LINEAR-ALGEBRA · Unit 9 · Video 1 · Interactive Practice

Linear Independence: The Null Space Test

IKey Formulas

Formula	Name	When to Use
$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n = \mathbf{0}$	Independence equation	Independent iff only $c_1 = c_2 = \cdots = c_n = 0$ works
$A\mathbf{c} = \mathbf{0}$	Matrix form	$A = [\mathbf{v}_1 \mid \cdots \mid \mathbf{v}_n]$ ; solve for coefficient vector $\mathbf{c}$
$\operatorname{Null}(A) = \{\mathbf{0}\}$	Null-space test	Vectors are independent iff the null space is trivial
$n > m \Rightarrow \text{dependent}$	Counting argument	More vectors than dimensions guarantees dependence

IIVisualization 1 — Collinear Means Dependent

Two vectors in $\mathbb{R}^2$ are dependent exactly when they line up — when $\det[\mathbf{v}_1\ \mathbf{v}_2] = 0$ .

v₁ x 1 v₁ y 2 v₂ x 2 v₂ y 4

IIIVisualization 2 — A Whole Line of Dependencies

Which coefficients $(c_1, c_2)$ make $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}$ for $\mathbf{v}_1 = (1,2),\ \mathbf{v}_2 = (2,4)$ ?

c₁ 2 c₂ -1

💡 That line of solutions is $\operatorname{Null}(A)$ for $A = [\mathbf{v}_1 \mid \mathbf{v}_2]$ — any nonzero point on it is a dependency, so the pair is independent only when the line shrinks to the single point $\mathbf{0}$ .

IVVisualization 3 — Counting Forces Dependence

With $\mathbf{v}_1 = (1,0)$ and $\mathbf{v}_2 = (0,1)$ fixed, can any third vector $\mathbf{v}_3$ in $\mathbb{R}^2$ escape dependence?

v₃ x 2 v₃ y 3

A = [ v₁ | v₂ | v₃ ]

💡 The matrix is $2 \times 3$ : at most $2$ pivots cover $3$ columns, so one column is always free. This is the counting argument — $n > m$ forces a nonzero null-space vector, hence dependence.

VQuiz Questions

Question 1

Consider the vectors $\mathbf{v}_1 = (3, 6)$ and $\mathbf{v}_2 = (1, 2)$ in $\mathbb{R}^2$ .

Are these vectors linearly independent or linearly dependent?

✅ Correct! Since v₁ = 3·v₂, the non-trivial solution c₁ = −1, c₂ = 3 proves dependence.

❌ Not quite. Check whether one vector is a scalar multiple of the other.

Show solution

Solution:

Check whether one vector is a scalar multiple of the other:

$\mathbf{v}_1 = (3, 6) = 3 \cdot (1, 2) = 3\,\mathbf{v}_2$

So we can write:

$(-1)\,\mathbf{v}_1 + 3\,\mathbf{v}_2 = (-3, -6) + (3, 6) = (0, 0) = \mathbf{0}$

This is a non-trivial solution ( $c_1 = -1 \neq 0$ , $c_2 = 3 \neq 0$ ), so the vectors are dependent.

⚠️ Different magnitudes do not imply independence — only direction matters for two vectors.

Question 2

True or False: The set $\{(1, 3, 5),\; (2, -1, 0),\; (0, 0, 0)\}$ is linearly independent.

✅ Correct! Any set containing the zero vector is automatically dependent — you can always put a nonzero coefficient on it.

❌ Not quite. Think about what happens when you assign a nonzero coefficient to the zero vector.

Show solution

Solution:

Any set containing the zero vector is automatically dependent.

We can always assign a nonzero coefficient to the zero vector:

$0 \cdot \mathbf{v}_1 + 0 \cdot \mathbf{v}_2 + 7 \cdot \mathbf{0} = \mathbf{0}$

The coefficient $7 \neq 0$ makes this a non-trivial solution. It doesn't matter what $\mathbf{v}_1$ and $\mathbf{v}_2$ are.

Answer: False — the set is dependent.

Question 3

You row reduce a $4 \times 4$ matrix $A$ (whose columns are four vectors in $\mathbb{R}^4$ ) and find that every column is a pivot column. What can you conclude about the column vectors?

✅ Correct! Every column being a pivot column means zero free variables, so the null space is trivial and the vectors are independent.

❌ Not quite. Remember: pivot in every column → no free variables → only the trivial solution exists.

Show solution

Solution:

Apply the row-reduction decision procedure:

The matrix $A$ is $4 \times 4$ (4 vectors in $\mathbb{R}^4$ , so $n = m = 4$ ).
Row reduce to RREF.
Every column is a pivot column $\Rightarrow$ 4 pivots in 4 columns $\Rightarrow$ 0 free variables.

No free variables means the only solution to $A\mathbf{c} = \mathbf{0}$ is $\mathbf{c} = \mathbf{0}$ (the trivial solution).

$\operatorname{Null}(A) = \{\mathbf{0}\} \quad \Longrightarrow \quad \text{INDEPENDENT}$

The counting argument ( $n > m$ forces dependence) does not apply here because $n = m$ .

Question 4

You have 5 vectors in $\mathbb{R}^3$ . Without performing any row reduction, what can you conclude?

✅ Correct! With 5 vectors in R³, the counting argument (n > m) guarantees dependence — at least 2 free variables must exist.

❌ Not quite. Think about the counting argument: what happens when you have more vectors than the dimension of the space?

Show solution

Solution:

Apply the counting argument: we have $n = 5$ vectors in $\mathbb{R}^m$ with $m = 3$ .

Since $n > m$ :

The matrix $A$ is $3 \times 5$ (3 rows, 5 columns).
Row reduction produces at most 3 pivots (one per row).
That leaves at least $5 - 3 = 2$ columns without pivots → at least 2 free variables.
Free variables produce non-trivial solutions to $A\mathbf{c} = \mathbf{0}$ .

$n > m \;\Longrightarrow\; \text{DEPENDENT (guaranteed)}$

No row reduction needed — the dimension of the space ( $m = 3$ ) sets a hard ceiling on the maximum number of independent vectors.

Answer: The vectors are dependent, guaranteed by the counting argument.

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