From Basis Coordinates to a Map
A basis gives each vector a unique description using its basis vectors. This makes a basis an effective set of instructions for defining a linear map: choose where each basis vector goes, then send every linear combination to the corresponding linear combination of the chosen images. When the images themselves form a basis, the instructions can be reversed, so the resulting map is an isomorphism.
Work over a common scalar field \(\mathbb{F}\), which may be \(\mathbb{R}\) or \(\mathbb{C}\). A basis of a vector space is an ordered list that is linearly independent and spans the space. Thus, if \(\mathcal{B}=(\mathbf{v}_1,\dots,\mathbf{v}_n)\) is a basis of \(V\), every \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=a_1\mathbf{v}_1+\cdots+a_n\mathbf{v}_n\), with \(a_i\in\mathbb{F}\). The scalars \(a_1,\dots,a_n\) are its coordinates in that basis.
Suppose \(\mathcal{B}=(\mathbf{v}_1,\dots,\mathbf{v}_n)\) is a basis of \(V\), and \(\mathcal{C}=(\mathbf{w}_1,\dots,\mathbf{w}_n)\) is a basis of \(W\). Matching the bases means prescribing \(T(\mathbf{v}_i)=\mathbf{w}_i\) for each index \(i\). The order matters: the first vector of \(\mathcal{B}\) is matched with the first vector of \(\mathcal{C}\), and so on.
Proof. First suppose \(n\geq 1\). Every \(\mathbf{v}\in V\) has a unique representation \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{v}_i\), so the formula assigns exactly one output to each input. To verify linearity, write \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{v}_i\) and \(\mathbf{u}=\sum_{i=1}^n b_i\mathbf{v}_i\). For scalars \(r,s\in\mathbb{F}\), the coordinates of \(r\mathbf{v}+s\mathbf{u}\) in \(\mathcal{B}\) are \(ra_i+sb_i\). Therefore $$ T(r\mathbf{v}+s\mathbf{u}) =\sum_{i=1}^n(ra_i+sb_i)\mathbf{w}_i =r\sum_{i=1}^n a_i\mathbf{w}_i+s\sum_{i=1}^n b_i\mathbf{w}_i =rT(\mathbf{v})+sT(\mathbf{u}). $$ Define \(S:W\to V\) by \(S(\sum_{i=1}^n c_i\mathbf{w}_i)=\sum_{i=1}^n c_i\mathbf{v}_i\). The unique coordinates in \(\mathcal{C}\) make this definition well-defined, and the same calculation shows that \(S\) is linear. If \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{v}_i\), then \(S(T(\mathbf{v}))=S(\sum_{i=1}^n a_i\mathbf{w}_i)=\sum_{i=1}^n a_i\mathbf{v}_i=\mathbf{v}\). If \(\mathbf{w}=\sum_{i=1}^n c_i\mathbf{w}_i\), then \(T(S(\mathbf{w}))=T(\sum_{i=1}^n c_i\mathbf{v}_i)=\sum_{i=1}^n c_i\mathbf{w}_i=\mathbf{w}\). Thus \(S\) is the inverse of \(T\), so \(T\) is a linear isomorphism, and the formula for \(S\) gives the stated inverse basis matches. Finally, any linear map with \(T(\mathbf{v}_i)=\mathbf{w}_i\) must, by linearity, send \(\sum_{i=1}^n a_i\mathbf{v}_i\) to \(\sum_{i=1}^n a_i\mathbf{w}_i\); this proves uniqueness. If \(n=0\), both spaces consist only of the zero vector, and their unique map is linear, bijective, and its own inverse, so the statement holds in that case as well. \(\square\)
The theorem’s formula is a coordinate rule, not a claim that the coordinates of a vector are necessarily its entries in some other representation. To calculate \(T(\mathbf{v})\), first express \(\mathbf{v}\) in the source basis \(\mathcal{B}\). Then use those same coefficients with the matched vectors in \(\mathcal{C}\).
