Subspaces in Linear Algebra - Examples with Solutions

A subspace is a subset of a vector space that satisfies specific closure and structural properties. This guide covers the formal definition of subspaces, the essential test criteria, and detailed step-by-step examples.

Table of Contents

Definition of Subspaces

If \( W \) is a subset of a vector space \( V \) and \( W \) is itself a vector space under the inherited operations of addition and scalar multiplication from \( V \), then \( W \) is called a subspace.

To show that \( W \) is a subspace of \( V \), it is sufficient to verify the following conditions:

  1. \( W \) is a non-empty subset of \( V \).
  2. Zero vector: The zero vector \( \textbf{0} \) of \( V \) is in \( W \).
  3. Closure under addition: For any vectors \( \mathbf{u} \) and \( \mathbf{v} \) in \( W \), the sum \( \mathbf{u} + \mathbf{v} \) is in \( W \).
  4. Closure under scalar multiplication: For any vector \( \mathbf{u} \) in \( W \) and scalar \( r \), the product \( r \cdot \mathbf{u} \) is in \( W \).

Note: Conditions 2, 3, and 4 can be combined into a single test: \( r\mathbf{u} + s\mathbf{v} \in W \) for all \( \mathbf{u}, \mathbf{v} \in W \) and all scalars \( r, s \).

Geometric illustration of a subspace W contained within a vector space V

Properties of Subspaces

Subspaces have several important properties that follow directly from the definition:

  1. Always contains the zero vector: Every subspace must contain the zero vector of the parent space.
  2. Every vector space is its own subspace: \( V \) is always a subspace of itself (called the trivial subspace).
  3. The zero subspace: \( \{\textbf{0}\} \) — the set containing only the zero vector — is always a subspace of any vector space.
  4. Intersection of subspaces: The intersection of two subspaces of \( V \) is also a subspace of \( V \).
  5. Union of subspaces: The union of two subspaces is generally NOT a subspace (unless one is contained in the other).
  6. Spans are subspaces: The span of any set of vectors in \( V \) is a subspace of \( V \).

Geometric Intuition in \( \mathbb{R}^3 \): The subspaces of \( \mathbb{R}^3 \) are: the origin \( \{\textbf{0}\} \), all lines through the origin, all planes through the origin, and \( \mathbb{R}^3 \) itself.

Examples of Subspaces

Example 1: A Subspace of \( \mathbb{R}^2 \) on the Coordinate Axis

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Show that the set \( W \) of vectors of the form \( (x,0) \) where \( x \in \mathbb{R} \) is a subspace of \( \mathbb{R}^2 \).

Solution

We check the subspace criteria:

  • \( W \) is a non-empty subset of \( \mathbb{R}^2 \), whose vectors are of the general form \( (x,y) \) where \( x, y \in \mathbb{R} \).
  • The zero vector \( (0,0) \) is in \( W \) (since setting \( x = 0 \) gives \( (0,0) \)).
  • Closure under addition: Let \( \mathbf{u} = (x_1, 0) \) and \( \mathbf{v} = (x_2, 0) \) be in \( W \). Then: \[ (x_1, 0) + (x_2, 0) = (x_1 + x_2, 0) \] Since \( x_1 + x_2 \) is a real number, the sum is in \( W \).
  • Closure under scalar multiplication: Let \( \mathbf{u} = (x, 0) \) be in \( W \) and \( r \) be a scalar. Then: \[ r \cdot (x, 0) = (rx, 0) \] Since \( rx \) is a real number, the product is in \( W \).

Therefore, \( W \) is a subspace of \( \mathbb{R}^2 \). Geometrically, \( W \) is the \( x \)-axis.

Example 2: A Non-Subspace Defined by Inequality Constraints

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Show that the set \( W \) of vectors of the form \( (x,y) \) such that \( x \ge 0 \) and \( y \ge 0 \) is NOT a subspace of \( \mathbb{R}^2 \).

Solution

To prove that \( W \) is not a subspace, we can find a counterexample demonstrating that it fails at least one of the conditions (in this case, closure under scalar multiplication).

  • Consider the vector \( \mathbf{u} = (2,2) \). Both components are non-negative, so \( \mathbf{u} \in W \).
  • Now multiply \( \mathbf{u} \) by the scalar \( r = -1 \): \[ -1 \cdot (2,2) = (-2, -2) \]
  • The resulting vector \( (-2, -2) \) has negative components, meaning it is not in \( W \).

Because \( W \) is not closed under scalar multiplication, it is not a subspace of \( \mathbb{R}^2 \). Geometrically, \( W \) is the first quadrant, which is not closed under scalar multiplication by negative numbers.

Example 3: A Plane Through the Origin in \( \mathbb{R}^3 \)

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Show that the set \( W = \{ (x,y,z) \mid x + y + z = 0 \} \) is a subspace of \( \mathbb{R}^3 \).

