LINEAR-ALGEBRA · Unit 5 · Video 3 · Interactive Practice

Vector Spaces and Subspaces

IKey Formulas

Requirement	Formula	What It Means
Closure (addition)	$\mathbf{u}, \mathbf{v} \in V \Rightarrow \mathbf{u} + \mathbf{v} \in V$	Sum of any two vectors stays in the set
Closure (scalar mult.)	$\mathbf{u} \in V,\; c \in \mathbb{R} \Rightarrow c\,\mathbf{u} \in V$	Any scalar multiple stays in the set
Linear combination	$c\,\mathbf{u} + d\,\mathbf{v} \in V$ for all $c, d \in \mathbb{R}$	Combines both closure rules into one
Zero vector test	$0 \cdot \mathbf{v} = \mathbf{0} \Rightarrow \mathbf{0} \in V$	Zero vector must belong to every vector space

IIVisualization 1 — Scalar Multiplication Breaks $Q$

Adding two vectors in $Q$ never leaves $Q$ — but does scaling one of them?

Scalar c −1

💡 A subspace must satisfy every axiom, so the single scalar-multiplication failure at $c = -1$ is enough to disqualify $Q$ .

IIIVisualization 2 — Where Does $\mathbf{u}+\mathbf{v}$ Land?

A line is a subspace only when it passes through the origin — so where does $\mathbf{u}+\mathbf{v}$ land?

Slope m 1 Intercept b 0

💡 On an off-origin line the sum absorbs the intercept twice, landing at height $m(u_x + v_x) + 2b$ — off the line unless $b = 0$ .

IVVisualization 3 — The Three Axioms at Once

Every line through the origin is a subspace; sliding it off the origin breaks all three axioms at once.

Angle 45° Shift off origin 0

💡 The complete catalog of subspaces of $\mathbb{R}^2$ : $\{\mathbf{0}\}$ , every line through the origin, and all of $\mathbb{R}^2$ . In $\mathbb{R}^3$ there is one more type — planes through the origin.

VQuiz Questions

Question 1

The first quadrant $Q = \{(x,y) : x \geq 0,\; y \geq 0\}$ is closed under addition. Why is it not a vector space?

✅ Correct! Multiplying by a negative scalar sends vectors outside the first quadrant.

❌ Not quite. Think about what happens when you multiply a vector in Q by a negative number like -1.

Show solution

Solution:

The first quadrant does contain the zero vector $(0,0)$ and is closed under addition. However, it fails closure under scalar multiplication.

Take $\mathbf{v} = (1, 1) \in Q$ and $c = -1$ :

$c \cdot \mathbf{v} = (-1)(1, 1) = (-1, -1) \notin Q$

The result has negative components, landing in the third quadrant — outside $Q$ . Closure under scalar multiplication requires the result to stay in the set for all real scalars, including negative ones. One failure is enough: $Q$ is not a vector space.

Question 2

The vector $(3, 2, 0)$ has a zero in its third component. Which space does it belong to?

✅ Correct! Count the components: 3 entries means R³, regardless of any zeros.

❌ Not quite. Membership in R^n depends on the number of components, not their values. Count carefully!

Show solution

Solution:

Membership in $\mathbb{R}^n$ is determined by how many components a vector has, not by the values of those components.

$(3, 2)$ has 2 components → belongs to $\mathbb{R}^2$
$(3, 2, 0)$ has 3 components → belongs to $\mathbb{R}^3$

The zero in the third position does not remove a dimension — it still counts as a component. $(3, 2)$ is a point on a flat 2D plane, while $(3, 2, 0)$ sits on the $xy$ -plane within three-dimensional space. Same $x$ and $y$ values, but they are different vectors in different spaces.

Question 3

Consider the line $y = 2x + 3$ in $\mathbb{R}^2$ . Is this line a subspace of $\mathbb{R}^2$ ?

✅ Correct! The zero vector test is the fastest check: (0,0) is not on y = 2x + 3 since 0 ≠ 3.

❌ Not quite. Start with the zero vector test: does the origin (0, 0) satisfy y = 2x + 3?

Show solution

Solution:

Apply the zero vector test — the fastest disqualification test in linear algebra.

Is $(0, 0)$ on the line $y = 2x + 3$ ? Substitute $x = 0$ :

$y = 2(0) + 3 = 3 \neq 0$

The origin is not on the line, so this is not a subspace. We can stop here!

Only lines of the form $y = mx$ (with $b = 0$ ) pass through the origin and can be subspaces. The y-intercept $b = 3$ guarantees this line misses the origin. Both addition and scalar multiplication also fail, but the zero vector test catches the problem instantly.

Question 4

True or False: If a set in $\mathbb{R}^n$ is closed under addition, it must be a vector space.

✅ Correct! Closure under addition alone is not enough — you also need closure under scalar multiplication. The first quadrant is the classic counterexample.

❌ Not quite. Think about the first quadrant: it's closed under addition, but is it a vector space?

Show solution

Solution: False.

Closure under addition alone is not sufficient. A vector space must be closed under both addition and scalar multiplication. Equivalently, it must be closed under all linear combinations $c\,\mathbf{u} + d\,\mathbf{v}$ .

The first quadrant is the classic counterexample:

✅ Closed under addition: adding two non-negative vectors gives a non-negative vector
❌ Not closed under scalar multiplication: $(-1) \cdot (1, 1) = (-1, -1) \notin Q$

You need both closures. One out of two is not enough!

Solved: 0 / 4

Vector Spaces and Subspaces

IKey Formulas

IIVisualization 1 — Scalar Multiplication Breaks QQQ

IIIVisualization 2 — Where Does u+v\mathbf{u}+\mathbf{v}u+v Land?

IVVisualization 3 — The Three Axioms at Once

VQuiz Questions

IIVisualization 1 — Scalar Multiplication Breaks $Q$

IIIVisualization 2 — Where Does $\mathbf{u}+\mathbf{v}$ Land?