Two Questions About a Linear Map
A linear map \(T:V\to W\) has a domain \(V\) and a codomain \(W\). Two basic questions concern how inputs and outputs relate. Is each output produced by at most one input? Is each element of the codomain produced by some input? The first question is injectivity; the second is surjectivity. The kernel and image turn these questions into precise statements about subsets of \(V\) and \(W\).
Throughout, \(V\) and \(W\) are vector spaces over the same field \(\mathbb{F}\), which may be \(\mathbb{R}\) or \(\mathbb{C}\). No finite-dimensional assumption is needed for the criteria proved here. A linear map preserves addition and scalar multiplication: \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) and \(T(c\mathbf{u})=cT(\mathbf{u})\). Write \(\mathbf{0}_V\) and \(\mathbf{0}_W\) for the zero vectors in the domain and codomain, respectively.
These definitions ask different questions. To test injectivity, compare two inputs with the same output. To test surjectivity, start with an arbitrary target in \(W\) and ask whether an input maps to it. The kernel and image give compact ways to carry out these tests.
The Kernel Criterion for Injectivity
Proof. Suppose first that \(T\) is injective. If \(\mathbf{v}\in\ker T\), then \(T(\mathbf{v})=\mathbf{0}_W\). Linearity gives \(T(\mathbf{0}_V)=\mathbf{0}_W\), so \(T(\mathbf{v})=T(\mathbf{0}_V)\). Injectivity implies \(\mathbf{v}=\mathbf{0}_V\). Also \(\mathbf{0}_V\in\ker T\), since \(T(\mathbf{0}_V)=\mathbf{0}_W\). Therefore \(\ker T=\{\mathbf{0}_V\}\).
Conversely, suppose \(\ker T=\{\mathbf{0}_V\}\). Let \(\mathbf{u},\mathbf{v}\in V\) satisfy \(T(\mathbf{u})=T(\mathbf{v})\). By linearity, \(T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})=\mathbf{0}_W\). Thus \(\mathbf{u}-\mathbf{v}\in\ker T\), so \(\mathbf{u}-\mathbf{v}=\mathbf{0}_V\). Hence \(\mathbf{u}=\mathbf{v}\), which proves injectivity. \(\square\)
The subtraction in the converse direction is the key step: equal outputs imply that the difference of the inputs lies in the kernel. The theorem depends on linearity because it uses \(T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})\). For an arbitrary function, a condition involving the preimage of zero need not determine whether distinct inputs share an output.
The Image Criterion for Surjectivity
Proof. Suppose \(T\) is surjective. Every \(\mathbf{w}\in W\) has an input \(\mathbf{v}\in V\) such that \(T(\mathbf{v})=\mathbf{w}\). By the definition of image, this says \(\mathbf{w}\in\operatorname{im}T\). Hence \(W\subseteq\operatorname{im}T\). Every value \(T(\mathbf{v})\) belongs to \(W\), because \(W\) is the codomain, so \(\operatorname{im}T\subseteq W\). The two inclusions give \(\operatorname{im}T=W\).
Conversely, suppose \(\operatorname{im}T=W\). Take any \(\mathbf{w}\in W\). The assumed equality gives \(\mathbf{w}\in\operatorname{im}T\). By the definition of image, there is a \(\mathbf{v}\in V\) with \(T(\mathbf{v})=\mathbf{w}\). This holds for every \(\mathbf{w}\in W\), so \(T\) is surjective. \(\square\)
This criterion also emphasizes why the codomain matters. The values of a map determine its image, but surjectivity asks whether that image fills the specified codomain. The same formula can be surjective with one codomain and fail to be surjective with a larger one.
Working Directly from the Definitions
For a matrix \(\mathbf{A}\in\mathbb{F}^{m\times n}\), the rule \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\) defines a linear map from \(\mathbb{F}^n\) to \(\mathbb{F}^m\): matrix multiplication distributes over vector addition and scalar multiplication. Its kernel can be found by solving \(\mathbf{A}\mathbf{x}=\mathbf{0}\); its image consists of the outputs \(\mathbf{A}\mathbf{x}\). The following examples use these equations without relying on dimension counts.
Worked Example: Injective but Not Surjective
Define \(T:\mathbb{R}^2\to\mathbb{R}^3\) by \(T(x,y)=(x+y,2x+2y,x-y)\). Determine whether \(T\) is injective and whether it is surjective.
For injectivity, find the kernel by setting all three output coordinates equal to zero:
The first and third equations give \(2x=0\), so \(x=0\), and then \(y=0\). These values satisfy the middle equation as well. Thus \(\ker T=\{(0,0)\}\), and the kernel criterion proves that \(T\) is injective. For a direct check, \(T(0,0)=(0,0,0)\).
To test surjectivity, consider the codomain vector \((0,1,0)\). If it were an output, we would need \(x+y=0\) and \(2x+2y=1\). But the left side of the second equation is \(2(x+y)\), which would be \(0\) by the first equation. This would require \(0=1\), which is impossible. Therefore \((0,1,0)\notin\operatorname{im}T\), so the image is not all of \(\mathbb{R}^3\) and \(T\) is not surjective.
