Why Prescribed Images Need a Compatibility Check
Suppose a finite list of vectors spans a space \(V\), and we want a linear map to send each listed vector to a chosen vector in another space \(W\). Every vector in \(V\) can be expressed as a linear combination of the generators, so it is tempting to define the image of that vector by applying the same combination to the assigned images. The difficulty is that a vector may have more than one such expression. If two expressions lead to different proposed images, there is no well-defined map.
Throughout, \(V\) and \(W\) are vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is either \(\mathbb{R}\) or \(\mathbb{C}\). A list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) spans \(V\) if each \(\mathbf{x}\in V\) can be written as \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\) for scalars \(c_i\in\mathbb{F}\). A linear map \(T:V\to W\) preserves linear combinations: \(T(\sum_i c_i\mathbf{x}_i)=\sum_i c_iT(\mathbf{x}_i)\). A linear relation among the generators is a choice of scalars \(c_i\) for which \(\sum_i c_i\mathbf{v}_i=\mathbf{0}\).
Proof. First suppose such a linear map \(T\) exists. If \(\sum_i c_i\mathbf{v}_i=\mathbf{0}\), then linearity and the assigned values give \[ \sum_i c_i\mathbf{w}_i =\sum_i c_iT(\mathbf{v}_i) =T\!\left(\sum_i c_i\mathbf{v}_i\right) =T(\mathbf{0}) =\mathbf{0}. \] Here \(T(\mathbf{0})=\mathbf{0}\), since linearity gives \(T(\mathbf{0})=T(0\mathbf{0})=0T(\mathbf{0})=\mathbf{0}\). Thus every relation is preserved.
Conversely, assume every relation among the \(\mathbf{v}_i\) is preserved. Because the generators span \(V\), for each \(\mathbf{x}\in V\) there is at least one expression \(\mathbf{x}=\sum_i c_i\mathbf{v}_i\). Try to define \(T(\mathbf{x})=\sum_i c_i\mathbf{w}_i\). To show this value does not depend on the chosen expression, suppose also \(\mathbf{x}=\sum_i d_i\mathbf{v}_i\). Subtracting gives \(\sum_i(c_i-d_i)\mathbf{v}_i=\mathbf{0}\). The assumed condition then gives \(\sum_i(c_i-d_i)\mathbf{w}_i=\mathbf{0}\), so \(\sum_i c_i\mathbf{w}_i=\sum_i d_i\mathbf{w}_i\). The definition therefore gives exactly one value for each \(\mathbf{x}\).
It remains to check linearity. If \(\mathbf{x}=\sum_i c_i\mathbf{v}_i\) and \(\mathbf{y}=\sum_i d_i\mathbf{v}_i\), then \(\mathbf{x}+\mathbf{y}=\sum_i(c_i+d_i)\mathbf{v}_i\), and the definition gives \(T(\mathbf{x}+\mathbf{y})=\sum_i(c_i+d_i)\mathbf{w}_i=T(\mathbf{x})+T(\mathbf{y})\). For \(a\in\mathbb{F}\), \(a\mathbf{x}=\sum_i(ac_i)\mathbf{v}_i\), so \(T(a\mathbf{x})=\sum_i(ac_i)\mathbf{w}_i=aT(\mathbf{x})\). Finally, \(\mathbf{v}_j\) has the expression with coefficient \(1\) at position \(j\) and \(0\) elsewhere, so \(T(\mathbf{v}_j)=\mathbf{w}_j\). This constructs the required map. \(\square\)
Applying the Criterion to a Redundant List
Worked Example: Consistent Images for Three Generators
Let \(V=\mathbb{R}^2\) and take \(\mathbf{v}_1=\begin{bmatrix}1\\0\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}0\\1\end{bmatrix}\), and \(\mathbf{v}_3=\begin{bmatrix}1\\1\end{bmatrix}\). Prescribe \(\mathbf{w}_1=\begin{bmatrix}1\\1\end{bmatrix}\), \(\mathbf{w}_2=\begin{bmatrix}2\\-1\end{bmatrix}\), and \(\mathbf{w}_3=\begin{bmatrix}3\\0\end{bmatrix}\) in \(W=\mathbb{R}^2\). Decide whether these assignments define a linear map.
First find every relation, not just one. The equation \(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\) becomes \[ \begin{bmatrix}c_1+c_3\\c_2+c_3\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}. \] Thus \(c_1=-c_3\) and \(c_2=-c_3\). Setting \(c_3=t\), every relation has coefficients \((-t,-t,t)\), or equivalently is a scalar multiple of \(\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3=\mathbf{0}\).
Check the corresponding relation among the assigned images: \[ \mathbf{w}_1+\mathbf{w}_2-\mathbf{w}_3 =\begin{bmatrix}1\\1\end{bmatrix} +\begin{bmatrix}2\\-1\end{bmatrix} -\begin{bmatrix}3\\0\end{bmatrix} =\begin{bmatrix}1+2-3\\1-1-0\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}. \] Every relation is a multiple of this one, so every relation is preserved. The theorem guarantees a linear map. In fact, writing \(\begin{bmatrix}x\\y\end{bmatrix}=x\mathbf{v}_1+y\mathbf{v}_2\), the constructed map is \(T\begin{bmatrix}x\\y\end{bmatrix}=x\mathbf{w}_1+y\mathbf{w}_2=\begin{bmatrix}x+2y\\x-y\end{bmatrix}\).
