Determine Whether the Following Sets Are Subspaces of ℝ³
When studying linear algebra, one of the first tasks is to decide if a given subset of ℝ³ satisfies the three conditions that define a subspace. This article explains the underlying theory, outlines a clear step‑by‑step method, and works through several concrete examples so you can determine whether the following sets are subspaces of ℝ³ with confidence And it works..
Introduction
A subset (W) of ℝ³ is a subspace if it is closed under vector addition and scalar multiplication and if it contains the zero vector. Because of that, these requirements are not arbitrary; they are the minimal conditions that guarantee (W) itself behaves like a three‑dimensional vector space. Understanding how to verify them is essential for later topics such as bases, dimension, and linear transformations.
The Subspace Test
To determine whether the following sets are subspaces of ℝ³, apply the following three criteria:
- Zero Vector – The set must contain (\mathbf{0} = (0,0,0)).
- Closed under Addition – If (\mathbf{u}, \mathbf{v} \in W), then (\mathbf{u}+\mathbf{v} \in W).
- Closed under Scalar Multiplication – If (\mathbf{u} \in W) and (c) is any scalar, then (c\mathbf{u} \in W).
If any of these conditions fails, the set is not a subspace. The test is straightforward, but careful algebraic manipulation is required to avoid hidden pitfalls Worth keeping that in mind. No workaround needed..
Common Sets to Test
In practice, textbooks often present sets described by equations, parametric forms, or explicit lists of vectors. Examples include:
- The set of all vectors ((x,y,z)) satisfying a linear equation such as (x+2y-3z=0).
- The set of all linear combinations of given vectors, e.g., ({,a(1,2,0)+b(0,1,1)\mid a,b\in\mathbb{R},}).
- Sets defined by inequalities, such as ({(x,y,z)\mid x\ge 0,;y\ge 0,;z\ge 0}).
- Finite collections of vectors, like ({(1,0,0),(0,1,0)}).
Each of these can be examined with the subspace test Took long enough..
Step‑by‑Step Procedure
- Identify the defining condition of the set. Write it in a clear mathematical form.
- Check the zero vector: substitute ((0,0,0)) into the condition. If it satisfies the condition, the set passes this test.
- Test closure under addition: take two arbitrary vectors that satisfy the condition, add them, and verify that the result still satisfies the condition.
- Test closure under scalar multiplication: multiply an arbitrary vector by an arbitrary scalar and check that the product meets the condition.
- Conclude: if all three tests succeed, the set is a subspace; otherwise, it is not.
When performing steps 3 and 4, use generic symbols (e.Think about it: g. , (\mathbf{u} = (u_1,u_2,u_3)), (\mathbf{v} = (v_1,v_2,v_3)), (c\in\mathbb{R})) to keep the proof general.
Worked Examples
Example 1: The Plane Through the Origin
Consider the set
[ W_1={(x,y,z)\in\mathbb{R}^3\mid x+2y-3z=0}. ]
Zero vector: Substituting ((0,0,0)) gives (0+0-0=0), so (\mathbf{0}\in W_1).
Addition: Let (\mathbf{u}=(u_1,u_2,u_3)) and (\mathbf{v}=(v_1,v_2,v_3)) be in (W_1). Then
[ u_1+2u_2-3u_3=0,\qquad v_1+2v_2-3v_3=0. ]
Adding the two equations yields
[ (u_1+v_1)+2(u_2+v_2)-3(u_3+v_3)=0, ]
so (\mathbf{u}+\mathbf{v}\in W_1).
Scalar multiplication: For any scalar (c),
[ c u_1+2(cu_2)-3(cu_3)=c(u_1+2u_2-3u_3)=c\cdot0=0, ]
hence (c\mathbf{u}\in W_1) Worth keeping that in mind..
Since all three criteria are satisfied, (W_1) is a subspace of ℝ³.
Example 2: A Translated Plane
Let
[ W_2={(x,y,z)\in\mathbb{R}^3\mid x+2y-3z=5}. ]
Zero vector: Substituting ((0,0,0)) gives (0\neq5); thus (\mathbf{0}\notin W_2).
Because the zero vector is missing, (W_2) cannot be a subspace, regardless of the other two tests. This illustrates how a single failure is enough to reject a set.
Example 3: Span of Two Vectors
Define
[ W_3={,a(1,0,2)+b(0,1,-1)\mid a,b\in\mathbb{R},}. ]
Zero vector: Choose (a=b=0); then (0(1,0,2)+0(0,1,-1)=(0,0,0)), so (\mathbf{0}\in W_3).
Addition: Let (\mathbf{u}=a_1(1,0,2)+b_1(0,1,-1)) and (\mathbf{v}=a_2(1,0,2)+b_2(0,1,-1)). Their sum is
[ (a_1+a_2)(1,0,2)+(b_1+b_2)(0,1,-1), ]
which is again of the same form, so (\mathbf{u}+\mathbf{v}\in W_3).
Scalar multiplication: For any scalar (c),
[c\mathbf{u}= (ca_1)(1,0,2)+(cb_1)(0,1,-1), ]
still belonging to (W_3).
So, (W_3) is a subspace; in fact, it is the span of the two given vectors, a standard example of a subspace Most people skip this — try not to..
Example 4: Set Defined by Inequalities
Consider
Example 4: Set Defined by Inequalities
Consider the set
[ W_4 = {(x,y,z) \in \mathbb{R}^3 \mid x^2 + y^2 \leq 1}. ]
Zero vector: Substituting ((0,0,0)) gives (0^2 + 0^2 = 0 \leq 1), so (\mathbf{0} \in W_4).
