Three Sets That Should Not Be Confused
Arguments about a linear map can go wrong even when the calculations are correct. A common source of error is treating different sets as though they were the same: the codomain is not automatically the image, a fiber is not automatically a subspace, and a theorem that requires finite dimension cannot be used just because the spaces are vector spaces. These distinctions matter when deciding whether a map is onto, describing all solutions of an equation, or using dimensions to infer injectivity or surjectivity.
Let \(T:V\to W\) be a linear map between vector spaces over \(\mathbb{R}\). Linearity means \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\) and all \(a,b\in\mathbb{R}\). The codomain is \(W\), the space named as the target of the map. The image is the set of outputs that actually occur:
The kernel is the set of inputs mapped to the zero vector of \(W\): \(\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}\). For a fixed \(\mathbf{w}\in W\), the fiber over \(\mathbf{w}\) is the set of inputs that map to \(\mathbf{w}\): \(T^{-1}(\mathbf{w})=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{w}\}\). This notation describes a set of preimages; it does not assert that \(T\) has an inverse map.
Why the Image Is a Subspace, but May Be Proper
The image is always a subspace of the codomain. This fact does not say that it equals the codomain; equality is the additional condition of surjectivity.
Proof. First, \(T(\mathbf{0})=\mathbf{0}\), since linearity gives \(T(\mathbf{0})=T(0\mathbf{0})=0T(\mathbf{0})=\mathbf{0}\). Thus \(\mathbf{0}\in\operatorname{im}T\). Next, take any \(\mathbf{y}_1,\mathbf{y}_2\in\operatorname{im}T\) and scalars \(a,b\in\mathbb{R}\). By the definition of image, there are \(\mathbf{v}_1,\mathbf{v}_2\in V\) such that \(T(\mathbf{v}_1)=\mathbf{y}_1\) and \(T(\mathbf{v}_2)=\mathbf{y}_2\). Linearity gives \(a\mathbf{y}_1+b\mathbf{y}_2=aT(\mathbf{v}_1)+bT(\mathbf{v}_2)=T(a\mathbf{v}_1+b\mathbf{v}_2)\), which belongs to \(\operatorname{im}T\). The image contains zero and is closed under linear combinations, so it is a subspace of \(W\). \(\square\)
Worked Example: A Proper Image Inside the Codomain
Define \(T:\mathbb{R}^2\to\mathbb{R}^3\) by \(T(x,y)=(x+y,2x+2y,0)\). Find its image and kernel, and decide whether it is onto.
Rewrite the output using the two input coordinates:
As \(x,y\) vary, \(x+y\) can be any real number, so \(\operatorname{im}T=\{s(1,2,0):s\in\mathbb{R}\}=\operatorname{span}\{(1,2,0)\}\). The kernel consists of inputs satisfying \((x+y,2x+2y,0)=(0,0,0)\). The first coordinate gives \(x+y=0\), and that condition also makes the second coordinate zero. Therefore \(\ker T=\{(t,-t):t\in\mathbb{R}\}=\operatorname{span}\{(1,-1)\}\).
The vector \((0,0,1)\) belongs to the codomain \(\mathbb{R}^3\), but it is not in the image: every output has third coordinate \(0\). Thus \(T\) is not onto. Check: \(T(1,0)=(1,2,0)\), \(T(0,1)=(1,2,0)\), and \(T(1,-1)=(0,0,0)\), as the descriptions of the image and kernel predict. The domain has dimension \(2\), the kernel has dimension \(1\), and the image has dimension \(1\), consistent with rank-nullity.
A mistaken argument might say, “The formula for \(T\) produces vectors in \(\mathbb{R}^3\), so the image is \(\mathbb{R}^3\).” The correction is that being an element of the codomain is necessary for an output, but does not mean every codomain vector occurs. To prove onto, one must show that each vector in the codomain has a preimage, or equivalently that the image equals the codomain.
A Nonempty Fiber Is a Translate of the Kernel
A fiber over \(\mathbf{w}\) can be empty if \(\mathbf{w}\notin\operatorname{im}T\). If it is nonempty, choose one preimage \(\mathbf{v}_0\), so \(T(\mathbf{v}_0)=\mathbf{w}\). Every other preimage differs from \(\mathbf{v}_0\) by an element of the kernel. This gives the precise description below.
Proof. Suppose \(\mathbf{v}\in T^{-1}(\mathbf{w})\). Then \(T(\mathbf{v})=\mathbf{w}=T(\mathbf{v}_0)\), so linearity gives \(T(\mathbf{v}-\mathbf{v}_0)=T(\mathbf{v})-T(\mathbf{v}_0)=\mathbf{0}\). Hence \(\mathbf{v}-\mathbf{v}_0\in\ker T\), and \(\mathbf{v}=\mathbf{v}_0+\mathbf{k}\) for some \(\mathbf{k}\in\ker T\). Conversely, if \(\mathbf{k}\in\ker T\), then \(T(\mathbf{v}_0+\mathbf{k})=T(\mathbf{v}_0)+T(\mathbf{k})=\mathbf{w}+\mathbf{0}=\mathbf{w}\), so \(\mathbf{v}_0+\mathbf{k}\in T^{-1}(\mathbf{w})\). Both inclusions hold, proving the equality. \(\square\)
This description also identifies exactly when a nonempty fiber is a subspace. If \(\mathbf{w}=\mathbf{0}\), then the fiber is \(\ker T\), which is a subspace. If \(\mathbf{w}\ne\mathbf{0}\), every vector in the fiber maps to \(\mathbf{w}\), so the zero vector cannot belong to the fiber because \(T(\mathbf{0})=\mathbf{0}\ne\mathbf{w}\). Since every subspace must contain zero, that fiber is not a subspace.
