From a Codomain Condition to a Set of Inputs
A linear map sends vectors in a domain \(V\) to vectors in a codomain \(W\). Often the question is not simply which outputs occur, but which inputs produce an output with a specified property. For example, we might ask which vectors map into a line in \(\mathbb{R}^2\), or which polynomials have derivatives in a chosen polynomial subspace. The set of all such inputs is a preimage.
Let \(T:V\to W\) be a map and let \(S\) be any subset of \(W\). The preimage of \(S\) consists of the elements of \(V\) whose images under \(T\) belong to \(S\). The word “preimage” describes the direction of the search: start with a set in the codomain and find the inputs that land in it. This definition does not require \(T\) to have an inverse.
A subspace is a subset that contains the zero vector and is closed under vector addition and scalar multiplication. The key fact is that linearity carries these closure properties back through a map: if two inputs land in a subspace, then their sum does too, and any scalar multiple of one of those inputs does too.
Proof. First, linearity gives \(T(\mathbf{0})=\mathbf{0}\): since \(T(\mathbf{0})=T(\mathbf{0}+\mathbf{0})=T(\mathbf{0})+T(\mathbf{0})\), subtracting \(T(\mathbf{0})\) from both sides gives \(T(\mathbf{0})=\mathbf{0}\). Because \(U\) is a subspace, \(\mathbf{0}\in U\), so \(\mathbf{0}\in T^{-1}(U)\). Now take any \(\mathbf{v}_1,\mathbf{v}_2\in T^{-1}(U)\). By the definition of preimage, \(T(\mathbf{v}_1)\in U\) and \(T(\mathbf{v}_2)\in U\). Since \(U\) is closed under addition, \(T(\mathbf{v}_1)+T(\mathbf{v}_2)\in U\). Linearity gives \(T(\mathbf{v}_1+\mathbf{v}_2)=T(\mathbf{v}_1)+T(\mathbf{v}_2)\), so \(\mathbf{v}_1+\mathbf{v}_2\in T^{-1}(U)\). Finally, for any scalar \(c\in\mathbb{F}\) and \(\mathbf{v}_1\in T^{-1}(U)\), closure of \(U\) under scalar multiplication gives \(cT(\mathbf{v}_1)\in U\). Since \(T(c\mathbf{v}_1)=cT(\mathbf{v}_1)\), we have \(c\mathbf{v}_1\in T^{-1}(U)\). Thus the preimage contains zero and is closed under addition and scalar multiplication, so it is a subspace of \(V\). \(\square\)
Turning the Definition into Equations
For coordinate maps, membership in a target subspace usually translates into one or more equations on the output coordinates. Substituting the formula for \(T\) then gives equations on the input coordinates. Solving those equations describes the preimage. To check the result, verify both directions: every input in the stated set maps into the target subspace, and every input whose image lies there satisfies the equations used to describe the set.
Express what it means for \(T(\mathbf{v})\) to belong to the specified subspace.
Replace the output coordinates by their formulas in terms of the input.
Describe all inputs satisfying the resulting conditions, then check that the description is equivalent to the original membership condition.
Worked Example: The Preimage of a Line in \(\mathbb{R}^2\)
Define \(T:\mathbb{R}^3\to\mathbb{R}^2\) by \(T(x,y,z)=(x+y,y+z)\), and let \(U=\operatorname{span}\{(1,1)\}\). Find \(T^{-1}(U)\).
The map is linear because each output coordinate is a linear combination of the input coordinates. A pair \((r,s)\) belongs to \(U\) exactly when \(r=s\). Therefore \((x,y,z)\) belongs to \(T^{-1}(U)\) exactly when \(x+y=y+z\). Subtracting \(y\) from both sides gives \(x=z\), while \(y\) is unrestricted. Write \(x=z=a\) and \(y=b\). Then
Check: For any \(a,b\in\mathbb{R}\), \(T(a,b,a)=(a+b,b+a)=(a+b)(1,1)\in U\). Conversely, if \(T(x,y,z)\in U\), its two coordinates must be equal, so \(x+y=y+z\), which forces \(x=z\). Thus the description includes exactly the desired inputs.
