Subspaces: Column Space and Null Space

Quick answer

A subspace of ℝⁿ is a set that contains the zero vector and is closed under addition and scalar multiplication: lines and planes through the origin, and every span. Each m × n matrix A has two. The column space Col A, the span of the columns, lives in ℝᵐ and holds every b for which Ax = b is solvable. The null space Nul A, the solutions of Ax = 0, lives in ℝⁿ. Row reduction gives a basis for both: the pivot columns of A for Col A, and the direction vectors of the parametric solution for Nul A.

What you'll learn

Decide whether a set of vectors is a subspace
Describe the column space and null space of a matrix
Find a basis for the null space from parametric vector form
Find a basis for the column space from the pivot columns

Sets that stay closed

The span of a set of vectors has a useful property: combining vectors inside it never leads outside it. Sets with that property are subspaces.

A subspace of ℝⁿ is a set $H$ of vectors in ℝⁿ such that

the zero vector is in $H$ ;

if $\mathbf{u}$ and $\mathbf{v}$ are in $H$ , so is $\mathbf{u} + \mathbf{v}$ ;

if $\mathbf{u}$ is in $H$ and $c$ is a scalar, $c\mathbf{u}$ is in $H$ .

Lines and planes through the origin are subspaces; so is every span, and so are $\{\mathbf{0}\}$ and ℝⁿ itself. A line that misses the origin is not a subspace, and neither is the first quadrant of the plane, which contains $(1, 1)$ but not $-1 \cdot (1, 1)$ .

Every $m \times n$ matrix $A$ comes with two subspaces.

The column space $\text{Col}\,A$ is the span of the columns of $A$ . It sits in ℝᵐ, and it is exactly the set of $\mathbf{b}$ for which $A\mathbf{x} = \mathbf{b}$ has a solution.
The null space $\text{Nul}\,A$ is the set of solutions of $A\mathbf{x} = \mathbf{0}$ . It sits in ℝⁿ.

Why the null space is a subspace

The zero vector solves $A\mathbf{x} = \mathbf{0}$ . 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} \qquad A(c\mathbf{u}) = cA\mathbf{u} = \mathbf{0}

so sums and multiples of solutions are solutions. Linearity of $A$ is exactly what keeps the solutions of $A\mathbf{x} = \mathbf{0}$ closed under both operations. The same argument fails for $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \ne \mathbf{0}$ . There, two solutions add to a solution of $A\mathbf{x} = 2\mathbf{b}$ , and the zero vector is not a solution at all.

Bases for both spaces from one row reduction

A basis of a subspace is a linearly independent set that spans it: enough vectors to reach everything, with none to spare. One row reduction of $A$ produces a basis for each of its subspaces.

Null space. Write the solution of $A\mathbf{x} = \mathbf{0}$ in parametric vector form. The vectors multiplying the free variables form a basis.
Column space. The columns of $A$ that hold pivots form a basis. Take them from $A$ itself: row operations change the column space, but they keep every linear relation among the columns, so they show which columns to keep.

Worked examples

Example 2 — Both bases from one reduction

Find bases for $\text{Col}\,B$ and $\text{Nul}\,B$ , where

B = \begin{bmatrix} 1 & 2 & 0 & 1\\ 2 & 4 & 1 & 4\\ 3 & 6 & 1 & 5 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 2 & 0 & 1\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}

The pivots are in columns $1$ and $3$ , so $\text{Col}\,B$ has basis $\{(1, 2, 3),\ (0, 1, 1)\}$ , the first and third columns of $B$ . The reduced form also shows the relations: column $2$ is $2$ times column $1$ , and column $4$ is column $1$ plus $2$ times column $3$ .

For the null space, $x_2$ and $x_4$ are free, $x_1 = -2x_2 - x_4$ and $x_3 = -2x_4$ :

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

so $\{(-2, 1, 0, 0),\ (-1, 0, -2, 1)\}$ is a basis for $\text{Nul}\,B$ , a subspace of ℝ⁴.

Common mistakes

Practice problems

Is the set of vectors $(x, y)$ with $x \ge 0$ a subspace of ℝ²?

Answer

No

Full solution

It contains $(1, 0)$ but not $-1 \cdot (1, 0) = (-1, 0)$ , so it is not closed under scalar multiplication.
Is the set of vectors $(x, y, z)$ with $x + y + z = 0$ a subspace of ℝ³?

