LINEAR-ALGEBRA · Unit 6 · Video 1 · Interactive Practice

Vector Spaces and Subspaces Explained

IKey Formulas

Property	Condition	Meaning
Closure under addition	$\mathbf{v}, \mathbf{w} \in S \Rightarrow \mathbf{v} + \mathbf{w} \in S$	Sum stays in the set
Closure under scalar mult.	$\mathbf{v} \in S \Rightarrow c\mathbf{v} \in S,\ \forall\, c \in \mathbb{R}$	Scaled vector stays in the set
Combined (linear combinations)	$c\mathbf{v} + d\mathbf{w} \in S,\ \forall\, c, d \in \mathbb{R}$	All linear combinations stay in $S$
Zero vector test	$\mathbf{0} \in S$	Every subspace must contain the origin

IIVisualization 1 — Closure Under Linear Combinations

A line through the origin is a subspace — can any combination $c\mathbf{v} + d\mathbf{w}$ of its vectors escape it?

Scalar c 1 Scalar d 1/2

💡 Because every $c\mathbf{v} + d\mathbf{w}$ stays on the line, the line is the span of its vectors — and that closure is exactly what makes it a subspace.

IIIVisualization 2 — The Zero Vector Test

A subspace must contain $\mathbf{0}$ — for which intercept $b$ does $y = x + b$ pass through the origin?

Intercept b 1

💡 Setting $c = 0$ gives $0 \cdot \mathbf{v} = \mathbf{0}$ , so every subspace is forced to contain the origin — a failed origin check disqualifies a set instantly.

IVVisualization 3 — Why Unions of Subspaces Fail

Each line through the origin is a subspace — but does the sum $\mathbf{v} + \mathbf{w}$ stay inside their union?

Angle of Line 2 90°

💡 The union fails closure, yet the intersection of two subspaces is always a subspace — here the two lines meet only at $\{\mathbf{0}\}$ .

VQuiz Questions

Question 1

Which of the following sets is not a subspace of $\mathbb{R}^3$ ?

✅ Correct! The plane $z = 1$ doesn't contain the zero vector $(0, 0, 0)$ , so it cannot be a subspace.

❌ Not quite. Hint: which of these sets does not contain the zero vector $(0, 0, 0)$ ?

Show solution

Solution:

The plane $z = 1$ is not a subspace because it does not contain the zero vector.

Check: Is $(0, 0, 0)$ in the set $\{(x, y, 1)\}$ ? No — the $z$ -component must be $1$ , but the zero vector has $z = 0$ .
Every subspace must contain the zero vector. Since $z = 1$ doesn't pass through the origin, it fails immediately.

The other three sets all contain $(0, 0, 0)$ and are closed under addition and scalar multiplication:

$xy$ -plane: $(x_1, y_1, 0) + (x_2, y_2, 0) = (x_1+x_2,\, y_1+y_2,\, 0)$ — still has $z = 0$ ✓
$z$ -axis: $(0, 0, z_1) + (0, 0, z_2) = (0, 0, z_1+z_2)$ — still has $x = y = 0$ ✓
$\{\mathbf{0}\}$ : $\mathbf{0} + \mathbf{0} = \mathbf{0}$ ✓

Question 2

True or False: The union of two subspaces of $\mathbb{R}^3$ is always a subspace of $\mathbb{R}^3$ .

✅ Correct! The union of two subspaces generally fails closure under addition.

❌ Not quite. Think about a vector from the $xy$ -plane plus a vector from the $z$ -axis — where does their sum land?

Show solution

Solution: False.

Counterexample: Let $P$ = the $xy$ -plane and $L$ = the $z$ -axis in $\mathbb{R}^3$ .

Both are subspaces. Now consider their union $P \cup L$ :

$\mathbf{v} = (1, 0, 0) \in P$ ✓
$\mathbf{w} = (0, 0, 1) \in L$ ✓
$\mathbf{v} + \mathbf{w} = (1, 0, 1)$

Is $(1, 0, 1)$ in $P$ ? No — $z = 1 \neq 0$ . Is $(1, 0, 1)$ in $L$ ? No — $x = 1 \neq 0$ .

The sum $(1, 0, 1) \notin P \cup L$ , so the union is not closed under addition and therefore not a subspace.

Question 3

Consider the set $S = \{(x, y, z) \in \mathbb{R}^3 : x + y + z = 0\}$ . Is $S$ a subspace of $\mathbb{R}^3$ ?

✅ Correct! The equation $x + y + z = 0$ defines a plane through the origin, and it passes all three subspace checks.

❌ Not quite. Try checking: does $(0, 0, 0)$ satisfy $x + y + z = 0$ ? If you add two vectors whose components sum to zero, does the result also sum to zero?

Show solution

Solution: Yes — it is a plane through the origin.

Zero vector test: $(0, 0, 0)$ : $\;0 + 0 + 0 = 0$ ✓

Closure under addition: Let $\mathbf{v} = (v_1, v_2, v_3)$ and $\mathbf{w} = (w_1, w_2, w_3)$ with $v_1 + v_2 + v_3 = 0$ and $w_1 + w_2 + w_3 = 0$ .

$\mathbf{v} + \mathbf{w} = (v_1 + w_1,\; v_2 + w_2,\; v_3 + w_3)$

$(v_1 + w_1) + (v_2 + w_2) + (v_3 + w_3) = \underbrace{(v_1 + v_2 + v_3)}_{=\,0} + \underbrace{(w_1 + w_2 + w_3)}_{=\,0} = 0 \;\checkmark$

Closure under scalar multiplication: For $c \in \mathbb{R}$ :

$c\mathbf{v} = (cv_1, cv_2, cv_3),\quad cv_1 + cv_2 + cv_3 = c(v_1 + v_2 + v_3) = c \cdot 0 = 0 \;\checkmark$

Geometrically, $x + y + z = 0$ is a plane through the origin in $\mathbb{R}^3$ — a dimension 2 subspace.

Question 4

If $S$ and $T$ are both subspaces of $\mathbb{R}^3$ , which of the following is always true?

✅ Correct! The intersection of any two subspaces is always a subspace. This is a fundamental theorem in linear algebra.

❌ Not quite. Remember: every subspace contains the zero vector, so $S \cap T$ always contains at least $\mathbf{0}$ — it's never empty!

❌ Not quite. Recall from the video: unions generally fail, but intersections always preserve the subspace property.

Show solution

Solution: $S \cap T$ is always a subspace.

Proof sketch:

Let $\mathbf{v}, \mathbf{w} \in S \cap T$ . Then:

$\mathbf{v}, \mathbf{w} \in S$ and $S$ is a subspace $\Rightarrow$ $\mathbf{v} + \mathbf{w} \in S$
$\mathbf{v}, \mathbf{w} \in T$ and $T$ is a subspace $\Rightarrow$ $\mathbf{v} + \mathbf{w} \in T$
Therefore $\mathbf{v} + \mathbf{w} \in S \cap T$ ✓

Same reasoning for scalar multiplication: $c\mathbf{v} \in S$ and $c\mathbf{v} \in T$ , so $c\mathbf{v} \in S \cap T$ ✓

Why the other options fail:

Option	Why it fails
$S \cup T$ is a subspace	Counterexample: $xy$ -plane $\cup$ $z$ -axis is not closed under addition
$S \cup T = S \cap T$	Generally false (e.g., $xy$ -plane ∪ $z$ -axis ≠ $xy$ -plane ∩ $z$ -axis)
$S \cap T$ is empty	Both subspaces contain $\mathbf{0}$ , so $\mathbf{0} \in S \cap T$ — it's never empty!

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