Equal Dimensions Give a Construction
Two vector spaces need not consist of the same kinds of objects to have the same linear structure. For example, one may be a space of coordinate vectors and the other a space of polynomials. When both spaces have the same finite dimension, we can build a linear bijection by choosing an ordered basis in each space and pairing their vectors in order. The key is that every vector has a unique expression as a linear combination of a basis, so the pairing also tells us exactly where every vector must go.
Throughout, the spaces are over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A linear isomorphism from \(V\) to \(W\) is a linear map \(T:V\to W\) that is both one-to-one and onto. We assume \(V\) and \(W\) are finite-dimensional. By definition, their dimensions are the numbers of vectors in their respective bases.
The construction works because it transfers coefficients from one basis to the other. It is important that the coefficients are those of \(\mathbf{v}\) in \(\mathcal{B}\), not necessarily its standard coordinates. Also, the chosen bases are ordered: the coefficient of \(\mathbf{b}_i\) is paired with \(\mathbf{c}_i\).
Why the Construction Is Well Defined
A definition of a map must assign exactly one output to each input. For a basis, every vector has a unique expansion in that basis: existence follows from spanning, and uniqueness follows from linear independence. Thus the coefficients \(a_1,\dots,a_n\) used in the construction are determined by \(\mathbf{v}\). The definition cannot assign two different outputs to the same input.
Proof. For each \(\mathbf{v}\in V\), its unique expansion \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{b}_i\) defines \(T(\mathbf{v})=\sum_{i=1}^n a_i\mathbf{c}_i\). To prove linearity, let \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{b}_i\) and \(\mathbf{u}=\sum_{i=1}^n d_i\mathbf{b}_i\), and let \(r,s\in\mathbb{F}\). Then \(r\mathbf{v}+s\mathbf{u}=\sum_{i=1}^n(ra_i+sd_i)\mathbf{b}_i\), so the definition gives \(T(r\mathbf{v}+s\mathbf{u})=\sum_{i=1}^n(ra_i+sd_i)\mathbf{c}_i=rT(\mathbf{v})+sT(\mathbf{u})\).
To prove that \(T\) is one-to-one, suppose \(T(\mathbf{v})=\mathbf{0}\). Write \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{b}_i\). Then \(\sum_{i=1}^n a_i\mathbf{c}_i=\mathbf{0}\). Since \(\mathcal{C}\) is linearly independent, every \(a_i=0\), and hence \(\mathbf{v}=\mathbf{0}\). Linearity now shows that if \(T(\mathbf{v})=T(\mathbf{u})\), then \(T(\mathbf{v}-\mathbf{u})=\mathbf{0}\), so \(\mathbf{v}=\mathbf{u}\). To prove that \(T\) is onto, take any \(\mathbf{w}\in W\) and write \(\mathbf{w}=\sum_{i=1}^n a_i\mathbf{c}_i\). The vector \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{b}_i\) satisfies \(T(\mathbf{v})=\mathbf{w}\). Finally, if \(S:V\to W\) is another linear map with \(S(\mathbf{b}_i)=\mathbf{c}_i\) for every \(i\), then \(S(\mathbf{v})=\sum_{i=1}^n a_iS(\mathbf{b}_i)=\sum_{i=1}^n a_i\mathbf{c}_i=T(\mathbf{v})\) for every \(\mathbf{v}\). Thus the map is unique. If \(n=0\), both spaces contain only their zero vector; the unique map between them is linear, one-to-one, and onto, and the same statement holds with empty bases. \(\square\)
The proof gives a useful coordinate description as well. If \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients of \(\mathbf{v}\) in \(\mathcal{B}\), then the coefficients of \(T(\mathbf{v})\) in \(\mathcal{C}\) are the same column. In the notation for a matrix of a linear map, this says \([T]_{\mathcal{C}\leftarrow\mathcal{B}}=\mathbf{I}_n\). This identity describes the map in the selected bases; it does not assert that the spaces contain identical vectors.
Worked Constructions
Worked Example: From \(\mathbb{R}^3\) to Polynomials of Degree at Most Two
Construct an isomorphism from \(\mathbb{R}^3\) to \(P_2\), the real polynomials of degree at most two. Use the ordered basis \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3)\), where \(\mathbf{b}_1=(1,0,1)\), \(\mathbf{b}_2=(0,1,1)\), and \(\mathbf{b}_3=(1,1,0)\), and the ordered basis \(\mathcal{C}=(1,x,x^2)\) of \(P_2\).
First verify that \(\mathcal{B}\) is a basis. To express an arbitrary \((r,s,t)\in\mathbb{R}^3\) as \(a\mathbf{b}_1+b\mathbf{b}_2+c\mathbf{b}_3\), equate coordinates: \(a+c=r\), \(b+c=s\), and \(a+b=t\). Solving gives \(a=(r-s+t)/2\), \(b=(-r+s+t)/2\), and \(c=(r+s-t)/2\). These formulas give one expression for every \((r,s,t)\). They also give the only possible coefficients, so the expression is unique and the vectors form a basis.
Matching the chosen basis vectors gives \(T(a\mathbf{b}_1+b\mathbf{b}_2+c\mathbf{b}_3)=a+bx+cx^2\). Substituting the coefficients just found gives the map directly in standard coordinates:
For instance, \((2,1,3)=2\mathbf{b}_1+\mathbf{b}_2+0\mathbf{b}_3\), so \(T(2,1,3)=2+x\). To check, the coefficient formulas give \(a=(2-1+3)/2=2\), \(b=(-2+1+3)/2=1\), and \(c=(2+1-3)/2=0\). The inverse sends \(a+bx+cx^2\) to \(a\mathbf{b}_1+b\mathbf{b}_2+c\mathbf{b}_3=(a+c,b+c,a+b)\). Applying it to \(2+x\) gives \((2,1,3)\), verifying the example in the reverse direction too.
