Why These Sets Have Structure
A linear map does more than send vectors in one space to vectors in another. It preserves the operations that define a vector space: addition and scalar multiplication. Consequently, the vectors sent to zero can be combined without leaving that collection, and outputs of the map can be combined to produce another output. These closure properties are exactly what make the kernel and image subspaces.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), and let \(T:V\to W\) be linear. The kernel of \(T\) is the set of vectors in the domain sent to the zero vector of the codomain. The image of \(T\) is the set of outputs the map actually attains.
A subset of a vector space is not automatically a vector space. To be a subspace, it must contain the zero vector and inherit the vector operations from its containing space. Instead of checking all vector-space properties separately, we can use the linear-combination test.
Proof. First suppose \(S\) is a subspace. A subspace contains \(\mathbf{0}\), so it is nonempty. It is closed under scalar multiplication and addition: if \(\mathbf{u},\mathbf{v}\in S\), then \(a\mathbf{u}\in S\) and \(b\mathbf{v}\in S\), so \(a\mathbf{u}+b\mathbf{v}\in S\). Conversely, suppose \(S\) is nonempty and is closed under the stated linear combinations. Choose \(\mathbf{s}\in S\). Taking \(\mathbf{u}=\mathbf{v}=\mathbf{s}\) and \(a=b=0\) gives \(\mathbf{0}\in S\). For any \(\mathbf{u}\in S\) and \(c\in\mathbb{F}\), use \(\mathbf{u}\) and \(\mathbf{0}\) with coefficients \(c\) and \(0\) to get \(c\mathbf{u}\in S\). For any \(\mathbf{u},\mathbf{v}\in S\), use coefficients \(1,1\) to get \(\mathbf{u}+\mathbf{v}\in S\). Thus \(S\) contains zero and is closed under addition and scalar multiplication, so it is a subspace. \(\square\)
The nonempty condition matters: the empty set satisfies the closure statement vacuously, because it contains no pair of vectors to test, but it is not a subspace. The combination \(a\mathbf{u}+b\mathbf{v}\) is useful because one test checks addition and scalar multiplication together. In a proof, the essential task is to start with arbitrary members of the set and show that this combination remains in it.
The Kernel and Image Theorem
For a linear map, the key calculation is that linear combinations pass through the map. If \(\mathbf{u},\mathbf{v}\in V\) and \(a,b\in\mathbb{F}\), linearity gives \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\). This follows by applying homogeneity to each vector and then additivity. The same identity will establish closure for both the kernel and the image, though the vectors we start with belong to different sets.
Proof. First note that \(T(\mathbf{0})=\mathbf{0}\): by additivity, \(T(\mathbf{0})=T(\mathbf{0}+\mathbf{0})=T(\mathbf{0})+T(\mathbf{0})\), and subtracting \(T(\mathbf{0})\) from both sides gives the claim. Thus \(\mathbf{0}\in\ker T\), so the kernel is nonempty. Let \(\mathbf{u},\mathbf{v}\in\ker T\), and let \(a,b\in\mathbb{F}\). Then \(T(\mathbf{u})=T(\mathbf{v})=\mathbf{0}\), and therefore \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})=a\mathbf{0}+b\mathbf{0}=\mathbf{0}\). By the definition of the kernel, \(a\mathbf{u}+b\mathbf{v}\in\ker T\). The linear-combination test proves that \(\ker T\) is a subspace of \(V\).
For the image, \(\mathbf{0}=T(\mathbf{0})\in\operatorname{im}T\), so it is nonempty. Let \(\mathbf{y}_1,\mathbf{y}_2\in\operatorname{im}T\). By the definition of the image, there are \(\mathbf{u},\mathbf{v}\in V\) such that \(\mathbf{y}_1=T(\mathbf{u})\) and \(\mathbf{y}_2=T(\mathbf{v})\). For any \(a,b\in\mathbb{F}\), linearity gives \(a\mathbf{y}_1+b\mathbf{y}_2=aT(\mathbf{u})+bT(\mathbf{v})=T(a\mathbf{u}+b\mathbf{v})\). Since \(a\mathbf{u}+b\mathbf{v}\in V\), its output belongs to \(\operatorname{im}T\) by definition. Thus the image is closed under linear combinations, and the test proves that \(\operatorname{im}T\) is a subspace of \(W\). \(\square\)
The two halves have a similar shape but use different definitions. For the kernel, we start with vectors already known to map to zero and show their combination still maps to zero. For the image, we start with outputs and choose preimages for them; linearity then constructs a preimage for their combination. The preimages need not be unique. The argument only needs one preimage for each output.
Worked Examples
Worked Example: The Zero Map
Consider the zero map \(Z:\mathbb{R}^2\to\mathbb{R}^2\), defined by \(Z(\mathbf{x})=\mathbf{0}\) for every \(\mathbf{x}\in\mathbb{R}^2\). It is linear because every output, including the output on any linear combination, is zero. Determine its kernel and image, then check closure directly.
Every vector in the domain maps to zero, so \(\ker Z=\mathbb{R}^2\). The only output is the zero vector, so \(\operatorname{im}Z=\{\mathbf{0}\}\). For a direct kernel check, take \(\mathbf{u}=\begin{bmatrix}1\\-2\end{bmatrix}\), \(\mathbf{v}=\begin{bmatrix}3\\4\end{bmatrix}\), \(a=2\), and \(b=-1\). Then
This particular calculation illustrates closure, and the arbitrary combination also works: \(Z(a\mathbf{u}+b\mathbf{v})=\mathbf{0}=aZ(\mathbf{u})+bZ(\mathbf{v})\). For the image, \(2\mathbf{0}-\mathbf{0}=\mathbf{0}\) is still in \(\{\mathbf{0}\}\). The kernel is the whole domain in this example; the image need not be the whole codomain.