Calculating a Map and Its Inverse
Worked Example: Matching Two Bases of R²
Let \(\mathcal{B}=((1,1),(1,-1))\) and \(\mathcal{C}=((2,1),(1,2))\), both ordered bases of \(\mathbb{R}^2\). Define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by matching the first vector to the first and the second to the second. Find \(T(x,y)\) and \(T^{-1}(s,t)\).
To find the source coordinates, write \((x,y)=a(1,1)+b(1,-1)=(a+b,a-b)\). Hence \(a+b=x\) and \(a-b=y\). Adding these equations gives \(2a=x+y\), while subtracting the second from the first gives \(2b=x-y\). Thus \(a=(x+y)/2\) and \(b=(x-y)/2\). By the matching rule,
For the inverse, start with a vector \((s,t)\) in the target and find its \(\mathcal{C}\)-coordinates: \((s,t)=c(2,1)+d(1,2)=(2c+d,c+2d)\). The equations \(2c+d=s\) and \(c+2d=t\) give \(c=(2s-t)/3\) and \(d=(-s+2t)/3\). The inverse sends these coefficients to the corresponding vectors of \(\mathcal{B}\):
Check: Substituting \(T(x,y)\) into the inverse gives first coordinate \(\bigl((3x+y)/2+(3x-y)/2\bigr)/3=x\) and second coordinate \((3x+y)/2-(3x-y)/2=y\). Conversely, substituting \(T^{-1}(s,t)\) into \(T\) gives first coordinate \(\bigl(3(s+t)/3+(s-t)\bigr)/2=s\) and second coordinate \(\bigl(3(s+t)/3-(s-t)\bigr)/2=t\). The maps undo one another.
This computation also checks that the proposed target list is a basis: every \((s,t)\) has the displayed \(\mathcal{C}\)-coordinate representation. Those coordinates are unique, since the equations \(2c+d=s\), \(c+2d=t\) determine \(c\) and \(d\) uniquely. Likewise, the source-coordinate equations determine \(a\) and \(b\) uniquely.
Worked Example: An Isomorphism onto a Plane
Let \(V=\mathbb{R}^2\), and let \(W=\operatorname{span}\{(1,0,1),(0,1,1)\}\), a subspace of \(\mathbb{R}^3\). Match the ordered basis \(\mathcal{B}=((1,1),(1,-1))\) of \(V\) with \(\mathcal{C}=((1,0,1),(0,1,1))\) of \(W\). Find the map and its inverse on \(W\).
The vectors in \(\mathcal{C}\) span \(W\) by its definition. They are linearly independent: if \(a(1,0,1)+b(0,1,1)=(0,0,0)\), the first coordinate gives \(a=0\), and the second gives \(b=0\). Thus \(\mathcal{C}\) is a basis of \(W\). For \((x,y)\in V\), its \(\mathcal{B}\)-coordinates are \(a=(x+y)/2\) and \(b=(x-y)/2\), as found by solving \(a+b=x,\ a-b=y\). Matching those coefficients gives
Every \((r,s,t)\in W\) has the form \(r(1,0,1)+s(0,1,1)=(r,s,r+s)\). Its \(\mathcal{C}\)-coordinates are therefore \(r,s\). Reversing the basis match yields
Check: Applying \(T\) to \((r+s,r-s)\) gives \(\bigl(((r+s)+(r-s))/2,\ ((r+s)-(r-s))/2,\ r+s\bigr)=(r,s,r+s)\). Applying \(T^{-1}\) to \(T(x,y)\) gives \(\bigl((x+y)/2+(x-y)/2,\ (x+y)/2-(x-y)/2\bigr)=(x,y)\). The codomain is \(W\), not all of \(\mathbb{R}^3\); the map is onto \(W\) because each vector \(r(1,0,1)+s(0,1,1)\) is the image of \((r+s,r-s)\).
Reading the Inverse from the Matched Bases
It is often faster to construct the inverse before simplifying a formula. If \(T(\mathbf{v}_i)=\mathbf{w}_i\), then \(T^{-1}(\mathbf{w}_i)=\mathbf{v}_i\). For a general target vector, find its coordinates in the target basis and rebuild the output using the source basis. The following example illustrates this procedure in three coordinates.