Worked Example: A Coordinate Map with One Output Constraint
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T(x,y)=(x+y,2x-y)\), and let \(U=\operatorname{span}\{(1,0)\}\). Find the preimage.
The map is linear, and \((r,s)\in U\) exactly when \(s=0\). Thus \(T(x,y)\in U\) exactly when \(2x-y=0\), so \(y=2x\). Setting \(x=t\), the input is \((t,2t)=t(1,2)\), giving
Check: \(T(t,2t)=(3t,0)=3t(1,0)\), which belongs to \(U\) for every \(t\). In the other direction, if \(T(x,y)\in U\), its second coordinate must vanish, so \(2x-y=0\) and \((x,y)=(x,2x)\). This proves that no other inputs belong to the preimage.
Polynomial Preimages
The same method applies when the domain or codomain consists of polynomials. Write a general input polynomial with unknown coefficients, apply the map, and translate membership in the target subspace into conditions on those coefficients. Here \(\mathcal{P}_n(\mathbb{R})\) denotes the vector space of real polynomials of degree at most \(n\), including the zero polynomial.
Worked Example: Derivatives That Are Linear Polynomials
Consider the derivative map \(D:\mathcal{P}_3(\mathbb{R})\to\mathcal{P}_2(\mathbb{R})\), defined by \(D(p)=p'\), and the subspace \(U=\operatorname{span}\{1,x\}\) of \(\mathcal{P}_2(\mathbb{R})\). Find \(D^{-1}(U)\).
Write an arbitrary polynomial in the domain as \(p(x)=a+bx+cx^2+dx^3\), where \(a,b,c,d\in\mathbb{R}\). Its derivative is
This derivative belongs to \(U\) exactly when its \(x^2\) coefficient is zero, so \(3d=0\) and therefore \(d=0\). There is no restriction on \(a,b,c\). Hence
Check: If \(p(x)=a+bx+cx^2\), then \(p'(x)=b+2cx\in U\). Conversely, if \(p'(x)\in U\), the coefficient \(3d\) of \(x^2\) in its derivative must be zero, forcing \(d=0\). The description is therefore exact. Differentiation is linear because \((c_1p+c_2q)'=c_1p'+c_2q'\) for all polynomials \(p,q\) and scalars \(c_1,c_2\in\mathbb{R}\).
Worked Example: Equal Values at Two Points
Define \(E:\mathcal{P}_2(\mathbb{R})\to\mathbb{R}^2\) by \(E(p)=(p(0),p(1))\), and let \(U=\operatorname{span}\{(1,1)\}\). Find \(E^{-1}(U)\).
Evaluation is linear: for scalars \(c_1,c_2\) and polynomials \(p,q\), evaluation of \(c_1p+c_2q\) at either point is \(c_1\) times the evaluation of \(p\) plus \(c_2\) times the evaluation of \(q\). Write \(p(x)=a+bx+cx^2\). Then \(E(p)=(a,a+b+c)\). This pair belongs to \(U\) exactly when its coordinates are equal. Thus \(a=a+b+c\), or \(b+c=0\). Setting \(c=-b\) gives \(p(x)=a+b(x-x^2)\), so
Check: For \(p(x)=a+b(x-x^2)\), we have \(p(0)=a\) and \(p(1)=a\), so \(E(p)=(a,a)=a(1,1)\in U\). Conversely, if \(E(p)\in U\), then \(p(0)=p(1)\), which gives \(a=a+b+c\) and hence \(c=-b\). This recovers exactly the stated family.
How the Preimage Relates to the Kernel and Image
The kernel of \(T\) is \(\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}\), and the image is \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). Since every subspace \(U\) contains \(\mathbf{0}\), every vector in \(\ker T\) belongs to \(T^{-1}(U)\). The following identity gives a more precise description of the outputs obtained from that preimage.
