LINEAR-ALGEBRA · Unit 6 · Video 3 · Interactive Practice

Null Space of a Matrix: Finding All Solutions to Ax = 0

IKey Formulas

Formula	Name	Description
$N(A) = \{x \in \mathbb{R}^n : Ax = 0\}$	Null space definition	All vectors mapped to zero by $A$
$A(v + w) = Av + Aw$	Distributive property	Key to proving closure under addition
$A(cv) = c(Av)$	Scalar factoring	Key to proving closure under scalar multiplication
Null space $\subseteq \mathbb{R}^n$ , Column space $\subseteq \mathbb{R}^m$	Dimension rule	$n$ = number of columns, $m$ = number of rows

IIVisualization 1 — The Matrix Machine

Feed a vector into $A$ : which nonzero $x$ comes out as $\mathbf{0}$ ?

x₁ 1 x₂ 0 x₃ 0

💡 Challenge: the columns satisfy $\mathbf{c}_3 = \mathbf{c}_1 + \mathbf{c}_2$ . What weights make $x_1\mathbf{c}_1 + x_2\mathbf{c}_2 + x_3\mathbf{c}_3 = \mathbf{0}$ ?

IIIVisualization 2 — Row-Reducing to the Null Space

One free column yields one special solution — and its multiples fill $N(A)$ .

Step 0

💡 The special solution $(-1,-1,1)$ is just $-(1,1,-1)$ — same direction, so $N(A) = \operatorname{span}\{(1,1,-1)\}$ .

IVVisualization 3 — A Subspace and Its Shift

Solutions of $Ax=\mathbf{0}$ pass through the origin; solutions of $Ax=\mathbf{b}$ are the same line, pushed off it.

t 1

💡 The two lines share direction $(1,1,-1)$ and never meet: the $Ax=\mathbf{b}$ line is $N(A)$ slid onto any one particular solution $\mathbf{p}=(1,0,0)$ .

VQuiz Questions

Question 1

Consider the matrix $A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix}$ . Which of the following vectors is in the null space $N(A)$ ?

✅ Correct! (2, 2, -2) = 2·(1, 1, -1) is a scalar multiple of the null space basis vector, so it's in N(A).

❌ Not quite. The null space consists of all multiples of (1, 1, -1). Check which option is a scalar multiple of this vector.

Show solution

Solution:

The null space is $N(A) = \{c(1, 1, -1) : c \in \mathbb{R}\}$ . We check each option:

(1, 0, 0): $Ax = (1, 2, 3, 4) \neq 0$ ✗
(2, 2, -2): This is $2 \cdot (1, 1, -1)$ . Check: $1(2)+1(2)+2(-2)=0$ , $2(2)+1(2)+3(-2)=0$ , etc. $Ax = (0,0,0,0)$ ✓
(1, -1, 0): $Ax = (0, 1, 2, 3) \neq 0$ ✗
(0, 1, -1): $Ax = (-1, -2, -3, -4) \neq 0$ ✗

The answer is (2, 2, -2) since it equals $2 \cdot (1, 1, -1)$ .

Question 2

True or False: The null space of a $4 \times 3$ matrix is a subspace of $\mathbb{R}^4$ .

✅ Correct! The null space lives in R³ (determined by the 3 columns), not R⁴. The column space is what lives in R⁴.

❌ Not quite. Think carefully: which dimension — rows or columns — determines where the null space lives?

Show solution

Solution:

False. The null space of a $4 \times 3$ matrix is a subspace of $\mathbb{R}^3$ , not $\mathbb{R}^4$ .

The rule: $N(A) \subseteq \mathbb{R}^n$ where $n$ is the number of columns. A $4 \times 3$ matrix has 3 columns, so $x$ must have 3 entries, meaning $N(A) \subseteq \mathbb{R}^3$ .

The column space lives in $\mathbb{R}^4$ (from the 4 rows). The null space and column space occupy entirely different spaces.

Question 3

Suppose $v$ and $w$ are both solutions to $Ax = b$ where $b \neq 0$ . What does $A(v + w)$ equal?

✅ Correct! A(v+w) = Av + Aw = b + b = 2b. Since 2b ≠ b when b ≠ 0, the sum of two solutions is not a solution.

❌ Not quite. Apply the distributive property: A(v+w) = Av + Aw. You know Av = b and Aw = b.

Show solution

Solution:

Since $Av = b$ and $Aw = b$ , use the distributive property:

$A(v + w) = Av + Aw = b + b = 2b$

This is exactly why the solution set of $Ax = b$ (with $b \neq 0$ ) is not a subspace: the sum of two solutions gives $2b \neq b$ , so $v + w$ is not itself a solution. Closure under addition fails.

Compare with the null space: if $Av = 0$ and $Aw = 0$ , then $A(v+w) = 0 + 0 = 0$ . Closure works when $b = 0$ .

Question 4

The column space and null space are two subspaces attached to every matrix. Which statement best describes how they differ?

✅ Correct! This is a deep insight: building from generators vs carving from constraints are the two complementary ways subspaces appear throughout linear algebra.

❌ Not quite. Think about the conceptual difference: one starts with pieces and builds up, the other starts with conditions and filters down.

Show solution

Solution:

The correct answer is: Column space is built from generators; null space is carved out by constraints.

These are the two fundamental ways subspaces arise in linear algebra:

Column space — start with generators (the columns of $A$ ), take all linear combinations. "Here are the pieces; what can I build?"
Null space — start with a constraint ( $Ax = 0$ ), collect everything satisfying it. "Here are the rules; what fits?"

The other options are incorrect:

Size comparison depends on the specific matrix
They live in different spaces: $C(A) \subseteq \mathbb{R}^m$ , $N(A) \subseteq \mathbb{R}^n$
Both can be found using elimination

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