Math Core

Every matrix $A$ comes with two subspaces attached. One collects the inputs that $A$ sends to zero, and the other collects every output $A$ can produce. These two subspaces, the null space and the column space, answer the two basic questions about $A\mathbf{x} = \mathbf{b}$ : are solutions unique, and do solutions exist?

The null space

Definition

Null space

The null space of an $m \times n$ matrix $A$ is the set of all solutions of the homogeneous equation $A\mathbf{x} = \mathbf{0}$ :

\operatorname{Nul} A = \lbrace \mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \mathbf{0} \rbrace.

$\operatorname{Nul} A$ is a subspace of $\mathbb{R}^n$ , where $n$ is the number of columns of $A$ . The check uses only the linearity of matrix multiplication: $A\mathbf{0} = \mathbf{0}$ ; if $A\mathbf{u} = \mathbf{0}$ and $A\mathbf{v} = \mathbf{0}$ then $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} = \mathbf{0}$ ; and $A(c\mathbf{u}) = cA\mathbf{u} = \mathbf{0}$ .

Deciding whether a particular vector is in $\operatorname{Nul} A$ is easy: multiply. For example, with

A = \begin{bmatrix} 1 & 3 & 1 \\ 2 & 1 & -3 \end{bmatrix}, \qquad \mathbf{u} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix},

you get $A\mathbf{u} = (2 - 3 + 1,\; 4 - 1 - 3) = (0, 0)$ , so $\mathbf{u}$ is in $\operatorname{Nul} A$ .

The definition of $\operatorname{Nul} A$ is implicit: it gives a test that vectors must pass, not a list of the vectors. To make it explicit, solve $A\mathbf{x} = \mathbf{0}$ and write the solution in parametric vector form. The vectors that multiply the free variables span $\operatorname{Nul} A$ .

Worked example: A spanning set for a null space

Find a spanning set for $\operatorname{Nul} A$ , where

A = \begin{bmatrix} 1 & 2 & 0 & -1 \\ 2 & 4 & 1 & 1 \\ 3 & 6 & 1 & 0 \end{bmatrix}.

Row reduce $A$ (the augmented column of zeros never changes, so leave it off). $R_2 - 2R_1$ and $R_3 - 3R_1$ give rows $(0, 0, 1, 3)$ and $(0, 0, 1, 3)$ ; then $R_3 - R_2$ clears the last row:

\begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}.

The pivots are in columns $1$ and $3$ , so $x_2$ and $x_4$ are free. Reading the rows:

x_1 = -2x_2 + x_4, \qquad x_3 = -3x_4.

So every solution has the form

\mathbf{x} = \begin{bmatrix} -2x_2 + x_4 \\ x_2 \\ -3x_4 \\ x_4 \end{bmatrix} = x_2 \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 1 \\ 0 \\ -3 \\ 1 \end{bmatrix}.

Hence $\operatorname{Nul} A = \operatorname{Span}\lbrace (-2, 1, 0, 0),\; (1, 0, -3, 1) \rbrace$ .

Two features of this method matter later. The spanning set it produces is automatically linearly independent, because each vector has a $1$ in the position of its own free variable and $0$ in the positions of the other free variables. And the number of vectors equals the number of free variables. If $A\mathbf{x} = \mathbf{0}$ has only the trivial solution, then $\operatorname{Nul} A = \lbrace \mathbf{0} \rbrace$ .

The column space

Definition

Column space

The column space of an $m \times n$ matrix $A = [\,\mathbf{a}_1 \;\cdots\; \mathbf{a}_n\,]$ is the set of all linear combinations of its columns:

\operatorname{Col} A = \operatorname{Span}\lbrace \mathbf{a}_1, \dots, \mathbf{a}_n \rbrace = \lbrace A\mathbf{x} : \mathbf{x} \in \mathbb{R}^n \rbrace.

As a span, $\operatorname{Col} A$ is automatically a subspace, and it lives in $\mathbb{R}^m$ , where $m$ is the number of rows. The second description says $\mathbf{b}$ is in $\operatorname{Col} A$ exactly when $A\mathbf{x} = \mathbf{b}$ is consistent. In particular, $\operatorname{Col} A = \mathbb{R}^m$ exactly when $A\mathbf{x} = \mathbf{b}$ has a solution for every $\mathbf{b}$ , which happens when $A$ has a pivot in every row.

The column space is the opposite of the null space: it is defined explicitly (you can write down its vectors instantly) but testing membership takes work (you must row reduce $[\,A \mid \mathbf{b}\,]$ ).

Worked example: Is b in the column space?