Addition: Let (\mathbf{u} = (u_1, u_2, u_3)) and (\mathbf{v} = (v_1, v_2, v_3)) be in (W_4), so (u_1^2 + u_2^2 \leq 1) and (v_1^2 + v_2^2 \leq 1). Their sum is (\mathbf{u}+\mathbf{v} = (u_1+v_1, u_2+v_2, u_3+v_3)). We must check whether ((u_1+v_1)^2 + (u_2+v_2)^2 \leq 1). By the triangle inequality,
[
(u_1+v_1)^2 + (u_2+v_2)^2 \leq (|u_1|+|v_1|)^2 + (|u_2|+|v_2|)^2.
]
Even if each original sum of squares is at most 1, the combined sum can exceed 1. Here's a good example: take (\mathbf{u} = (1,0,0)) and (\mathbf{v} = (1,0,0)); both satisfy the condition (since (1^2+0^2=1 \leq 1)), but (\mathbf{u}+\mathbf{v} = (2,0,0)) gives (2^2+0^2=4 > 1). Thus, (\mathbf{u}+\mathbf{v} \notin W_4), and closure under addition fails.
Scalar multiplication: For any scalar (c), consider (c\mathbf{u} = (c u_1, c u_2, c u_3)). We need ((c u_1)^2 + (c u_2)^2 = c^2(u_1^2+u_2^2) \leq 1). If (c = 2) and (\mathbf{u} = (1,0,0)), then (c^2(1^2+0^2)=4 > 1), so (c\mathbf{u} \notin W_4). Closure under scalar multiplication also fails That's the whole idea..
Since two of the three criteria are not satisfied
Example 4 (continued): Set Defined by Inequalities
Since two of the three criteria are not satisfied, (W_{4}) is not a subspace. This example illustrates how a set that looks “nice” geometrically—here a solid cylinder extending infinitely in the (z)-direction—fails to be linear because the linear combination of two points can leave the set Took long enough..
6. Common Pitfalls and How to Avoid Them
| Pitfall | Why It Happens | Remedy |
|---|---|---|
| Forgetting the zero vector | Beginners sometimes overlook the first test, assuming that closure under addition and scalar multiplication will automatically bring the zero vector in. | |
| Thinking inequalities preserve subspace structure | Sets defined by inequalities (e.g. | Test closure with simple counterexamples (e.Day to day, , ((1,0,0)+(1,0,0))). |
| Misreading the definition of a span | The span of a set is always a subspace, but one might think any linear combination of vectors from a set is a subspace. | Remember that the span is defined as all finite linear combinations; the set itself need not be a subspace. g. |
| Confusing “vector space” with “subspace” | A vector space must satisfy the same axioms in its own right; a subspace inherits the operations from a larger space. | |
| Assuming “closed under addition” implies “closed under subtraction” | Closure under addition only guarantees that (\mathbf{u}+\mathbf{v}\in W); it does not guarantee that (\mathbf{u}-\mathbf{v}\in W). | When verifying a subspace, always keep the ambient space’s operations in mind. |
7. Quick‑Check Checklist
- Zero vector: Is (\mathbf{0}) in the set?
- Closure under addition: For any (\mathbf{u},\mathbf{v}) in the set, is (\mathbf{u}+\mathbf{v}) in the set?
- Closure under scalar multiplication: For any scalar (c) and (\mathbf{u}) in the set, is (c\mathbf{u}) in the set?
If yes to all three, you have a subspace. If no to any, the set fails to be a subspace.
8. Conclusion
Determining whether a given subset of (\mathbb{R}^n) is a subspace boils down to a handful of concrete verifications. The zero vector guarantees a base point; closure under addition and scalar multiplication ensures the set is stable under the vector operations inherited from the surrounding space. By systematically applying these tests—and being mindful of the common pitfalls—you can confidently classify any candidate set as a subspace or not.
Remember: the elegance of linear algebra lies in its simplicity. Think about it: once you master these three checks, you’ll find that many seemingly complicated sets either fit neatly into the subspace framework or reveal their non‑linear nature with a single counterexample. Happy exploring!
8. Conclusion
Determining whether a given subset of (\mathbb{R}^n) is a subspace boils down to a handful of concrete verifications. That's why once you master these three checks, you’ll find that many seemingly complicated sets either fit neatly into the subspace framework or reveal their non-linear nature with a single counterexample. The zero vector guarantees a base point; closure under addition and scalar multiplication ensures the set is stable under the vector operations inherited from the surrounding space. By systematically applying these tests—and being mindful of the common pitfalls—you can confidently classify any candidate set as a subspace or not. That's why remember: the elegance of linear algebra lies in its simplicity. Happy exploring!
Final Thoughts:
Subspaces are the building blocks of linear algebra, and their properties underpin concepts like linear independence, bases, and dimension. By rigorously verifying the three criteria—zero vector inclusion, closure under addition, and closure under scalar multiplication—you develop a disciplined approach to abstract problems. Avoiding common mistakes, such as assuming closure under subtraction or misinterpreting inequalities, sharpens your critical thinking. With practice, these checks become second nature, empowering you to tackle more advanced topics with confidence. Keep refining your intuition, and let the structure of vector spaces guide your mathematical journey That's the part that actually makes a difference..