Answer

Yes

Full solution

It is the null space of $\begin{bmatrix} 1 & 1 & 1 \end{bmatrix}$ , and every null space is a subspace.
Find a basis for $\text{Nul}\begin{bmatrix} 1 & 3\\ 2 & 6 \end{bmatrix}$ .

Answer

$\{(-3, 1)\}$

Full solution

The reduced form is $\begin{bmatrix} 1 & 3\\ 0 & 0 \end{bmatrix}$ , so $x_1 = -3x_2$ with $x_2$ free.
Find a basis for $\text{Nul}\begin{bmatrix} 1 & 0 & 2\\ 0 & 1 & -1 \end{bmatrix}$ .

Answer

$\{(-2, 1, 1)\}$

Full solution

The matrix is already reduced: $x_3$ is free, $x_1 = -2x_3$ and $x_2 = x_3$ .
For $H = \begin{bmatrix} 1 & 1 & 2\\ 1 & 2 & 3\\ 2 & 1 & 3 \end{bmatrix}$ , find a basis for $\text{Col}\,H$ .

Answer

$\{(1, 1, 2),\ (1, 2, 1)\}$

Full solution

Row reduction gives $\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}$ , with pivots in columns $1$ and $2$ . Take those columns of $H$ . The third column is their sum.
Find a basis for $\text{Nul}\,H$ for the same matrix.

Answer

$\{(-1, -1, 1)\}$

Full solution

From the reduced form, $x_3$ is free, $x_1 = -x_3$ and $x_2 = -x_3$ .
A $4 \times 7$ matrix $A$ is given. Is $\text{Nul}\,A$ a subspace of ℝ⁴ or of ℝ⁷?

Answer

ℝ⁷

Full solution

Solutions of $A\mathbf{x} = \mathbf{0}$ need $7$ entries, one per column.
If $A\mathbf{x} = \mathbf{b}$ has no solution, what does that say about $\mathbf{b}$ and $\text{Col}\,A$ ?

Answer

$\mathbf{b}$ is not in $\text{Col}\,A$ .

Full solution

$A\mathbf{x}$ is always a combination of the columns, so a solution exists exactly when $\mathbf{b}$ is one of those combinations.
For an invertible $n \times n$ matrix, what are $\text{Col}\,A$ and $\text{Nul}\,A$ ?

Answer

$\text{Col}\,A =$ ℝⁿ and $\text{Nul}\,A = \{\mathbf{0}\}$

Full solution

By the Invertible Matrix Theorem the columns span ℝⁿ and $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.
A student reduces $B$ from Example 2 and reports $\{(1, 0, 0),\ (0, 1, 0)\}$ as a basis for $\text{Col}\,B$ , the pivot columns of the reduced matrix. What went wrong?

Hint

Is $(1, 0, 0)$ a combination of the columns of $B$ ?

Answer

The basis must come from $B$ itself: $\{(1, 2, 3),\ (0, 1, 1)\}$ .

Full solution

Every column of $B$ has third entry equal to the sum of its first two entries, so every vector in $\text{Col}\,B$ does too. $(1, 0, 0)$ breaks that pattern and is not in $\text{Col}\,B$ at all. Row operations change the column space; they only preserve the relations among columns.

Frequently asked questions

What is a subspace?

A set of vectors in ℝⁿ that contains the zero vector and is closed under addition and scalar multiplication. Lines and planes through the origin are examples.

What is the column space of a matrix?

The span of its columns, written Col A. It is exactly the set of vectors b for which Ax = b has a solution.

What is the null space of a matrix?

The set of all solutions of Ax = 0, written Nul A. For an m × n matrix it is a subspace of ℝⁿ.

How do you find a basis for the column space?

Row reduce A to find the pivot columns, then take the corresponding columns of the original matrix A, not of its reduced form.

How do you find a basis for the null space?

Solve Ax = 0 in parametric vector form. The vectors multiplying the free variables form a basis.

What to learn next

Basis, Dimension and the Rank TheoremA basis spans a subspace with no redundancy, and every basis has the same size: the dimension. The rank theorem splits a matrix's columns between Col A and Nul A.

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