \(c_1,\dots,c_n\), so \(\Phi_{\mathcal{B}}(\mathbf{v})=\mathbf{c}\). Every element of \(\mathbb{R}^n\) is therefore an output. The map is linear, one-to-one, and onto, so it is an isomorphism.

If \(n=0\), the basis is the empty list and \(V\) consists only of \(\mathbf{0}\). The space \(\mathbb{R}^0\) consists of the single empty column. The coordinate map sends \(\mathbf{0}\) to that column; it is linear, one-to-one, and onto. Thus the theorem includes this case as well. \(\square\)

The proof uses the two defining properties of a basis in different directions. Uniqueness gives injectivity: equal coordinates force equal vectors. Existence gives surjectivity: any real coefficient column can be used to form a vector. Linearity then says that this correspondence respects the vector-space operations, not just the underlying sets.

Reading the Isomorphism in Examples

The Inverse Correspondence

The isomorphism theorem also gives a precise description of how to recover a vector from its coordinates. For a fixed ordered basis, the inverse correspondence takes a coefficient column and forms the vector with those basis coefficients. This observation does not make a vector identical to its coordinate column: the vector belongs to \(V\), while its coordinate column belongs to \(\mathbb{R}^n\). The map relates two different spaces.

Proof. Let \(\mathbf{c}=\begin{bmatrix}c_1\\ \vdots\\ c_n\end{bmatrix}\). By definition of the coordinate map, the vector \(\sum_{i=1}^{n}c_i\mathbf{b}_i\) has coordinate column \(\mathbf{c}\). Hence applying \(\Phi_{\mathcal{B}}\) to that vector returns \(\mathbf{c}\), so the displayed rule is the inverse of \(\Phi_{\mathcal{B}}\). To verify linearity directly, take columns \(\mathbf{c}=\begin{bmatrix}c_1\\ \vdots\\ c_n\end{bmatrix}\), \(\mathbf{d}=\begin{bmatrix}d_1\\ \vdots\\ d_n\end{bmatrix}\), and scalars \(r,s\in\mathbb{R}\). The rule sends \(r\mathbf{c}+s\mathbf{d}\) to \(\sum_{i=1}^{n}(rc_i+sd_i)\mathbf{b}_i=r\sum_{i=1}^{n}c_i\mathbf{b}_i+s\sum_{i=1}^{n}d_i\mathbf{b}_i\), which is \(r\Phi_{\mathcal{B}}^{-1}(\mathbf{c})+s\Phi_{\mathcal{B}}^{-1}(\mathbf{d})\). Thus the inverse is linear. \(\square\)

This inverse description explains why each part of the theorem matters. If the list spans \(V\), every vector can be assigned coordinates. If it is linearly independent, those coordinates are unambiguous. And because every coefficient column defines a linear combination of the basis vectors, every possible column in \(\mathbb{R}^n\) corresponds to a vector in \(V\).

Common Pitfall: A List Must Really Be a Basis

A list of \(n\) vectors in an \(n\)-dimensional space does not define the coordinate isomorphism merely because its length is right. It must be a basis. For a concrete failure, take \(V=\mathbb{R}^2\) and the list \(\mathbf{w}_1=\begin{bmatrix}1\\0\end{bmatrix}\), \(\mathbf{w}_2=\begin{bmatrix}2\\0\end{bmatrix}\). A tempting but wrong claim is that coefficients \((a,b)\) give coordinates for every vector in \(\mathbb{R}^2\). In fact, the corresponding combination is \(a\mathbf{w}_1+b\mathbf{w}_2=\begin{bmatrix}a+2b\\0\end{bmatrix}\), whose second entry is always zero. For example, no values of \(a,b\) can produce \(\begin{bmatrix}0\\1\end{bmatrix}\). The list does not span \(\mathbb{R}^2\), so it is not a basis and the proposed coordinate assignment is not onto.

There is also a uniqueness failure: \(\mathbf{w}_2=2\mathbf{w}_1\), so the zero vector has at least two coefficient descriptions, \(0\mathbf{w}_1+0\mathbf{w}_2=\mathbf{0}\) and \(2\mathbf{w}_1-1\mathbf{w}_2=\mathbf{0}\). Thus the list is linearly dependent. The correction is to check both basis properties—spanning and linear independence—before interpreting coefficients as a unique coordinate vector.

The basis determines the coordinate map. A different ordered basis can assign different coordinates to the same vector, but each such basis still gives its own linear isomorphism between \(V\) and \(\mathbb{R}^n\). The vector itself has not changed; only its representation has.