Worked Example: A Fiber That Is Not a Subspace
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T(x,y)=(x+y,2x+2y)\). Find the fiber over \(\mathbf{w}=(3,6)\) and determine whether it is a subspace.
The equation \(T(x,y)=(3,6)\) requires \(x+y=3\); its second coordinate then reads \(2x+2y=6\), which is the same condition. A particular solution is \(\mathbf{v}_0=(3,0)\). The kernel condition is \(x+y=0\), so \(\ker T=\{(t,-t):t\in\mathbb{R}\}\). The fiber proposition gives
This is a line not passing through the origin, so it is not a subspace. In particular, \((3,0)\) belongs to the fiber, but its zero vector does not: \(T(0,0)=(0,0)\ne(3,6)\). Check: for every real \(t\), \(T(3+t,-t)=(3+t-t,\,2(3+t)-2t)=(3,6)\). Conversely, every solution of \(x+y=3\) has \(y=-t\) and \(x=3+t\) for \(t=-y\), so the displayed set includes every point of the fiber.
A common incorrect calculation is to conclude that \(T^{-1}(3,6)\) is a subspace because it is a line and it is closed under subtraction of its points. The displayed fiber contains \((3,0)\), but not \((0,0)\); it is a translated line, not a subspace through the origin. Closure under differences between points does not replace the subspace requirement that zero belong to the set.
Rank-Nullity Requires a Finite-Dimensional Domain
The Rank-Nullity Theorem applies when the domain is finite-dimensional. The codomain need not be finite-dimensional. In that case, the image is finite-dimensional because it is the image of a finite-dimensional domain, and the theorem states:
This formula is not a license to apply finite-dimensional dimension tests to arbitrary spaces. In particular, the result that an injective linear map between finite-dimensional spaces of equal dimension is onto relies on those finite dimensions. When the spaces are infinite-dimensional, that implication can fail.
Worked Example: An Injective Map on an Infinite-Dimensional Space That Is Not Onto
Let \(V=\mathbb{R}[x]\), the vector space of real polynomials, and define \(T:V\to V\) by \(T(p)=xp\). Show that \(T\) is injective but not onto, and identify the invalid dimension argument.
If \(T(p)=0\), then \(xp(x)\) is the zero polynomial. Writing \(p(x)=a_0+a_1x+\cdots+a_nx^n\) gives \(xp(x)=a_0x+a_1x^2+\cdots+a_nx^{n+1}\). Every coefficient of \(xp\) is therefore zero only when every \(a_i=0\), so \(p=0\). Thus \(\ker T=\{0\}\), and \(T\) is injective.
Every output \(xp(x)\) has constant coefficient zero. The constant polynomial \(1\) does not, so \(1\notin\operatorname{im}T\); therefore \(T\) is not onto. Check: \(T(1)=x\), not \(1\), while \(T(x^2+2)=x^3+2x\) has constant coefficient zero, as every output must.
The vector space \(V\) is infinite-dimensional: the polynomials \(1,x,x^2,\dots\) are linearly independent, since a finite linear combination equal to the zero polynomial has every coefficient equal to zero. It is invalid to use the finite-dimensional equal-dimensions test to conclude that injectivity forces surjectivity here. Likewise, the finite-dimensional Rank-Nullity Theorem cannot be invoked with \(V\) as its domain. The calculation \(\ker T=\{0\}\) is valid; the unsupported step would be using finite-dimensional dimension reasoning to turn that fact into surjectivity.
A Short Diagnostic Before You Conclude
When reviewing an argument about a linear map, check what each object actually represents and which hypotheses have been verified. The following distinctions catch many errors:
| Claim being considered | Question to check |
|---|---|
| The map is onto | Has every vector in the codomain been shown to lie in the image? |
| A fiber is a subspace | Does it contain zero? If the target is nonzero, it cannot. |
| Rank-nullity gives a dimension equation | Is the domain finite-dimensional? |
| Injectivity implies surjectivity | Are both spaces finite-dimensional and of equal dimension? |
| A fiber has a particular form | Is it empty, or has a preimage been chosen so it can be written as a kernel translate? |
These checks do not replace a proof, but they help locate the exact step that needs justification. For an onto claim, establish image equals codomain. For a fiber, first check whether the target is attained; if so, use one preimage and the kernel. For a dimension argument, state finite-dimensionality before applying rank-nullity or a dimension test. A correct conclusion depends not only on the calculation, but also on applying the right result under its actual hypotheses.
Check Your Understanding
For each question, identify the relevant set and state the hypothesis that supports your conclusion.
- For \(T:\mathbb{R}^2\to\mathbb{R}^3\) given by \(T(x,y)=(x,y,0)\), describe the image and give a codomain vector that is not in it.
- Explain why a nonempty fiber \(T^{-1}(\mathbf{w})\) with \(\mathbf{w}\ne\mathbf{0}\) cannot be a subspace.
- If \(T(\mathbf{v}_0)=\mathbf{w}\), express every vector in \(T^{-1}(\mathbf{w})\) using \(\mathbf{v}_0\) and \(\ker T\).
- State the finite-dimensionality hypothesis required to use rank-nullity for \(T:V\to W\).
- For \(T:\mathbb{R}[x]\to\mathbb{R}[x]\) defined by \(T(p)=xp\), explain why injectivity does not imply surjectivity.