Let $A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \\ 3 & 7 \end{bmatrix}$ . For which $h$ is $\mathbf{b} = (1, h, 5)$ in $\operatorname{Col} A$ ?

You need $c_1, c_2$ with $c_1 + 2c_2 = 1$ , $2c_1 + 5c_2 = h$ and $3c_1 + 7c_2 = 5$ . The first and third equations involve no $h$ . Subtracting $3$ times the first from the third gives $c_2 = 2$ , and then $c_1 = 1 - 4 = -3$ . The middle equation must also hold:

h = 2(-3) + 5(2) = 4.

So $\mathbf{b}$ is in $\operatorname{Col} A$ only when $h = 4$ .

Comparing the two

For an $m \times n$ matrix $A$ :

	$\operatorname{Nul} A$	$\operatorname{Col} A$
Lives in	$\mathbb{R}^n$	$\mathbb{R}^m$
Description	implicit: $A\mathbf{x} = \mathbf{0}$	explicit: span of the columns
Membership test	compute $A\mathbf{v}$	row reduce $[\,A \mid \mathbf{v}\,]$
Finding a spanning set	solve $A\mathbf{x} = \mathbf{0}$	read off the columns
Trivial case	$\lbrace \mathbf{0} \rbrace$ iff columns are independent	$\mathbb{R}^m$ iff a pivot in every row

Common mistake

Keep track of where each subspace lives. For a $3 \times 5$ matrix, null space vectors have $5$ entries and column space vectors have $3$ . A vector can be in one only if it has the right size, so a quick size check catches many errors.

Kernel and range

The same ideas apply to any linear transformation $T : V \to W$ between vector spaces. The kernel of $T$ is the set of $\mathbf{u}$ in $V$ with $T(\mathbf{u}) = \mathbf{0}$ , and the range of $T$ is the set of all outputs $T(\mathbf{u})$ . The kernel is a subspace of $V$ and the range is a subspace of $W$ . When $T(\mathbf{x}) = A\mathbf{x}$ , the kernel is $\operatorname{Nul} A$ and the range is $\operatorname{Col} A$ .

Worked example: The kernel of a transformation

Let $T : \mathbb{R}^3 \to \mathbb{R}^2$ be $T(x, y, z) = (x - y,\; y - z)$ . Find the kernel.

$T(x, y, z) = \mathbf{0}$ means $x = y$ and $y = z$ . So the kernel is all vectors $(t, t, t)$ , the line $\operatorname{Span}\lbrace (1, 1, 1) \rbrace$ . The range is all of $\mathbb{R}^2$ , since any $(a, b)$ equals $T(a + b, b, 0)$ .

Tip

A nonzero vector in $\operatorname{Nul} A$ is a dependence relation among the columns of $A$ : if $A\mathbf{x} = \mathbf{0}$ , then $x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n = \mathbf{0}$ . In the first example, $(-2, 1, 0, 0)$ says column $2$ is twice column $1$ .

Practice

Practice 1

$A$ is a $4 \times 6$ matrix. $\operatorname{Col} A$ is a subspace of $\mathbb{R}^k$ . What is $k$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

$A$ is a $5 \times 7$ matrix. $\operatorname{Nul} A$ is a subspace of $\mathbb{R}^k$ . What is $k$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Let $A = \begin{bmatrix} 3 & 1 & -2 \\ 1 & 4 & 5 \end{bmatrix}$ and $\mathbf{v} = (1, -2, 1)$ . Is $\mathbf{v}$ in $\operatorname{Nul} A$ ?

Practice 4

Let $A = \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 3 \end{bmatrix}$ . $\operatorname{Nul} A$ is spanned by a single vector of the form $(a, b, 1)$ . Find it.

Enter a point like (2, -3)

Practice 5

Let $A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & -4 & 8 \\ -1 & 2 & -4 \end{bmatrix}$ . The standard method (solving $A\mathbf{x} = \mathbf{0}$ in parametric vector form) produces a spanning set for $\operatorname{Nul} A$ . How many vectors are in it?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Let $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 3 \end{bmatrix}$ . For what value of $h$ is $\mathbf{b} = (2, 1, h)$ in $\operatorname{Col} A$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Let $T : \mathbb{R}^3 \to \mathbb{R}^2$ be $T(x, y, z) = (x + z,\; y - 2z)$ . The kernel of $T$ is a line spanned by a vector whose third entry is $1$ . Find that vector.

Enter a point like (2, -3)

Practice 8

$A$ is a $3 \times 3$ matrix and $\operatorname{Nul} A = \lbrace \mathbf{0} \rbrace$ . Which statement must be true?