Worked Example: Inclusion of a Plane
Let \(U=\{(s,t,s+t):s,t\in\mathbb{R}\}\subseteq\mathbb{R}^3\), and let \(I:U\to\mathbb{R}^3\) be the inclusion map \(I(\mathbf{u})=\mathbf{u}\). First, \(U\) is a subspace: it contains \(\mathbf{0}=(0,0,0)\), and for \(\mathbf{u}=(s,t,s+t)\), \(\mathbf{v}=(r,q,r+q)\), and \(a,b\in\mathbb{R}\),
The map \(I\) is linear because it leaves each vector unchanged: \(I(a\mathbf{u}+b\mathbf{v})=a\mathbf{u}+b\mathbf{v}=aI(\mathbf{u})+bI(\mathbf{v})\). Its kernel is \(\{\mathbf{0}\}\), since the only vector unchanged by inclusion that equals the zero vector is zero. Its image is \(U\), since every vector in \(U\) is an output and no vector outside \(U\) is in the domain.
For a concrete closure check, choose \(\mathbf{u}=(1,2,3)\) and \(\mathbf{v}=(-2,1,-1)\), both in \(U\), and take \(a=2,b=-3\). Then
Applying the map gives \(I(2\mathbf{u}-3\mathbf{v})=(8,1,9)=2I(\mathbf{u})-3I(\mathbf{v})\), as required. Also, the kernel check is direct: \(I(\mathbf{0})=\mathbf{0}\), and if \(I(\mathbf{u})=\mathbf{0}\), then \(I(\mathbf{u})=\mathbf{u}\) forces \(\mathbf{u}=\mathbf{0}\).
Worked Example: Projection onto the First Two Coordinates
Define \(P:\mathbb{R}^3\to\mathbb{R}^3\) by \(P(x,y,z)=(x,y,0)\). It is linear: for any \(\mathbf{u},\mathbf{v}\in\mathbb{R}^3\) and \(a,b\in\mathbb{R}\), the first two coordinates of \(a\mathbf{u}+b\mathbf{v}\) are the corresponding linear combinations of the first two coordinates, and the third output coordinate is zero. The formula shows that \(P(x,y,z)=\mathbf{0}\) exactly when \(x=y=0\). Hence \(\ker P=\{(0,0,z):z\in\mathbb{R}\}\). Its outputs have the form \((x,y,0)\), and every such vector is \(P(x,y,0)\), so \(\operatorname{im}P=\{(x,y,0):x,y\in\mathbb{R}\}\).
For the kernel, take \(\mathbf{k}_1=(0,0,2)\), \(\mathbf{k}_2=(0,0,-3)\), \(a=2\), and \(b=-1\). Their combination is \(2\mathbf{k}_1-\mathbf{k}_2=(0,0,4)-(0,0,-3)=(0,0,7)\), and \(P(0,0,7)=(0,0,0)\). Thus this combination remains in the kernel.
For the image, let \(\mathbf{u}=(1,-1,2)\) and \(\mathbf{v}=(3,2,-4)\). Their outputs are \(P(\mathbf{u})=(1,-1,0)\) and \(P(\mathbf{v})=(3,2,0)\). Using \(a=2,b=-1\),
The result is in the image and equals \(2P(\mathbf{u})-P(\mathbf{v})\). In the general proof, this same equality follows from linearity for every pair of inputs and every pair of scalars.
Common Pitfalls
Do not confuse the ambient spaces. The kernel consists of inputs, so \(\ker T\subseteq V\). The image consists of outputs, so \(\operatorname{im}T\subseteq W\). For the projection above, the kernel is the vertical axis in the domain, while the image is the horizontal plane in the codomain. Both are subsets of \(\mathbb{R}^3\) here, but they describe different sets.
Linearity is a necessary hypothesis. A tempting but incorrect argument is to use the kernel-and-image theorem for any function, just because its inputs and outputs are vectors. For example, \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=x^2\) is not linear: \(f(-1)=1\), whereas \(-f(1)=-1\). Its image is \([0,\infty)\), which is not a subspace because it contains \(1\) but not \(-1=(-1)\cdot1\). The closure proof fails because the equality \(f(a u+b v)=a f(u)+b f(v)\) is not available. Always establish that the map is linear before applying the theorem.
These results are useful because they let us recognize structure without listing every vector in a kernel or image. Once a map is known to be linear, the subspace conclusion follows from one general proof rather than a separate verification of every vector-space axiom. The key argument is always the same: begin with arbitrary members of the set, form an arbitrary linear combination, and use the definition and linearity to show that the result remains in the set.
Check Your Understanding
For each question, identify the relevant domain or codomain and use arbitrary scalars when checking closure.
- In the proof that \(\ker T\) is nonempty, why does \(T(\mathbf{0})=\mathbf{0}\) follow from linearity?
- In the image proof, why do vectors \(\mathbf{y}_1,\mathbf{y}_2\in\operatorname{im}T\) have preimages \(\mathbf{u},\mathbf{v}\in V\)?
- For the inclusion map in the example, explain why its kernel is \(\{\mathbf{0}\}\) but its image is \(U\).
- For the projection \(P(x,y,z)=(x,y,0)\), state the ambient space of its kernel and of its image.
- Give a reason the kernel-and-image theorem cannot be applied to \(f(x)=x^2\) on \(\mathbb{R}\).