Proof. Suppose \(\mathbf{w}\in T(T^{-1}(U))\). By the definition of the image of a set, there is a vector \(\mathbf{v}\in T^{-1}(U)\) with \(\mathbf{w}=T(\mathbf{v})\). The definition of preimage gives \(T(\mathbf{v})\in U\), so \(\mathbf{w}\in U\). Also \(\mathbf{w}=T(\mathbf{v})\) belongs to \(\operatorname{im}T\). Thus \(T(T^{-1}(U))\subseteq U\cap\operatorname{im}T\). Conversely, suppose \(\mathbf{w}\in U\cap\operatorname{im}T\). Since \(\mathbf{w}\in\operatorname{im}T\), some \(\mathbf{v}\in V\) satisfies \(T(\mathbf{v})=\mathbf{w}\). Since \(\mathbf{w}\in U\), this \(\mathbf{v}\) belongs to \(T^{-1}(U)\). Therefore \(\mathbf{w}=T(\mathbf{v})\in T(T^{-1}(U))\), proving the reverse inclusion and the identity. \(\square\)
One consequence distinguishes a preimage from the kernel. The preimage is always at least as large as the kernel, but it equals the kernel only when \(U\) contains no nonzero vector that is also an output of \(T\).
Proof. First suppose \(T^{-1}(U)=\ker T\). If \(\mathbf{w}\in U\cap\operatorname{im}T\), write \(\mathbf{w}=T(\mathbf{v})\) for some \(\mathbf{v}\in V\). Because \(\mathbf{w}\in U\), we have \(\mathbf{v}\in T^{-1}(U)=\ker T\). Hence \(\mathbf{w}=T(\mathbf{v})=\mathbf{0}\), so the intersection contains only \(\mathbf{0}\). Now suppose \(U\cap\operatorname{im}T=\{\mathbf{0}\}\). If \(\mathbf{v}\in T^{-1}(U)\), then \(T(\mathbf{v})\in U\), and it also belongs to \(\operatorname{im}T\). Thus \(T(\mathbf{v})=\mathbf{0}\), so \(\mathbf{v}\in\ker T\). Conversely, if \(\mathbf{v}\in\ker T\), then \(T(\mathbf{v})=\mathbf{0}\in U\), so \(\mathbf{v}\in T^{-1}(U)\). The two sets are equal. \(\square\)
Common Pitfall: Checking the Wrong Coordinate
A frequent error is to treat the coordinates of an input as if they were the coordinates of its output. In the second worked example, a tempting but incorrect answer is \(T^{-1}(U)=\operatorname{span}\{(1,0)\}\), on the mistaken idea that the line \(U=\operatorname{span}\{(1,0)\}\) requires the input’s second coordinate to be zero. But membership in \(U\) is a condition on \(T(x,y)\), not directly on \((x,y)\). In fact, \(T(1,0)=(1,2)\), which is not in \(U\), so \((1,0)\) cannot belong to the preimage. The correct test is to require the output’s second coordinate \(2x-y\) to be zero, giving \(y=2x\) and \(T^{-1}(U)=\operatorname{span}\{(1,2)\}\).
Another distinction is useful: \(\operatorname{im}T\) is a set of outputs in \(W\), while \(T^{-1}(U)\) is a set of inputs in \(V\). The proposition above says that mapping the preimage into \(W\) produces precisely the outputs in both \(U\) and \(\operatorname{im}T\). This also explains why not every element of \(U\) must come from an input: some elements of \(U\) may not be in the image at all.
Check Your Understanding
For each question, translate the target-subspace condition into an equation on the input, then verify the resulting set by applying the map.
- Let \(T(x,y,z)=(x-z,y+z)\) and \(U=\operatorname{span}\{(1,0)\}\). Find \(T^{-1}(U)\).
- Let \(T(x,y)=(x-y,x+y)\) and \(U=\operatorname{span}\{(1,1)\}\). Find the preimage and check both directions.
- For \(D:\mathcal{P}_2(\mathbb{R})\to\mathcal{P}_1(\mathbb{R})\), \(D(p)=p'\), find the preimage of \(\operatorname{span}\{1\}\).
- State why the preimage of a subspace is a subspace, identifying where linearity is used.
- For a linear map \(T:V\to W\) and subspace \(U\subseteq W\), explain why \(T^{-1}(U)=\ker T\) when \(U\cap\operatorname{im}T=\{\mathbf{0}\}\).
