Two Subsets Associated with a Linear Map
A linear map connects two vector spaces, but it need not use every vector in its codomain, and some vectors in its domain may map to zero. The kernel records the inputs with zero output; the image records the outputs that the map actually attains. These are defined as subsets of different spaces, so their membership questions have different forms. Linearity then gives a direct way to prove that each subset is a subspace.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is either \(\mathbb{R}\) or \(\mathbb{C}\). A map \(T:V\to W\) is linear if \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\) and all scalars \(a,b\in\mathbb{F}\). Write \(\mathbf{0}_V\) and \(\mathbf{0}_W\) for the zero vectors in the two spaces.
These definitions give the membership criteria directly: to test whether \(\mathbf{v}\in V\) belongs to \(\ker T\), compute \(T(\mathbf{v})\) and compare it with \(\mathbf{0}_W\). To test whether \(\mathbf{w}\in W\) belongs to \(\operatorname{im}T\), determine whether there is at least one input \(\mathbf{v}\in V\) satisfying \(T(\mathbf{v})=\mathbf{w}\). The input is called a preimage of that output. An output can have more than one preimage.
Proof. By definition, \(\ker T\) consists exactly of those elements \(\mathbf{v}\in V\) for which \(T(\mathbf{v})=\mathbf{0}_W\), proving the first equivalence in both directions. Also by definition, \(\operatorname{im}T\) consists exactly of those elements \(\mathbf{w}\in W\) for which there exists an input \(\mathbf{v}\in V\) with \(T(\mathbf{v})=\mathbf{w}\), proving the second equivalence in both directions. \(\square\)
Why Both Sets Are Subspaces
A subspace of a vector space is a subset that is itself a vector space under the same addition and scalar multiplication. In particular, it must contain the zero vector, be closed under addition, and be closed under scalar multiplication. For the kernel, these closure checks use outputs equal to zero. For the image, they use inputs that produce the outputs being added or scaled.
Proof. First consider the kernel. Linearity gives \(T(\mathbf{0}_V)=\mathbf{0}_W\), so \(\mathbf{0}_V\in\ker T\). Let \(\mathbf{u},\mathbf{v}\in\ker T\). Then \(T(\mathbf{u})=\mathbf{0}_W\) and \(T(\mathbf{v})=\mathbf{0}_W\). Therefore \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})=\mathbf{0}_W+\mathbf{0}_W=\mathbf{0}_W\), so \(\mathbf{u}+\mathbf{v}\in\ker T\). If \(c\in\mathbb{F}\) and \(\mathbf{u}\in\ker T\), then \(T(c\mathbf{u})=cT(\mathbf{u})=c\mathbf{0}_W=\mathbf{0}_W\), so \(c\mathbf{u}\in\ker T\). Thus the kernel contains zero and is closed under addition and scalar multiplication, hence is a subspace of \(V\).
Now consider the image. Since \(T(\mathbf{0}_V)=\mathbf{0}_W\), the zero vector of \(W\) belongs to \(\operatorname{im}T\). Let \(\mathbf{w}_1,\mathbf{w}_2\in\operatorname{im}T\). By the definition of image, there are \(\mathbf{v}_1,\mathbf{v}_2\in V\) with \(T(\mathbf{v}_1)=\mathbf{w}_1\) and \(T(\mathbf{v}_2)=\mathbf{w}_2\). Then \(T(\mathbf{v}_1+\mathbf{v}_2)=T(\mathbf{v}_1)+T(\mathbf{v}_2)=\mathbf{w}_1+\mathbf{w}_2\), so \(\mathbf{w}_1+\mathbf{w}_2\in\operatorname{im}T\). For any \(c\in\mathbb{F}\), \(T(c\mathbf{v}_1)=cT(\mathbf{v}_1)=c\mathbf{w}_1\), so \(c\mathbf{w}_1\in\operatorname{im}T\). The image therefore contains zero and is closed under addition and scalar multiplication, hence is a subspace of \(W\). \(\square\)
The proof for the image depends on choosing preimages for the outputs being considered. It does not assume that an output has a unique preimage: any preimage of each output is enough. The kernel, by contrast, is described by a single equation on domain vectors, namely \(T(\mathbf{v})=\mathbf{0}_W\).
Worked Examples: Finding Kernels and Images
Worked Example: A Kernel and an Image Equal to the Codomain
Let \(T:\mathbb{R}^3\to\mathbb{R}^2\) be defined by \(T(x,y,z)=(x-y,\;y+z)\). Find its kernel and determine which vectors belong to its image.
For the kernel, use the zero-output criterion. The equation \(T(x,y,z)=(0,0)\) gives the two scalar equations \(x-y=0\) and \(y+z=0\). The first gives \(x=y\), and the second gives \(z=-y\). Setting \(y=t\), where \(t\in\mathbb{R}\), gives \(\ker T=\{(t,t,-t):t\in\mathbb{R}\}\).
To check this description, substitute an arbitrary listed vector into the map: \(T(t,t,-t)=(t-t,\;t-t)=(0,0)\). Conversely, the two equations above show that every vector with zero output has this form.
For the image, take any proposed codomain vector \((a,b)\in\mathbb{R}^2\). Choose the input \((a,0,b)\). Then \(T(a,0,b)=(a-0,\;0+b)=(a,b)\). Every vector in the codomain is attained, so \(\operatorname{im}T=\mathbb{R}^2\). The displayed input also verifies membership for each \((a,b)\), rather than merely suggesting that the image is all of \(\mathbb{R}^2\).
