Kernel (linear map)

In linear algebra, the kernel of a linear map $T:V\to W$ is the set of vectors in $V$ that $T$ sends to the zero vector in $W$. Also called the null space, it is denoted by $\ker T$. The kernel describes which input differences are invisible to the map: two inputs have the same output precisely when their difference belongs to the kernel. It is therefore fundamental to questions of uniqueness and the solution of linear equations. (homepages.ucl.ac.uk)

Definition and subspace structure

Let $V$ and $W$ be vector spaces over the same field $F$. Then

The zero here is the zero vector of the codomain, not necessarily a scalar. The kernel belongs to the domain; by contrast, the image $\operatorname{im}T=\{T(v):v\in V\}$ belongs to the codomain. These definitions apply to both finite- and infinite-dimensional spaces. (homepages.ucl.ac.uk)

The kernel is always a linear subspace. Indeed, linearity gives $T(0_V)=0_W$, so it is nonempty. If $u,v\in\ker T$ and $a,b\in F$, then

Thus every linear combination of kernel vectors remains in the kernel. In particular, a kernel is never the empty set: the smallest possible kernel is $\{0_V\}$, called the trivial kernel. (homepages.ucl.ac.uk)

Injectivity and solution sets

A linear map is an injective function if and only if its kernel is trivial. The key identity is

If the kernel contains only zero, equal outputs force equal inputs. Conversely, any nonzero kernel vector and the zero vector have the same output, preventing injectivity. (math.ucla.edu)

More generally, suppose $T(x)=b$ has a particular solution $x_0$. Its complete solution set is

This is an affine space whose direction space is the kernel. If $b\notin\operatorname{im}T$, there is no solution. Consequently, the image determines whether an output is attainable, while the kernel determines the ambiguity among inputs producing that output. (math.dartmouth.edu)

Nullity and the rank–nullity theorem

The dimension of the kernel is called the nullity of $T$. For a finite-dimensional domain, the rank–nullity theorem states

The second term is the rank. A proof begins with a basis of the kernel, extends it to a basis of $V$, and shows that the images of the added basis vectors form a basis of the image. (math.ucla.edu)

For an $m\times n$ matrix $A$, this becomes

where matrix rank measures the number of independent output directions. In particular, more columns than rows guarantee a nontrivial kernel. (live.ocw.mit.edu)

Matrix representation and computation

After bases are chosen, a linear map between finite-dimensional spaces is represented by a matrix $A$, and its kernel is represented by the solutions of the homogeneous system of linear equations

Gaussian elimination preserves this solution set. In reduced row-echelon form, pivot variables are expressed in terms of free variables. Setting one free variable to $1$ and the others to $0$, in turn, produces a set of linearly independent solutions whose span is the entire kernel. (ocw.mit.edu)

For example, over the real numbers, take

The equations give $x_2=-x_3$ and $x_1=-x_3$. Hence

The kernel is a line through the origin, with basis $\{(-1,-1,1)\}$, and its nullity is $1$. This is a direct application of the free-variable construction. (ocw.mit.edu)

Although coordinate descriptions change with the chosen bases, the kernel itself is an intrinsic subspace of the original domain. (math.ucla.edu)

Quotient-space interpretation

The kernel defines an equivalence relation on $V$: inputs are equivalent when their difference lies in $\ker T$. The corresponding quotient vector space $V/\ker T$ groups together exactly the inputs having identical outputs. (math.dartmouth.edu)

The first isomorphism theorem gives a canonical isomorphism

This map is well-defined because changing a representative by a kernel vector does not change its output. It is injective because only the zero coset maps to zero, and surjective onto the image by definition. No finite-dimensional assumption is required. (math.dartmouth.edu)

Differentiation example

Kernels need not consist of coordinate vectors. On the space of real polynomials of degree at most $n$, the differentiation map $D(p)=p'$ has kernel equal to the constant polynomials. For $n\geq1$, its image consists of polynomials of degree at most $n-1$. Its nullity is therefore $1$, its rank is $n$, and the domain has dimension $n+1$, illustrating rank–nullity for a space of functions. (homepages.ucl.ac.uk)