Null Spaces, Column Spaces, and Linear Transformations

A matrix $A$ carries two subspaces with it. One lives in the domain of the map $x \mapsto A x$ and records what gets sent to zero; the other lives in the codomain and records what can be reached. The two are described in opposite ways. The same two subspaces reappear, under the names kernel and range, for any linear transformation between vector spaces.¹

The null space

A vector belongs to $Nul A$ exactly when the linear transformation $x \mapsto A x$ sends it to the zero vector of $R^{m}$ . Testing membership is a single matrix-vector product.

The proof checks the three subspace conditions using one property of matrix multiplication, that $A$ distributes over sums and scalars. The zero vector satisfies $A 0 = 0$ . If $A u = 0$ and $A v = 0$ , then $A (u + v) = A u + A v = 0$ , and $A (c u) = c (A u) = 0$ . All three hold, so $Nul A$ is a subspace. The homogeneity is essential: the solution set of a nonhomogeneous system $A x = b$ with $b \neq = 0$ omits the zero vector and is never a subspace.

The null space is defined implicitly. Membership is stated as a condition to be checked, not as a list. Producing an explicit spanning set means solving $A x = 0$ and reading the general solution.

An explicit spanning set from free variables

Row reduction converts the implicit description into a span.

Two features of this construction hold whenever the null space contains nonzero vectors:

The spanning set is automatically linearly independent. Each free variable is the weight on exactly one vector, and it appears as a $1$ in that vector's slot with $0$ there in the others. The combination $x_{2} u + x_{4} v + x_{5} w$ can equal $0$ only if every weight is $0$ .
The number of vectors equals the number of free variables. This count is the dimension of the null space, developed under dimension and rank.

Each free variable contributes one basis vector of the null space; the

1

's sit in the free-variable slots, forcing linear independence.

The column space

Because a product $A x$ is by definition a linear combination of the columns of $A$ , the column space collects the attainable outputs:

Col A = {b : b = A x for some x in R^{n}} .

This is the range of the map $x \mapsto A x$ . Since it is a span, the span-subspace theorem applies directly.

The column space is defined explicitly: the columns of $A$ are given directly as a spanning set, and any vector in $Col A$ is built from them. The solvability criterion from the matrix-equation theory restates as a statement about $Col A$ :

Col A = R^{m} \leftrightarrow A x = b has a solution for every b in R^{m} .

Building a matrix with a prescribed column space is immediate: to realize $W = Span {(6, 1, 7), (- 1, - 1, 0)}$ as a column space, use those two vectors as the columns of $A$ , and then $W = Col A$ .

The column space is the span of the columns of

A

: every output

A x

is a linear combination of

a_{1}, \dots, a_{n}

, so

Col A

is the flat they sweep out through the origin.

Contrasting the two subspaces

For a non-square $A$ the two subspaces live in different spaces entirely: $Nul A \subseteq R^{n}$ (the domain) while $Col A \subseteq R^{m}$ (the codomain). A vector in $R^{4}$ cannot even be compared with a vector in $R^{3}$ . When $A$ is square the two share the zero vector, and in special cases they can share more.

For an

m \times n

matrix, the null space sits in the domain

R^{n}

and the column space is the image inside the codomain

R^{m}

; the map

x \mapsto A x

sends

Nul A

0

	$Nul A$	$Col A$
Ambient space	subspace of $R^{n}$	subspace of $R^{m}$
How defined	implicitly, by the condition $A x = 0$	explicitly, as a span of the columns
Finding vectors in it	row-reduce $[A 0]$	read off the columns of $A$
Relation to entries of $A$	none obvious	each column is in it
Test $v$ for membership	compute $A v$ ; check $= 0$	row-reduce $[A v]$ ; check consistency
Whole space when	$Nul A = {0}$ iff $x \mapsto A x$ is one-to-one	$Col A = R^{m}$ iff $x \mapsto A x$ is onto

The last row connects the two subspaces to the injectivity and surjectivity of the underlying map, a link that linear transformations established for matrices and that the kernel-and-range view below carries to abstract spaces.

The two membership tests differ because the two subspaces are described in opposite ways.

Kernel and range of a linear transformation

Subspaces of abstract vector spaces are often described through a linear map rather than a matrix. The definition of a linear transformation carries over unchanged from $R^{n}$ .

When $T (x) = A x$ is a matrix transformation, its kernel is $Nul A$ and its range $Col A$ . In general the kernel is a subspace of the domain $V$ and the range is a subspace of the codomain $W$ ; the proofs mirror the null-space and column-space subspace theorems.

A linear map

T : V \to W

collapses its kernel onto the zero vector of

W

and fills out its range; the kernel is a subspace of

V

and the range a subspace of

W

Differentiation is a linear transformation on function spaces. Let $V$ be the space of functions on $[a, b]$ with continuous derivatives, let $W = C [a, b]$ be the continuous functions, and let $D (f) = f^{'}$ . The rules $D (f + g) = D (f) + D (g)$ and $D (c f) = cD (f)$ are the linearity conditions, so $D$ is linear. Its kernel is the set of constant functions — those with zero derivative — which is a subspace of $V$ . This is the mechanism behind difference equations: a solution set of a homogeneous linear equation is the kernel of a linear transformation, hence a subspace, and much of its structure follows from that fact alone.

Lay, Linear Algebra and Its Applications, §4.2 — Null Spaces, Column Spaces, and Linear Transformations: Theorem 2 ( $Nul A$ a subspace), Theorem 3 ( $Col A$ a subspace), the implicit/explicit contrast, and the kernel and range of $T : V \to W$ . ↩

The null space

An explicit spanning set from free variables

The column space

Contrasting the two subspaces

Kernel and range of a linear transformation

Footnotes