What a Fiber Records
A linear map can send several different vectors to the same output. For a fixed output \(\mathbf{w}\), the collection of all inputs that produce it is called a fiber. Understanding that collection is useful because it separates two questions: whether an output can be reached at all, and, if it can, how many inputs reach it. For a linear map, the kernel gives a precise description of the differences among those inputs.
Let \(V\) and \(W\) be vector spaces over \(\mathbb{R}\), and let \(T:V\to W\) be linear. Linearity means that \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\) and \(a,b\in\mathbb{R}\). The kernel is \(\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}\), and the image is \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). These definitions do not require the spaces to be finite-dimensional.
A fiber may be empty: if no input maps to \(\mathbf{w}\), then \(\mathbf{w}\notin\operatorname{im}T\). The key situation is when at least one input does map to \(\mathbf{w}\). Choose such an input and call it \(\mathbf{v}_0\), so \(T(\mathbf{v}_0)=\mathbf{w}\). The result below says that every other input in that fiber differs from \(\mathbf{v}_0\) by a vector in the kernel, and that adding any kernel vector to \(\mathbf{v}_0\) gives another input in the fiber.
The Fiber-Translate Theorem
Proof. Assume \(T(\mathbf{v}_0)=\mathbf{w}\). To prove equality of the two sets, prove both inclusions. First take any \(\mathbf{v}\in T^{-1}(\{\mathbf{w}\})\). By the definition of a fiber, \(T(\mathbf{v})=\mathbf{w}=T(\mathbf{v}_0)\). Linearity gives \(T(\mathbf{v}-\mathbf{v}_0)=T(\mathbf{v})-T(\mathbf{v}_0)=\mathbf{0}\), so \(\mathbf{v}-\mathbf{v}_0\in\ker T\). Thus \(\mathbf{v}=\mathbf{v}_0+(\mathbf{v}-\mathbf{v}_0)\in\mathbf{v}_0+\ker T\). For the reverse inclusion, take any \(\mathbf{z}\in\ker T\). Then \(T(\mathbf{z})=\mathbf{0}\), and linearity gives \(T(\mathbf{v}_0+\mathbf{z})=T(\mathbf{v}_0)+T(\mathbf{z})=\mathbf{w}+\mathbf{0}=\mathbf{w}\). Hence \(\mathbf{v}_0+\mathbf{z}\) belongs to the fiber. Both inclusions hold, proving the equality. If there is no such \(\mathbf{v}_0\), the fiber has no elements and is empty. \(\square\)
The two inclusions explain why the conclusion is an equality rather than merely a way to find some solutions. Subtracting one known preimage from any solution produces an element of the kernel; adding any kernel element to the known preimage produces a solution. The choice of \(\mathbf{v}_0\) is not unique, but choosing a different point in the same fiber does not change the fiber as a set.
Proof. If \(T(\mathbf{u})=T(\mathbf{v})\), linearity gives \(T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})=\mathbf{0}\), so \(\mathbf{u}-\mathbf{v}\in\ker T\). Conversely, if \(\mathbf{u}-\mathbf{v}\in\ker T\), then \(T(\mathbf{u}-\mathbf{v})=\mathbf{0}\). By linearity, \(T(\mathbf{u})-T(\mathbf{v})=\mathbf{0}\), which implies \(T(\mathbf{u})=T(\mathbf{v})\). Both implications hold. \(\square\)
This also gives a useful check on the theorem: a fiber contains exactly the inputs whose pairwise differences belong to the kernel. It is not generally a subspace, because it need not contain \(\mathbf{0}\); instead, it is a translated copy of a subspace.
Worked Examples
Worked Example: A Map from \(\mathbb{R}^3\) to \(\mathbb{R}^2\)
Define \(T:\mathbb{R}^3\to\mathbb{R}^2\) by \(T(x,y,z)=(x+y,y+z)\). Find the fiber over \(\mathbf{w}=(2,1)\).
First find one preimage. Setting \(y=0\), the equations \(x+y=2\) and \(y+z=1\) give \(x=2\) and \(z=1\). Thus \(\mathbf{v}_0=(2,0,1)\) and \(T(\mathbf{v}_0)=(2,1)\). To find the kernel, solve \(T(x,y,z)=(0,0)\): the equations are \(x+y=0\) and \(y+z=0\). Taking \(y=t\) gives \(x=-t\) and \(z=-t\), so $$ \ker T=\{(-t,t,-t):t\in\mathbb{R}\} =\{t(-1,1,-1):t\in\mathbb{R}\}. $$ The fiber-translate theorem now gives $$ T^{-1}(\{(2,1)\})=(2,0,1)+\ker T =\{(2-t,t,1-t):t\in\mathbb{R}\}. $$ Check: for every \(t\), \(T(2-t,t,1-t)=((2-t)+t,\;t+(1-t))=(2,1)\). At \(t=0\) the input is the chosen preimage, and all other values differ from it by a kernel vector.
Worked Example: A Fiber of Polynomial Differentiation
Let \(P_3\) be the real vector space of polynomials of degree at most \(3\), let \(P_2\) be the real vector space of polynomials of degree at most \(2\), and define \(D:P_3\to P_2\) by \(D(p)=p'\). Describe the fiber over \(q(x)=2x+3x^2\).