Solution

We verify the subspace requirements for \( W \):

  • \( W \) is a non-empty subset of \( \mathbb{R}^3 = \{ (x,y,z) \mid x, y, z \in \mathbb{R} \} \).
  • Zero vector check: The zero vector \( (0,0,0) \) is in \( W \) because \( 0 + 0 + 0 = 0 \).
  • Closure under addition: Let \( \mathbf{u} = (x_1, y_1, z_1) \) and \( \mathbf{v} = (x_2, y_2, z_2) \) be vectors in \( W \). By definition, \( x_1 + y_1 + z_1 = 0 \) and \( x_2 + y_2 + z_2 = 0 \). \[ \mathbf{u} + \mathbf{v} = (x_1 + x_2, y_1 + y_2, z_1 + z_2) \] Checking the equation for the sum vector: \[ (x_1 + x_2) + (y_1 + y_2) + (z_1 + z_2) = (x_1 + y_1 + z_1) + (x_2 + y_2 + z_2) = 0 + 0 = 0 \] Thus, \( \mathbf{u} + \mathbf{v} \in W \).
  • Closure under scalar multiplication: Let \( \mathbf{u} = (x_1, y_1, z_1) \) be in \( W \) (so \( x_1 + y_1 + z_1 = 0 \)) and let \( r \) be a real number. \[ r \cdot (x_1, y_1, z_1) = (rx_1, ry_1, rz_1) \] Checking the equation for the scaled vector: \[ rx_1 + ry_1 + rz_1 = r(x_1 + y_1 + z_1) = r \cdot 0 = 0 \] Thus, \( r\mathbf{u} \in W \).

Since all conditions are satisfied, \( W \) is a subspace of \( \mathbb{R}^3 \). Geometrically, \( W \) is a plane passing through the origin with normal vector \( (1,1,1) \).

Example 4: A Line Through the Origin in \( \mathbb{R}^3 \)

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Show that the set \( W = \{ t(1, 2, 3) \mid t \in \mathbb{R} \} \) is a subspace of \( \mathbb{R}^3 \).

Solution

We verify the subspace requirements for \( W \):

  • Zero vector check: Setting \( t = 0 \) gives \( (0, 0, 0) \in W \).
  • Closure under addition: Let \( \mathbf{u} = t_1(1,2,3) \) and \( \mathbf{v} = t_2(1,2,3) \) be in \( W \). Then: \[ \mathbf{u} + \mathbf{v} = t_1(1,2,3) + t_2(1,2,3) = (t_1 + t_2)(1,2,3) \] Since \( t_1 + t_2 \) is a real number, the sum is in \( W \).
  • Closure under scalar multiplication: Let \( \mathbf{u} = t(1,2,3) \) be in \( W \) and let \( r \) be a real number. Then: \[ r \cdot \mathbf{u} = r \cdot t(1,2,3) = (rt)(1,2,3) \] Since \( rt \) is a real number, the product is in \( W \).

Since all conditions are satisfied, \( W \) is a subspace of \( \mathbb{R}^3 \). Geometrically, \( W \) is the line through the origin in the direction of \( (1,2,3) \).

Example 5: A Non-Subspace — Plane Not Through the Origin

Show Solution

Show that the set \( W = \{ (x,y,z) \mid x + y + z = 1 \} \) is NOT a subspace of \( \mathbb{R}^3 \).

Solution

We test the zero vector condition first:

  • Consider the zero vector \( (0,0,0) \). Substituting into the defining equation: \( 0 + 0 + 0 = 0 \neq 1 \).
  • Therefore, the zero vector \( (0,0,0) \) is not in \( W \).

Since \( W \) fails the zero vector condition, it is not a subspace of \( \mathbb{R}^3 \). Geometrically, \( W \) is a plane that does not pass through the origin.

Additional Counterexample (Closure under addition): Let \( \mathbf{u} = (1,0,0) \) and \( \mathbf{v} = (0,1,0) \). Both satisfy \( x + y + z = 1 \), so both are in \( W \). But \( \mathbf{u} + \mathbf{v} = (1,1,0) \), and \( 1 + 1 + 0 = 2 \neq 1 \), so the sum is not in \( W \).

Frequently Asked Questions

What is a subspace in linear algebra?

A subspace is a subset of a vector space that is itself a vector space under the inherited operations of addition and scalar multiplication. It must contain the zero vector and be closed under both addition and scalar multiplication.

How do you test if a set is a subspace?

To test if a set \( W \) is a subspace of a vector space \( V \), verify three conditions: (1) the zero vector of \( V \) is in \( W \), (2) \( W \) is closed under addition, and (3) \( W \) is closed under scalar multiplication.

What is the difference between a subset and a subspace?

Every subspace is a subset, but not every subset is a subspace. A subset becomes a subspace only when it satisfies the closure properties and contains the zero vector, making it a vector space in its own right.

Is every plane in \( \mathbb{R}^3 \) a subspace?

No. Only planes that pass through the origin are subspaces of \( \mathbb{R}^3 \). A plane that does not contain the origin fails the zero vector condition and is not a subspace.

What are the subspaces of \( \mathbb{R}^3 \)?

The subspaces of \( \mathbb{R}^3 \) are: the zero subspace \( \{\textbf{0}\} \), all lines through the origin, all planes through the origin, and \( \mathbb{R}^3 \) itself.

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