Check: For example, \(T(1,0)=(1,2,1)\), since its coordinates are \(1+0=1\), \(2(1)+2(0)=2\), and \(1-0=1\). The proposed missing vector cannot be obtained, as the first two output coordinates of every \(T(x,y)\) always satisfy second coordinate \(=2\) times first coordinate.
Worked Example: Neither Injective Nor Surjective
Define \(S:\mathbb{R}^2\to\mathbb{R}^2\) by \(S(x,y)=(x+2y,2x+4y)\). Determine its kernel and image, then decide injectivity and surjectivity.
For the kernel, solve \(x+2y=0\) and \(2x+4y=0\). The second equation is twice the first, so the solutions satisfy \(x=-2y\). Writing \(y=t\), the kernel is \(\{(-2t,t):t\in\mathbb{R}\}\), which contains nonzero vectors. In particular, \(S(-2,1)=(-2+2,\,2(-2)+4)=(0,0)\), while \(S(0,0)=(0,0)\). Since \((-2,1)\ne(0,0)\), \(S\) is not injective.
For the image, let \(s=x+2y\). Then \(S(x,y)=(s,2s)\), so every output has second coordinate twice its first. Conversely, for any \(s\in\mathbb{R}\), choosing \((x,y)=(s,0)\) gives \(S(s,0)=(s,2s)\). Therefore \(\operatorname{im}S=\{(s,2s):s\in\mathbb{R}\}\), not all of \(\mathbb{R}^2\). For instance, \((0,1)\) is not in the image, since it would require \(1=2\cdot0\). Thus \(S\) is not surjective.
Check: The input \((3,0)\) gives \(S(3,0)=(3,6)\), which has the required form \((s,2s)\) with \(s=3\). The nonzero input \((-2,1)\) gives zero, confirming the nontrivial kernel found above.
Worked Example: Surjective but Not Injective
Define \(P:\mathbb{R}^3\to\mathbb{R}^2\) by \(P(x,y,z)=(x,y)\). Decide whether \(P\) is injective and surjective.
The kernel consists of inputs whose first two coordinates are zero: \(P(x,y,z)=(0,0)\) exactly when \(x=0\) and \(y=0\). Thus \(\ker P=\{(0,0,z):z\in\mathbb{R}\}\), which contains \((0,0,1)\ne(0,0,0)\). In fact, \(P(0,0,1)=(0,0)=P(0,0,0)\), so \(P\) is not injective.
Now take an arbitrary target \((a,b)\in\mathbb{R}^2\). The input \((a,b,0)\in\mathbb{R}^3\) satisfies \(P(a,b,0)=(a,b)\). Since this construction works for every target, \(\operatorname{im}P=\mathbb{R}^2\) and \(P\) is surjective.
Check: For the target \((3,-2)\), the input \((3,-2,0)\) gives \(P(3,-2,0)=(3,-2)\). Also \(P(0,0,1)=(0,0)\), confirming the stated nonzero kernel element.
Common Pitfall: A Zero Output Does Not Force a Zero Input
In the second example, the equations for a zero output are \(x+2y=0\) and \(2x+4y=0\). A tempting but incorrect calculation is to treat these as two independent equations and conclude \(x=0\) and \(y=0\). The correction is to notice that the second equation is exactly twice the first. Taking \(y=1\) in the first equation gives \(x=-2\), and then both equations hold: \(-2+2=0\) and \(2(-2)+4=0\). Thus a nonzero input maps to zero, so injectivity fails.
The kernel criterion does not say that every input with output zero must be zero for every map. It says that this property holds precisely for injective linear maps. Likewise, surjectivity is not established by finding several outputs; it requires an argument that every element of the codomain has a preimage, or a specific codomain element that has none.
Zero spaces cause no exception to the criteria. If \(V=\{\mathbf{0}_V\}\), then \(\ker T=\{\mathbf{0}_V\}\), so every map from \(V\) is injective; it is surjective only if \(W=\{\mathbf{0}_W\}\). If \(W=\{\mathbf{0}_W\}\), every map into \(W\) is surjective, while injectivity still depends on whether the kernel contains only \(\mathbf{0}_V\). These cases follow from the same definitions and proofs.
Check Your Understanding
Use the kernel and image definitions, and justify each conclusion from the relevant criterion.
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\) given by \(T(x,y)=(x-y,2x-2y)\), find \(\ker T\). Is \(T\) injective?
- For the map in question 1, describe its image and give a vector in the codomain that is not in the image.
- Prove directly that the identity map \(\operatorname{id}_V:V\to V\), defined by \(\operatorname{id}_V(\mathbf{v})=\mathbf{v}\), is both injective and surjective.
- For \(Q:\mathbb{R}^2\to\mathbb{R}\) given by \(Q(x,y)=x\), find its kernel and determine whether it is injective or surjective.
- In one sentence each, state the kernel test for injectivity and the image test for surjectivity, including the spaces to which the kernel and image belong.