One preimage is \(p_0(x)=x^2+x^3\), since \(p_0'(x)=2x+3x^2=q(x)\). To identify the kernel, write an arbitrary element of \(P_3\) as \(p(x)=a+bx+cx^2+dx^3\). Then \(p'(x)=b+2cx+3dx^2\). This derivative is the zero polynomial exactly when \(b=c=d=0\), while \(a\) can be any real number. Therefore \(\ker D\) consists of the constant polynomials, and $$ D^{-1}(\{q\})=p_0+\ker D =\{x^2+x^3+a:a\in\mathbb{R}\}. $$ Check: for every \(a\in\mathbb{R}\), \((x^2+x^3+a)'=2x+3x^2\). The constant changes the input but not its derivative, exactly as the kernel description predicts.
Worked Example: A Fiber of the Trace Map
Define \(T:\mathbb{R}^{2\times 2}\to\mathbb{R}\) by \(T(\mathbf{A})=\operatorname{tr}\mathbf{A}\), the sum of the diagonal entries. Describe the fiber over \(3\).
The matrix \(\mathbf{A}_0=\begin{bmatrix}3&0\\0&0\end{bmatrix}\) has trace \(3\), so it is one preimage. A general matrix has the form \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\), and its trace is zero exactly when \(d=-a\). Hence $$ \ker T= \left\{ \begin{bmatrix}a&b\\c&-a\end{bmatrix} :a,b,c\in\mathbb{R} \right\}. $$ Translating this kernel by \(\mathbf{A}_0\) gives $$ T^{-1}(\{3\})= \left\{ \begin{bmatrix}3+a&b\\c&-a\end{bmatrix} :a,b,c\in\mathbb{R} \right\}. $$ Check: each displayed matrix has trace \((3+a)+(-a)=3\). Conversely, every \(2\times2\) matrix with trace \(3\) has diagonal entries \(r\) and \(3-r\), and can be written in the displayed form by taking \(a=r-3\). Thus the set includes every matrix in the fiber.
Empty Fibers, Distinct Fibers, and Uniqueness
The fiber-translate theorem applies only after a preimage has been found. The exact condition for such a preimage to exist follows directly from the definition of the image.
Proof. If the fiber is nonempty, it contains some \(\mathbf{v}\) with \(T(\mathbf{v})=\mathbf{w}\). By the definition of the image, this means \(\mathbf{w}\in\operatorname{im}T\). Conversely, if \(\mathbf{w}\in\operatorname{im}T\), the definition of the image supplies a \(\mathbf{v}\in V\) with \(T(\mathbf{v})=\mathbf{w}\). That \(\mathbf{v}\) belongs to the fiber, so the fiber is nonempty. \(\square\)
For example, let \(S:\mathbb{R}^2\to\mathbb{R}^2\) be \(S(x,y)=(x,0)\). The fiber over \((1,2)\) is empty: the second coordinate of every output of \(S\) is \(0\), so it cannot equal \(2\). In contrast, \((1,0)\) is an output, and its fiber is \(\{(1,y):y\in\mathbb{R}\}\). The distinction matters: an empty set cannot be described as a translate based at a preimage, because no preimage exists.
If a fiber is nonempty, it has exactly one element precisely when the kernel contains only \(\mathbf{0}\). Indeed, if the kernel is \(\{\mathbf{0}\}\), the translate \(\mathbf{v}_0+\ker T\) is the singleton \(\{\mathbf{v}_0\}\). If the kernel contains some \(\mathbf{z}\ne\mathbf{0}\), then both \(\mathbf{v}_0\) and \(\mathbf{v}_0+\mathbf{z}\) belong to the fiber, and they are distinct because \(\mathbf{z}\ne\mathbf{0}\). Thus the kernel measures the freedom remaining once one preimage is known.
Common Pitfall: Confusing the Fiber with the Kernel
Consider \(T:\mathbb{R}^2\to\mathbb{R}^2\) given by \(T(x,y)=(x,0)\), and the output \((2,0)\). The kernel is \(\{(0,t):t\in\mathbb{R}\}\). A common incorrect calculation is to say that the fiber over \((2,0)\) is the kernel, and therefore consists of vectors whose first coordinate is \(0\). That set maps to \((0,0)\), not to \((2,0)\).
The correction is to choose a preimage, such as \(\mathbf{v}_0=(2,0)\), and translate the kernel by it: $$ T^{-1}(\{(2,0)\})=(2,0)+\ker T =\{(2,t):t\in\mathbb{R}\}. $$ Check: \(T(2,t)=(2,0)\) for every real \(t\). The kernel describes the allowable differences between inputs in one fiber; it is the fiber itself only when the chosen output is \(\mathbf{0}\), in which case the zero vector is a preimage.
Check Your Understanding
Use the definitions and the fiber-translate theorem to justify each answer.
- For \(T(x,y)=(x+y,y)\), find the kernel and describe the fiber over \((3,1)\).
- For differentiation \(D:P_3\to P_2\), describe the fiber over \(q(x)=1+2x\), or explain why it is empty.
- For \(S(x,y)=(x,0)\), explain why the fiber over \((4,1)\) is empty.
- Prove that if \(\mathbf{u}\) and \(\mathbf{v}\) belong to the same fiber of a linear map, then \(\mathbf{u}-\mathbf{v}\in\ker T\).
- If a nonempty fiber contains two distinct vectors, what can you conclude about \(\ker T\)? Justify your answer.