Check: \(T(\mathbf{v}_1)=\begin{bmatrix}1\\1\end{bmatrix}=\mathbf{w}_1\), \(T(\mathbf{v}_2)=\begin{bmatrix}2\\-1\end{bmatrix}=\mathbf{w}_2\), and \(T(\mathbf{v}_3)=T\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\0\end{bmatrix}=\mathbf{w}_3\).
This example illustrates the role of redundancy. The first two generators already span \(\mathbb{R}^2\), while the third satisfies \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\). Its assigned image cannot be chosen independently: it must equal \(\mathbf{w}_1+\mathbf{w}_2\).
Worked Example: An Assignment That Is Not Well-Defined
Keep the same generators, but assign \(\mathbf{w}_1=\begin{bmatrix}1\\0\end{bmatrix}\), \(\mathbf{w}_2=\begin{bmatrix}0\\1\end{bmatrix}\), and \(\mathbf{w}_3=\begin{bmatrix}0\\0\end{bmatrix}\). The generator relation is \(\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3=\mathbf{0}\), but its assigned images give \[ \mathbf{w}_1+\mathbf{w}_2-\mathbf{w}_3 =\begin{bmatrix}1\\0\end{bmatrix} +\begin{bmatrix}0\\1\end{bmatrix} -\begin{bmatrix}0\\0\end{bmatrix} =\begin{bmatrix}1\\1\end{bmatrix}\ne\begin{bmatrix}0\\0\end{bmatrix}. \] The condition fails, so no linear map can have all three prescribed values.
The conflict can also be seen directly from two expressions for the same vector. Since \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\), the vector \(\begin{bmatrix}1\\1\end{bmatrix}\) would have to map to \(\mathbf{w}_3=\begin{bmatrix}0\\0\end{bmatrix}\) using the first expression, but to \(\mathbf{w}_1+\mathbf{w}_2=\begin{bmatrix}1\\1\end{bmatrix}\) using the second. A function cannot assign two different outputs to the same input.
Checking All Relations in a Larger List
When there are several redundancies, a single relation check is not enough unless we have shown that every relation follows from the ones checked. The next example finds all relations by solving the vector equation, then verifies the image condition for every one of them.
Worked Example: Two Independent Relations to Check
In \(V=\mathbb{R}^3\), let \(\mathbf{v}_1=\begin{bmatrix}1\\0\\1\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}0\\1\\1\end{bmatrix}\), \(\mathbf{v}_3=\begin{bmatrix}1\\1\\2\end{bmatrix}\), and \(\mathbf{v}_4=\begin{bmatrix}2\\-1\\1\end{bmatrix}\). Prescribe scalar images in \(W=\mathbb{R}\): \(\mathbf{w}_1=2,\mathbf{w}_2=-1,\mathbf{w}_3=1,\mathbf{w}_4=5\). Determine whether these images are compatible.
For a relation, write \(\sum_{i=1}^4c_i\mathbf{v}_i=\mathbf{0}\). Comparing the three coordinates gives \(c_1+c_3+2c_4=0\), \(c_2+c_3-c_4=0\), and \(c_1+c_2+2c_3+c_4=0\). Choose \(c_3=s\) and \(c_4=t\). The first two equations give \(c_1=-s-2t\) and \(c_2=-s+t\). Substituting into the third gives \[ (-s-2t)+(-s+t)+2s+t=0, \] so the third equation imposes no additional restriction. Thus every relation has coefficients \((-s-2t,-s+t,s,t)\), and can be written as a combination of the two relations with coefficient lists \((-1,-1,1,0)\) and \((-2,1,0,1)\).
Check both corresponding image relations: \[ -\,\mathbf{w}_1-\,\mathbf{w}_2+\mathbf{w}_3 =-2-(-1)+1=0, \qquad -2\mathbf{w}_1+\mathbf{w}_2+\mathbf{w}_4 =-2(2)+(-1)+5=0. \] Every relation has coefficients built from these two lists, and the same combinations of their image sums are therefore zero. The assignment is compatible.
The construction can be written explicitly. Each vector in the span has the form \(a\mathbf{v}_1+b\mathbf{v}_2=\begin{bmatrix}a\\b\\a+b\end{bmatrix}\), and the assigned value is \(2a-b\). For example, \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\), so its image is \(2+(-1)=1=\mathbf{w}_3\); also \(\mathbf{v}_4=2\mathbf{v}_1-\mathbf{v}_2\), so its image is \(2(2)-(-1)=5=\mathbf{w}_4\).