Worked Example: Testing Membership in a Proper Image
Define \(S:\mathbb{R}^2\to\mathbb{R}^3\) by \(S(x,y)=(x+y,\;2x+2y,\;x-y)\). Describe the image and decide whether \((3,6,1)\) and \((3,5,1)\) belong to it.
Write a general output as \((a,b,c)=S(x,y)\). The coordinate equations are \(a=x+y\), \(b=2x+2y\), and \(c=x-y\). The first two equations require \(b=2a\), so any output must satisfy that condition. To show it is also sufficient, suppose \(b=2a\). Solving the first and third equations gives \(x=(a+c)/2\) and \(y=(a-c)/2\). Substituting these values gives \(x+y=a\), \(2x+2y=2a=b\), and \(x-y=c\). Thus the proposed vector is attained. Consequently, \(\operatorname{im}S=\{(a,2a,c):a,c\in\mathbb{R}\}\).
For \((3,6,1)\), the condition holds because \(6=2\cdot3\). Its preimage from the formulas is \(x=(3+1)/2=2\) and \(y=(3-1)/2=1\). Check: \(S(2,1)=(2+1,\;2\cdot2+2\cdot1,\;2-1)=(3,6,1)\). So \((3,6,1)\in\operatorname{im}S\).
For \((3,5,1)\), the condition fails because \(5\ne2\cdot3=6\). The coordinate equations would require both \(x+y=3\) and \(2x+2y=5\), but doubling the first equation gives \(2x+2y=6\), a contradiction. There is no preimage, so \((3,5,1)\notin\operatorname{im}S\).
Worked Example: A Kernel with a Free Coordinate and a Restricted Image
Let \(U:\mathbb{R}^3\to\mathbb{R}^2\) be \(U(x,y,z)=(x+2y,\;2x+4y)\). Find the kernel, describe the image, and test a member and a nonmember of the image.
The zero-output equations are \(x+2y=0\) and \(2x+4y=0\). The second equation is twice the first, so the condition is \(x=-2y\), while \(z\) is unrestricted. Writing \(y=t\) and \(z=s\) gives \(\ker U=\{(-2t,t,s):t,s\in\mathbb{R}\}\). For every such vector, \(U(-2t,t,s)=(-2t+2t,\;-4t+4t)=(0,0)\). Conversely, the first zero-output equation forces \(x=-2y\), and neither equation restricts \(z\), so this describes the entire kernel.
A general output has the form \((x+2y,\;2x+4y)=(a,2a)\), where \(a=x+2y\). Thus every output has second coordinate twice its first. Each pair \((a,2a)\) is attained by choosing \((x,y,z)=(a,0,0)\), since \(U(a,0,0)=(a,2a)\). Therefore \(\operatorname{im}U=\{(a,2a):a\in\mathbb{R}\}\).
For example, \(U(3,0,0)=(3,6)\), so \((3,6)\) belongs to the image. The vector \((1,3)\) does not: its second coordinate is not twice its first, since \(3\ne2\). The coordinate condition rules out every possible preimage.
Common Pitfall: Checking Only Part of the Output
For a map into \(\mathbb{R}^2\), a zero output requires both coordinates to be zero. For example, take \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(x+y,\;x-y)\). A mistaken calculation might note that \(1+(-1)=0\) and claim \((1,-1)\in\ker T\). This checks only the first coordinate. In fact, \(T(1,-1)=(1-1,\;1-(-1))=(0,2)\ne(0,0)\), so \((1,-1)\notin\ker T\). The correct kernel calculation solves both equations: \(x+y=0\) and \(x-y=0\). Adding them gives \(2x=0\), hence \(x=0\); then \(y=0\). Thus \(\ker T=\{(0,0)\}\).
A related mistake is to assume that every vector in the codomain belongs to the image. The second worked example shows why this is not justified: image membership requires an input that produces the proposed output. A coordinate condition such as \(b=2a\) can show that some codomain vectors cannot be attained.
Keeping the Domain and Codomain Straight
The kernel is a subset of \(V\), because its elements are inputs. The image is a subset of \(W\), because its elements are outputs. Their zero vectors belong to different spaces, even when both spaces happen to be \(\mathbb{R}^n\). Keeping these roles explicit prevents a common type of error: testing a domain vector using an image condition, or treating an output as though it were an input.
For any linear map, the kernel subspace proof uses the fact that sums and scalar multiples of zero-output inputs still have zero output. The image subspace proof uses the fact that sums and scalar multiples of attained outputs are attained by sums and scalar multiples of their preimages. These arguments require linearity, but they do not require the spaces to be finite-dimensional.
Check Your Understanding
For each question, distinguish domain vectors from codomain vectors and use the relevant membership criterion.
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x,y)=(x-y,\;x+y)\), find \(\ker T\) by solving both zero-output equations.
- Let \(S:\mathbb{R}^2\to\mathbb{R}^3\) be \(S(x,y)=(x,\;y,\;x+y)\). State a coordinate condition for image membership and justify its sufficiency by finding a preimage.
- In the proof that \(\operatorname{im}T\) is closed under addition, explain why the existence of preimages for two outputs is enough.
- For \(U:\mathbb{R}^3\to\mathbb{R}^2\), \(U(x,y,z)=(x+2y,\;2x+4y)\), give one vector in \(\ker U\) with \(z\ne0\), and verify its image.
- Explain why the statement “\(\mathbf{w}\in\operatorname{im}T\)” requires an input in the domain, rather than merely a calculation involving \(\mathbf{w}\) itself.
