Inputs Sent to Zero and Outputs That Can Occur
A linear map takes inputs from one vector space and produces outputs in another. Two sets help describe what the map does: the kernel records the inputs that are sent to zero, while the image records all outputs that can actually occur. These sets answer different questions. The kernel is a subset of the domain; the image is a subset of the codomain.
Let \(V\) and \(W\) be vector spaces over a field \(\mathbb{F}\), and let \(T:V\to W\) be a linear map. Write \(\mathbf{0}_V\) and \(\mathbf{0}_W\) for the zero vectors in the two spaces when it is useful to distinguish them. An input \(\mathbf{v}\in V\) belongs to the kernel precisely when its output is \(\mathbf{0}_W\). An output \(\mathbf{w}\in W\) belongs to the image precisely when at least one input in \(V\) maps to it.
The image can also be written \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). This description emphasizes that the image is the collection of values of the map, not necessarily the whole codomain. The kernel may contain many inputs, or just the zero input; its elements all have the same output, namely zero. The definitions apply to any linear map, whether or not the spaces are finite-dimensional.
Proof. Since \(\mathbf{0}_V=\mathbf{0}_V+\mathbf{0}_V\), linearity gives \(T(\mathbf{0}_V)=T(\mathbf{0}_V)+T(\mathbf{0}_V)\). Subtracting \(T(\mathbf{0}_V)\) from both sides yields \(T(\mathbf{0}_V)=\mathbf{0}_W\). The kernel definition then puts \(\mathbf{0}_V\) in \(\ker T\). Taking \(\mathbf{0}_V\) as an input whose output is \(\mathbf{0}_W\) puts \(\mathbf{0}_W\) in \(\operatorname{im}T\). \(\square\)
A Map Between Coordinate Spaces
Worked Example: A Map from R³ to R²
Define \(T:\mathbb{R}^3\to\mathbb{R}^2\) by \(T(x,y,z)=(x+y,\;y+z)\). Find its kernel and image.
To find the kernel, set the output equal to the zero vector in \(\mathbb{R}^2\). The two coordinates must each be zero:
The first equation gives \(x=-y\), and the second gives \(z=-y\). Let \(y=-t\), where \(t\in\mathbb{R}\). Then \(x=t\), \(z=t\), so every kernel input has the form \((t,-t,t)\). Conversely, substituting this form into the map gives \(T(t,-t,t)=(t-t,-t+t)=(0,0)\). Therefore \(\ker T=\{(t,-t,t):t\in\mathbb{R}\}\).
To find the image, take an arbitrary output \((r,s)\in\mathbb{R}^2\). Choose the input \((r,0,s)\). Then \(T(r,0,s)=(r+0,0+s)=(r,s)\), so every vector in the codomain is attainable. Since outputs are already required to lie in \(\mathbb{R}^2\), this proves \(\operatorname{im}T=\mathbb{R}^2\).
Check: A general kernel input \((t,-t,t)\) maps to \((0,0)\), and an arbitrary proposed output \((r,s)\) is produced by the displayed input \((r,0,s)\). Both descriptions follow directly from evaluating \(T\).
Polynomial and Matrix Spaces
The same definitions work when vectors are polynomials or matrices. The zero vector is then the zero polynomial or zero matrix, and the kernel and image remain sets of elements of the domain and codomain, respectively. For a polynomial space, let \(P_n(\mathbb{R})\) denote the real polynomials of degree at most \(n\), including the zero polynomial.
Worked Example: A Map on Polynomial Spaces
Define \(T:P_2(\mathbb{R})\to P_1(\mathbb{R})\) by \(T(a+bx+cx^2)=(a+c)+(b-c)x\), where \(a,b,c\in\mathbb{R}\). Determine the kernel and image.