Check: The displayed formula sends \(\mathbf{v}_1\) to \(2\), \(\mathbf{v}_2\) to \(-1\), \(\mathbf{v}_3\) to \(1\), and \(\mathbf{v}_4\) to \(5\). The two relations also verify directly: \(-\mathbf{v}_1-\mathbf{v}_2+\mathbf{v}_3=\mathbf{0}\) and \(-2\mathbf{v}_1+\mathbf{v}_2+\mathbf{v}_4=\mathbf{0}\).
When Arbitrary Images Are Allowed
A list is a basis of \(V\) if it is linearly independent and spans \(V\). Independence means that the only coefficients satisfying \(\sum_i c_i\mathbf{v}_i=\mathbf{0}\) are all zero. A basis therefore has no nontrivial relations to obstruct an assignment.
Proof. Consider any relation \(\sum_i c_i\mathbf{v}_i=\mathbf{0}\). Since the list is independent, \(c_i=0\) for every \(i\), and hence \(\sum_i c_i\mathbf{w}_i=\mathbf{0}\). The relation-preservation condition in the theorem holds, so a linear map with the prescribed images exists. \(\square\)
This proposition separates two situations. With a basis, each vector has a unique expression in terms of the basis vectors, so there are no competing expressions that could force a compatibility condition on the chosen images. With a spanning list that is not independent, the assignments must respect its redundancies.
Worked Example: Assigning Images on a Basis
Let \(V=\mathbb{R}^2\) with basis \(\mathbf{v}_1=\begin{bmatrix}1\\1\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}1\\-1\end{bmatrix}\). Prescribe \(\mathbf{w}_1=\begin{bmatrix}2\\0\end{bmatrix}\) and \(\mathbf{w}_2=\begin{bmatrix}0\\4\end{bmatrix}\) in \(W=\mathbb{R}^2\). Construct the map.
For \(\begin{bmatrix}x\\y\end{bmatrix}=a\mathbf{v}_1+b\mathbf{v}_2\), coordinate comparison gives \(a+b=x\) and \(a-b=y\). Adding the equations gives \(2a=x+y\), so \(a=(x+y)/2\); subtracting the second equation from the first gives \(2b=x-y\), so \(b=(x-y)/2\). Define the image using these coefficients: \[ T\begin{bmatrix}x\\y\end{bmatrix} =a\mathbf{w}_1+b\mathbf{w}_2 =\begin{bmatrix}2a\\4b\end{bmatrix} =\begin{bmatrix}x+y\\2x-2y\end{bmatrix}. \]
Check: Substituting \(\mathbf{v}_1\) gives \(T\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}2\\0\end{bmatrix}=\mathbf{w}_1\), and substituting \(\mathbf{v}_2\) gives \(T\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}0\\4\end{bmatrix}=\mathbf{w}_2\). The construction uses each vector’s basis coefficients, so it applies to every input in \(V\).
Common Pitfall: Checking Only a Convenient Relation
For the four generators in the earlier example, suppose instead that we assign \(\mathbf{w}_1=2\), \(\mathbf{w}_2=-1\), \(\mathbf{w}_3=1\), and \(\mathbf{w}_4=4\). Checking only \(\mathbf{v}_3-\mathbf{v}_1-\mathbf{v}_2=\mathbf{0}\) appears to work: \[ \mathbf{w}_3-\mathbf{w}_1-\mathbf{w}_2=1-2-(-1)=0. \] But the other relation is \(\mathbf{v}_4-2\mathbf{v}_1+\mathbf{v}_2=\mathbf{0}\). Its assigned images give \[ \mathbf{w}_4-2\mathbf{w}_1+\mathbf{w}_2=4-2(2)+(-1)=-1\ne0. \] So this assignment does not define a linear map. The correction is to find all relations, or to prove that the relations checked generate every relation. In this example, the second condition requires \(\mathbf{w}_4=2\mathbf{w}_1-\mathbf{w}_2=5\).
A practical proof can be organized as follows:
- Write a general combination of the generators and set it equal to \(\mathbf{0}\).
- Solve for all possible coefficients, or find a set of relations and prove every relation follows from them.
- Apply the same coefficients to the prescribed images and verify that the result is always \(\mathbf{0}\).
- When the condition holds, define the image of an arbitrary vector using any generator expression; the theorem proves that the choice of expression does not matter.
This method turns the phrase “define a map by its values on generators” into a precise proof obligation. The essential question is not whether the proposed images look reasonable individually, but whether they preserve every equality among the generators.
Check Your Understanding
For each question, identify the relations that matter before deciding whether the prescribed images are possible.
- For \(\mathbf{v}_1=\begin{bmatrix}1\\0\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}0\\1\end{bmatrix}\), and \(\mathbf{v}_3=\begin{bmatrix}2\\1\end{bmatrix}\), write one relation and state the required condition on \(\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3\).
- Explain why any prescribed images on a linearly independent spanning list satisfy the relation-preservation condition.
- For the four vectors in the third worked example, verify that \((-2,1,0,1)\) gives a relation among the generators.
- Give an example of two different expressions for one vector that would lead to conflicting assigned images.
- Why is checking one relation insufficient unless you also establish that every relation is generated by the ones checked?