Worked Example: A Three-Dimensional Basis Match
Take the standard basis \(\mathcal{B}=(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\) of \(\mathbb{R}^3\), and let \(\mathcal{C}=((1,1,0),(0,1,1),(1,0,1))\). Define \(T\) by matching these bases in order. Find \(T\) and \(T^{-1}\).
The coordinates of \((x,y,z)\) in the standard basis are \(x,y,z\), so the matching rule gives \(T(x,y,z)=x(1,1,0)+y(0,1,1)+z(1,0,1)=(x+z,x+y,y+z)\). To find the inverse, set \((r,s,t)=(x+z,x+y,y+z)\). Adding the first two equations and subtracting the third gives \(2x=r+s-t\). Subtracting the first equation from the sum of the second and third gives \(2y=-r+s+t\). Adding the first and third equations and subtracting the second gives \(2z=r-s+t\). Therefore
This solution also verifies that \(\mathcal{C}\) is a basis: the coefficients expressing any \((r,s,t)\) as a combination of its three vectors are exactly the three displayed coordinates, and the equations uniquely determine them. Check: Put \(x=(r+s-t)/2\), \(y=(-r+s+t)/2\), and \(z=(r-s+t)/2\). Then \(x+z=r\), \(x+y=s\), and \(y+z=t\), so \(T(T^{-1}(r,s,t))=(r,s,t)\). In the other direction, substituting \(r=x+z,\ s=x+y,\ t=y+z\) into the inverse gives coordinates \(x,y,z\), respectively. Thus both compositions give the appropriate identity map.
Uniqueness and a Basis-Order Pitfall
Proof. Suppose \(T_1,T_2:V\to W\) are linear maps with \(T_1(\mathbf{v}_i)=T_2(\mathbf{v}_i)\) for every \(i\). For any \(\mathbf{v}\in V\), write its basis representation as \(\mathbf{v}=\sum_{i=1}^n a_i\mathbf{v}_i\). Linearity gives \(T_1(\mathbf{v})=\sum_{i=1}^n a_iT_1(\mathbf{v}_i)=\sum_{i=1}^n a_iT_2(\mathbf{v}_i)=T_2(\mathbf{v})\). This holds for every \(\mathbf{v}\in V\), so \(T_1=T_2\). \(\square\)
Common mistake: treating source coordinates as output coordinates. In the first worked example, \((1,0)=\tfrac12(1,1)+\tfrac12(1,-1)\). A wrong calculation would report \(T(1,0)=(1/2,1/2)\), simply reusing the two coefficients as standard coordinates. But the target vectors are \((2,1)\) and \((1,2)\), so the matched-basis rule gives \(T(1,0)=\tfrac12(2,1)+\tfrac12(1,2)=(3/2,3/2)\). The coefficients stay the same; the vectors used to rebuild the output change.
Order is equally important. If the target basis in that example were listed as \(((1,2),(2,1))\) instead, matching in order would send the first source vector to \((1,2)\) and the second to \((2,1)\). That is a different map. To avoid this error, write both ordered lists explicitly, match index \(i\) to index \(i\), and derive the inverse by reversing those same matches.
Check Your Understanding
For each question, keep the order of the bases visible and check any proposed inverse by composing the maps.
- Let \(\mathcal{B}=((1,0),(1,1))\) and \(\mathcal{C}=((1,1),(0,1))\) be ordered bases of \(\mathbb{R}^2\). Find the matched-basis map \(T(x,y)\).
- For the map in Question 1, find \(T^{-1}(s,t)\) by expressing \((s,t)\) in the basis \(\mathcal{C}\).
- Let \(T:\mathbb{R}^2\to W\) match \(((1,1),(1,-1))\) with \(((1,0,0),(0,1,1))\). Describe \(W\) and calculate \(T(x,y)\).
- Explain why prescribing the images of all vectors in a basis determines at most one linear map.
- In a matched-basis construction, what changes if the order of the target basis is reversed?