For a polynomial to lie in the kernel, its output polynomial must have both coefficients equal to zero. Thus \(a+c=0\) and \(b-c=0\). These equations give \(a=-c\) and \(b=c\). Writing \(c=t\), every kernel polynomial is \(-t+tx+tx^2=t(-1+x+x^2)\). Substitution confirms the result: for \(p(x)=t(-1+x+x^2)\), the coefficients are \(a=-t\), \(b=t\), \(c=t\), and \(T(p)=(-t+t)+(t-t)x=0\). Hence \(\ker T=\{t(-1+x+x^2):t\in\mathbb{R}\}\).
For the image, let \(q(x)=u+vx\) be any polynomial in \(P_1(\mathbb{R})\). Choose \(p(x)=u+vx\) as an input in \(P_2(\mathbb{R})\), so \(a=u\), \(b=v\), and \(c=0\). Then \(T(p)=u+vx=q(x)\). Thus every polynomial in the codomain is attainable, and \(\operatorname{im}T=P_1(\mathbb{R})\).
Check: The kernel condition used the two coefficients of the output, not the coefficients of the input. Substituting \(a=-t,\;b=t,\;c=t\) makes both output coefficients zero; choosing \(a=u,\;b=v,\;c=0\) produces the requested output \(u+vx\).
Worked Example: A Map on 2×2 Matrix Spaces
Over \(\mathbb{R}\), define \(T:\mathbb{R}^{2\times2}\to\mathbb{R}^{2\times2}\) by \(T(\mathbf{A})=\mathbf{A}+\mathbf{A}^{\top}\). Find its kernel and image.
Write \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\). Then \(T(\mathbf{A})=\begin{bmatrix}2a&b+c\\b+c&2d\end{bmatrix}\). For this output to be the zero matrix, its entries require \(2a=0\), \(b+c=0\), and \(2d=0\). Over \(\mathbb{R}\), these equations give \(a=d=0\) and \(c=-b\). Therefore \(\ker T=\left\{\begin{bmatrix}0&t\\-t&0\end{bmatrix}:t\in\mathbb{R}\right\}\). Indeed, adding the transpose of each such matrix gives the zero matrix.
Every output has equal off-diagonal entries, so every output is symmetric. Conversely, take any symmetric matrix \(\mathbf{S}=\begin{bmatrix}r&s\\s&t\end{bmatrix}\). Choose \(\mathbf{A}=\tfrac12\mathbf{S}=\begin{bmatrix}r/2&s/2\\s/2&t/2\end{bmatrix}\). Since \(\mathbf{A}^{\top}=\mathbf{A}\), we obtain \(T(\mathbf{A})=\mathbf{A}+\mathbf{A}^{\top}=2\mathbf{A}=\mathbf{S}\). Thus the image is exactly the set of symmetric \(2\times2\) real matrices.
Check: For a kernel element, \(\begin{bmatrix}0&t\\-t&0\end{bmatrix}+\begin{bmatrix}0&-t\\t&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\). For any proposed image element \(\mathbf{S}\), the chosen input \(\mathbf{S}/2\) maps back to \(\mathbf{S}\).
Matrix Maps: Null Space and Column Space
For a matrix map, the kernel and image can be described using familiar matrix sets. Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\), and define \(T:\mathbb{R}^n\to\mathbb{R}^m\) by \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\). The null space \(\operatorname{Nul}\mathbf{A}\) is the set of solutions \(\mathbf{x}\in\mathbb{R}^n\) of \(\mathbf{A}\mathbf{x}=\mathbf{0}\). The column space \(\operatorname{Col}\mathbf{A}\) is the set of linear combinations of the columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\) of \(\mathbf{A}\).
Proof. By definition, \(\mathbf{x}\in\ker T\) exactly when \(T(\mathbf{x})=\mathbf{0}\). Since \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\), this is exactly the condition \(\mathbf{A}\mathbf{x}=\mathbf{0}\), which defines \(\operatorname{Nul}\mathbf{A}\). Therefore the two sets are equal. For the image, the matrix-vector product is \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). As \(\mathbf{x}\) ranges over \(\mathbb{R}^n\), its entries \(x_1,\dots,x_n\) range independently over all real scalars. Hence the possible outputs are exactly all linear combinations of the columns, which is \(\operatorname{Col}\mathbf{A}\). \(\square\)
This result identifies the relevant sets, but it also highlights the roles of the columns and variables: solve the homogeneous equation to find the kernel, and form combinations of the columns to describe the image. The image is not the row space. The row space consists of combinations of rows, whereas the image consists of outputs in \(\mathbb{R}^m\).
