Subspace Tester

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About This MicroSim

This visualization helps students understand what makes a set a subspace by testing the closure property. Students can select different sets, drag test vectors within those sets, and observe whether linear combinations stay within the set.

Learning Objective: Test whether sets are subspaces by checking closure under linear combinations.

How to Use

Select a Set: Use the dropdown to choose different sets (line through origin, line not through origin, first quadrant, circle, or entire plane)
Drag Test Vectors: Click and drag the endpoints of vectors u (red) and v (blue). Vectors are constrained to stay within the selected set
Adjust Scalars: Use the c and d sliders to change the scalar multipliers for the linear combination cu + dv
Observe the Result: The result vector is shown in green if it stays in the set, or orange if it leaves the set
Check Subspace: Click "Check if Subspace" to see an explanation of why the set is or is not a subspace

Key Concepts Demonstrated

Subspace Definition

A subset H of a vector space V is a subspace if:

Zero vector: The zero vector is in H
Closure under addition: For all u, v in H, u + v is in H
Closure under scalar multiplication: For all u in H and scalar c, cu is in H

Equivalently, H is a subspace if it is closed under linear combinations: for all u, v in H and scalars c, d, the vector cu + dv is also in H.

Examples Explored

Set	Is Subspace?	Reason
Line through origin	Yes	Contains zero, closed under linear combinations
Line NOT through origin	No	Does not contain zero vector
First quadrant	No	Not closed under scalar multiplication (try c = -1)
Circle	No	Does not contain zero, not closed under addition
Entire plane R^2	Yes	The whole space is always a subspace

Finding Counter-Examples

For non-subspaces, try these strategies to find counter-examples:

Line not through origin: Set c = 0, d = 0 to see that (0, 0) is not on the line
First quadrant: Set c = -1, d = 0 to see that -u has negative components
Circle: Set c = 1, d = 1 to see that u + v is not on the circle

Embedding This MicroSim

<iframe src="https://dmccreary.github.io/linear-algebra/sims/subspace-tester/main.html"
        height="502px"
        width="100%"
        scrolling="no">
</iframe>

Lesson Plan

Grade Level

Undergraduate introductory linear algebra

Duration

20-25 minutes

Prerequisites

Vector addition and scalar multiplication
Linear combinations
Basic set theory concepts

Learning Activities

Exploration with a True Subspace (5 min):
Select "Line through origin (y = 2x)"
Drag vectors u and v along the line
Adjust c and d sliders and observe that cu + dv always stays on the line
Click "Check if Subspace" to confirm
Finding Counter-Examples (10 min):
Select "Line not through origin (y = 2x + 1)"
Note that vectors can exist on this line
Set c = 0, d = 0. Where is the origin relative to the line?
Explain why this fails the subspace test
First Quadrant Investigation (5 min):
Select "First quadrant"
Place u at (1, 2)
Set c = -1, d = 0. Where does the result end up?
Discuss: even though (1, 2) is in the first quadrant, (-1, -2) is not
Circle Failure (5 min):
Select "Circle"
Place u at (2, 0) and v at (0, 2)
With c = 1, d = 1, where is u + v?
Verify it's not on the circle

Discussion Questions

Why must every subspace contain the zero vector?
If a set is closed under addition, is it necessarily closed under scalar multiplication?
Can you think of a set that contains zero but is not a subspace?
What is the smallest possible subspace of R^2?

Assessment Ideas

Given a description of a set, predict whether it's a subspace
For non-subspaces, provide specific counter-examples
Prove that the intersection of two subspaces is also a subspace

Mathematical Background

The Subspace Test

To verify H is a subspace, we can use the combined test: H is a subspace if and only if for all u, v in H and all scalars c, d:

\[c\mathbf{u} + d\mathbf{v} \in H\]

This single condition implies all three properties (zero, addition closure, scalar multiplication closure).

Why Lines Through Origin Are Subspaces

A line through the origin can be written as:

\[L = \{t\mathbf{v} : t \in \mathbb{R}\}\]

for some direction vector v. For any \(t_1\mathbf{v}\) and \(t_2\mathbf{v}\) in L:

\[c(t_1\mathbf{v}) + d(t_2\mathbf{v}) = (ct_1 + dt_2)\mathbf{v} \in L\]

Why the First Quadrant Fails

The first quadrant Q = {(x, y) : x >= 0, y >= 0} contains (1, 1). However:

\[(-1) \cdot (1, 1) = (-1, -1) \notin Q\]

So Q is not closed under scalar multiplication.

References

Strang, G. (2016). Introduction to Linear Algebra (5th ed.). Chapter 3.1 - Spaces of Vectors.
Lay, D. C. (2015). Linear Algebra and Its Applications (5th ed.). Section 4.1 - Vector Spaces and Subspaces.
3Blue1Brown - Span and Subspaces - Visual introduction to subspaces.
Khan Academy - Subspaces