LINEAR-ALGEBRA · Unit 7 · Video 2 · Interactive Practice

One Free Variable, One Special Solution

IKey Formulas

Formula	Name	Description
$\dim(N(A)) = n - r$	Null space dimension	Columns minus rank = number of special solutions
$\mathbf{x} = c_1 \mathbf{s}_1 + c_2 \mathbf{s}_2 + \cdots$	General null space element	Every solution is a combination of special solutions
Free rows of $N$ form $I$	Identity block property	Guarantees linear independence of special solutions

The Recipe: For each free variable, set it to 1 and all other free variables to 0, then back-substitute for the pivot variables.

IIVisualization 1 — Back-Substitution

Set the free variables $x_2, x_4$ ; back-substitution then forces the pivots $x_1, x_3$ so that $A\mathbf{x} = \mathbf{0}$ .

x₂ (free) 1 x₄ (free) 0

💡 Any assignment produces $\mathbf{x} = x_2\mathbf{s}_1 + x_4\mathbf{s}_2$ , so $(2, 0)$ merely rebuilds $2\mathbf{s}_1$ — the recipe's $(1, 0)$ and $(0, 1)$ are the two independent building blocks it never double-counts.

IIIVisualization 2 — Navigating the Null Space

Every null-space vector is $c_1\mathbf{s}_1 + c_2\mathbf{s}_2$ , so two coordinates $(c_1, c_2)$ reach the entire 2-D plane inside $\mathbb{R}^4$ .

c₁ 1 c₂ 1

💡 In this parameter space $\mathbf{s}_1$ and $\mathbf{s}_2$ act as the coordinate axes — the identity block is exactly what makes $(c_1, c_2)$ unambiguous coordinates for the plane.

IVVisualization 3 — The Identity Block in N

Stack the special solutions as the columns of $N$ — do the free-variable rows always come out as the identity $I_2$ ?

s₁: x₂ 1 s₁: x₄ 0 s₂: x₂ 0 s₂: x₄ 1

💡 Only the recipe's $(1, 0)$ and $(0, 1)$ force the free block to $I_2$ , which guarantees independence with no extra work — other assignments can still be independent, but only when their determinant is nonzero.

VQuiz Questions

Question 1

A $4 \times 7$ matrix $A$ has rank $r = 3$ . How many special solutions does the null space of $A$ have?

✅ Correct! The null space dimension is $n - r = 7 - 3 = 4$ .

❌ Not quite. The number of special solutions is columns minus rank: $n - r$ .

Show solution

Solution:

Number of special solutions $= n - r$ (columns minus rank).

$n - r = 7 - 3 = 4$

There are 4 special solutions, one for each free variable.

Common mistakes:

Choosing 3: that's the rank $r$ , not $n - r$
Choosing 7: that's $n$ , the total number of columns
Choosing 11: that's $n + r$ , which has no meaning here

Question 2

When using the recipe to find special solution $\mathbf{s}_k$ , which assignment do we make for the free variables?

✅ Correct! Activate one free variable, silence the rest.

❌ Not quite. The recipe activates exactly one free variable at a time.

Show solution

Solution:

The recipe: for special solution $\mathbf{s}_k$ , set free variable $k$ to 1 and all other free variables to 0. Then back-substitute for the pivot variables.

Why this works: This "activate one, silence the rest" pattern places an identity matrix $I$ in the free variable rows of the null space matrix $N$ , automatically guaranteeing that the special solutions are linearly independent.

Setting all free variables to 1 would give a valid null space vector, but it equals $\mathbf{s}_1 + \mathbf{s}_2 + \cdots$ — a combination of the special solutions, not one of the basis vectors itself.

Question 3

For the echelon form $U = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ , find the special solution with $x_2 = 0,\; x_4 = 1$ .

What is the value of $x_1$ ?

(Hint: first find $x_3$ from row 2, then find $x_1$ from row 1.)

✅ Correct! Back-substitution gives $x_3 = -2$ , then $x_1 = 2$ .

❌ Not quite. Careful with the signs — find $x_3$ first from row 2, then plug everything into row 1.

Show solution

Solution:

Step 1 — Row 2: $2x_3 + 4x_4 = 0$

$2x_3 + 4(1) = 0 \implies 2x_3 = -4 \implies x_3 = -2$

Step 2 — Row 1: $x_1 + 2x_2 + 2x_3 + 2x_4 = 0$

$x_1 + 2(0) + 2(-2) + 2(1) = 0$ $x_1 + 0 - 4 + 2 = 0$ $x_1 = 2$

The special solution is $\mathbf{s}_2 = (2,\; 0,\; {-2},\; 1)$ .

Common mistake: forgetting that $x_3 = -2$ contributes $2(-2) = -4$ in row 1, leading to a sign error.

Question 4

True or False: If $A$ is a $3 \times 4$ matrix and $A^T$ is the $4 \times 3$ transpose, and both have rank 2, then their null spaces have the same dimension.

✅ Correct! Same rank does not mean same null space dimension — the number of columns $n$ matters since $\dim = n - r$ .

❌ Think again. The null space dimension depends on $n - r$ . Do $A$ and $A^T$ have the same number of columns?

Show solution

Solution: False

The null space dimension is $n - r$ , where $n$ is the number of columns. While $\text{rank}(A) = \text{rank}(A^T)$ is always true, they have different numbers of columns:

$A$ is $3 \times 4$ : $\quad\dim(N(A)) = 4 - 2 = 2$ (a plane in $\mathbb{R}^4$ )
$A^T$ is $4 \times 3$ : $\quad\dim(N(A^T)) = 3 - 2 = 1$ (a line in $\mathbb{R}^3$ )

Same rank, different null space dimensions! The key is that more columns ( $n$ ) create more room for free variables.

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