What Composition Tells Us
Kernels and images also give immediate information about a composition. Suppose \(T:V\to W\) and \(S:W\to U\) are linear maps. The composition \(S\circ T:V\to U\) first applies \(T\), then applies \(S\). An input that \(T\) sends to zero is also sent to zero by the composition. Likewise, every output of the composition must be an output that \(S\) can produce.
Proof. If \(\mathbf{v}\in\ker T\), then \(T(\mathbf{v})=\mathbf{0}_W\). Linearity of \(S\) gives \(S(\mathbf{0}_W)=\mathbf{0}_U\), so \((S\circ T)(\mathbf{v})=\mathbf{0}_U\). This proves \(\mathbf{v}\in\ker(S\circ T)\) and the first inclusion. If \(\mathbf{u}\in\operatorname{im}(S\circ T)\), then for some \(\mathbf{v}\in V\), \(\mathbf{u}=S(T(\mathbf{v}))\). Since \(T(\mathbf{v})\in W\), this expresses \(\mathbf{u}\) as an output of \(S\), proving \(\mathbf{u}\in\operatorname{im}S\) and the second inclusion. \(\square\)
The inclusions need not be equalities. For example, let \(T:\mathbb{R}^2\to\mathbb{R}^2\) send \((x,y)\) to \((0,y)\), and let \(S:\mathbb{R}^2\to\mathbb{R}^2\) send \((x,y)\) to \((x,0)\). Then \((S\circ T)(x,y)=S(0,y)=(0,0)\). Thus \(\ker T=\{(x,0):x\in\mathbb{R}\}\), whereas \(\ker(S\circ T)=\mathbb{R}^2\). Also \(\operatorname{im}(S\circ T)=\{(0,0)\}\), whereas \(\operatorname{im}S=\{(x,0):x\in\mathbb{R}\}\). The direction of each inclusion matters.
Common Pitfall: Mixing Inputs and Outputs
A common error is to treat the kernel as a set of output vectors. For the map \(T:\mathbb{R}^3\to\mathbb{R}^2\) above, a mistaken calculation might say “\(T(1,0,0)=(1,0)\), so \((1,0)\) is in the kernel.” This is wrong twice over: the calculation shows that \((1,0)\) is an output, and it is not zero. The input \((1,0,0)\) is not in the kernel either, because its image is nonzero.
The correction is to keep the defining equations and the spaces in view. Kernel membership asks whether an input \(\mathbf{v}\in V\) satisfies \(T(\mathbf{v})=\mathbf{0}_W\). Image membership asks whether a proposed output \(\mathbf{w}\in W\) can be written as \(T(\mathbf{v})\) for some input \(\mathbf{v}\in V\). In the example, \((1,0)\) is in the image because \(T(1,0,0)=(1,0)\), while \((1,0,0)\) is not in the kernel.
Check Your Understanding
For each question, keep track of whether the proposed element is an input or an output.
- For \(T:\mathbb{R}^2\to\mathbb{R}\) defined by \(T(x,y)=x-y\), describe \(\ker T\) and check your description by substitution.
- For the polynomial map in the worked example, find an input that maps to \(3-2x\).
- For \(T(\mathbf{A})=\mathbf{A}+\mathbf{A}^{\top}\), explain why a matrix with unequal off-diagonal entries cannot lie in the image.
- State the relationship between the kernel of a matrix map and the null space of its matrix, and justify it from the definitions.
- If \(\mathbf{v}\in\ker T\), what can be concluded about \(\mathbf{v}\) and the kernel of \(S\circ T\)?