Worked Example: From \(P_2\) to Symmetric Matrices
Let \(\operatorname{Sym}_2(\mathbb{R})\) be the space of real symmetric \(2\times2\) matrices. Use the basis \((1,x,x^2)\) of \(P_2\) and the ordered list
Every symmetric \(2\times2\) matrix has the form \(\begin{bmatrix}u&v\\v&w\end{bmatrix}=u\mathbf{C}_1+v\mathbf{C}_2+w\mathbf{C}_3\). This expression is unique because the entries in positions \((1,1)\), \((1,2)\), and \((2,2)\) determine \(u,v,w\). Thus the list is a basis, and both spaces have dimension three.
Match \(1\) to \(\mathbf{C}_1\), \(x\) to \(\mathbf{C}_2\), and \(x^2\) to \(\mathbf{C}_3\). The resulting map is \(T(a+bx+cx^2)=a\mathbf{C}_1+b\mathbf{C}_2+c\mathbf{C}_3=\begin{bmatrix}a&b\\b&c\end{bmatrix}\). For \(p(x)=2-3x+4x^2\), this gives \(T(p)=\begin{bmatrix}2&-3\\-3&4\end{bmatrix}\). The inverse reads off the three independent entries: \(\begin{bmatrix}2&-3\\-3&4\end{bmatrix}\mapsto 2-3x+4x^2\). Applying the map to that polynomial returns the displayed matrix, and applying the inverse to the matrix returns the original polynomial.
Worked Example: From \(\mathbb{R}^2\) to a Plane in \(\mathbb{R}^3\)
Let \(W=\operatorname{span}\{(1,0,1),(0,1,1)\}\subseteq\mathbb{R}^3\). These two vectors are independent: if \(a(1,0,1)+b(0,1,1)=(0,0,0)\), the first coordinate gives \(a=0\), and the second gives \(b=0\). They span \(W\) by its definition, so they form a basis of \(W\). Match the standard basis of \(\mathbb{R}^2\) to them.
The resulting map is \(T(r,s)=r(1,0,1)+s(0,1,1)=(r,s,r+s)\). It is linear because each output coordinate is a linear expression in \(r,s\). For example, \(T(3,-2)=(3,-2,1)\), which belongs to \(W\). The inverse on \(W\) sends \((r,s,r+s)\) to \((r,s)\). Indeed, applying \(T\) to \((r,s)\) returns \((r,s,r+s)\); applying the inverse to \(T(r,s)\) returns \((r,s)\). Every vector in \(W\) has this form by the definition of span, so the map is onto \(W\), not onto all of \(\mathbb{R}^3\).
Choice of Bases and a Common Pitfall
The theorem constructs an isomorphism, but it does not select a uniquely preferred one independently of bases. Even from \(\mathbb{R}^2\) to itself, matching the standard basis to itself gives the identity map. Instead choose the source basis \(((1,0),(0,1))\) and the target basis \(((1,0),(1,1))\). Matching in order gives \(T(r,s)=r(1,0)+s(1,1)=(r+s,s)\), which is not the identity: \(T(0,1)=(1,1)\). Both maps are isomorphisms; the chosen pairing determines which one is constructed.
Common mistake: using standard coordinates as if they were basis coefficients. Suppose the source basis is \(\mathcal{B}=((1,0),(1,1))\) and the target basis is the standard basis of \(\mathbb{R}^2\). The vector \((1,1)\) is the second source basis vector, so the construction must send it to the second target basis vector, \((0,1)\). A mistaken calculation might read \((1,1)\) as standard coordinates and send it to \(1(1,0)+1(0,1)=(1,1)\). That uses the wrong coefficients: \((1,1)=0(1,0)+1(1,1)\) in \(\mathcal{B}\), not \(1(1,0)+1(1,1)\). The correction is to first express the input in the selected source basis, then transfer those coefficients to the target basis.
The construction is useful whenever two finite-dimensional spaces are presented in different forms. It gives a concrete map rather than only asserting that one exists, and it provides a practical inverse: transfer coefficients back from the target basis to the source basis. The checks in the examples follow the same pattern: the map preserves the selected coefficients, while the inverse recovers them.
Check Your Understanding
For each construction, identify the source and target bases and track the coefficients in the indicated order.
- Let \(\mathcal{B}=((1,1),(1,-1))\) in \(\mathbb{R}^2\), and let \(\mathcal{C}=(1,x)\) in \(P_1\). What polynomial is assigned to \((3,1)\)?
- For the map \(T(r,s)=(r+s,s)\), find the inverse formula and verify both compositions on an arbitrary pair \((r,s)\).
- Why does matching bases of different lengths not give the construction proved here?
- Let \(W=\operatorname{span}\{(1,1,0),(0,1,1)\}\). Define a map from \(\mathbb{R}^2\) to \(W\) by matching the standard basis to the displayed vectors. Is \((2,-1)\) mapped to a vector in \(W\)? Compute it.
- Explain why the isomorphism in the theorem is unique once both ordered bases and their pairing are fixed, even though different choices of bases may produce different isomorphisms.
