Subspace and Examples

by Kurious Fox

A subspace is a subset of a vector space, and it is itself a vector space with addition and scalar multiplication operations defined the same way as in the original vector space. This means that all characteristics and properties of the original vector space are preserved in the subspace, allowing us to apply mathematical theories and methods that have been developed for the original vector space. Furthermore, subspaces provide a powerful approach to studying problems related to geometry and algebra, opening up many possibilities for applications in fields such as physics, engineering, and computer science.

Here are examples of subspaces in linear algebra, which are subsets of a vector space that satisfy the properties of being a vector space themselves:

Example 1: Subspace of \mathbb{R}^3

Let V = \mathbb{R}^3 , the set of all 3-dimensional vectors.

Define a subset W as:
W = \{(x, y, z) \in \mathbb{R}^3 \mid x + y + z = 0\}.

Verifying W is a subspace:

  1. Contains the zero vector: The zero vector \mathbf{0} = (0, 0, 0) satisfies 0 + 0 + 0 = 0 , so \mathbf{0} \in W .
  2. Closed under addition: If \mathbf{u} = (x_1, y_1, z_1) \in W and \mathbf{v} = (x_2, y_2, z_2) \in W , then:
    \mathbf{u} + \mathbf{v} = (x_1 + x_2, y_1 + y_2, z_1 + z_2),
    and:
    (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.
    So \mathbf{u} + \mathbf{v} \in W .
  3. Closed under scalar multiplication: If \mathbf{u} = (x, y, z) \in W and c \in \mathbb{R} , then:
    c \mathbf{u} = (cx, cy, cz),
    and:
    cx + cy + cz = c(x + y + z) = c \cdot 0 = 0.
    So c \mathbf{u} \in W .

Since all conditions are satisfied, W is a subspace of \mathbb{R}^3 .

Example 2: Subspace of \mathbb{R}^2

Let V = \mathbb{R}^2 , the set of all 2-dimensional vectors.

Define a subset W as:
W = \{(x, y) \in \mathbb{R}^2 \mid y = 2x\}.

Verifying W is a subspace:

  1. Contains the zero vector: The zero vector \mathbf{0} = (0, 0) satisfies y = 2x because 0 = 2 \cdot 0 . Thus, \mathbf{0} \in W .
  2. Closed under addition: If \mathbf{u} = (x_1, y_1) \in W and \mathbf{v} = (x_2, y_2) \in W , then:
    \mathbf{u} + \mathbf{v} = (x_1 + x_2, y_1 + y_2),
    and since y_1 = 2x_1 and y_2 = 2x_2 :
    y_1 + y_2 = 2x_1 + 2x_2 = 2(x_1 + x_2).
    Thus, \mathbf{u} + \mathbf{v} \in W .
  3. Closed under scalar multiplication: If \mathbf{u} = (x, y) \in W and c \in \mathbb{R} , then:
    c \mathbf{u} = (cx, cy),
    and cy = c(2x) = 2(cx) . Thus, c \mathbf{u} \in W .

Since all conditions are satisfied, W is a subspace of \mathbb{R}^2 .

Example 3: Subspace of M_{2 \times 2} , the space of 2 \times 2 matrices

Let V = M_{2 \times 2} , the set of all 2 \times 2 matrices with real entries.

Define a subset W as:
W = \{A \in M_{2 \times 2} \mid \text{trace}(A) = 0\},
where \text{trace}(A) = a_{11} + a_{22} is the sum of the diagonal entries of A .

Verifying W is a subspace:

  1. Contains the zero matrix: The zero matrix \mathbf{0} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} satisfies \text{trace}(\mathbf{0}) = 0 , so \mathbf{0} \in W .
  2. Closed under addition: If A, B \in W , then \text{trace}(A + B) = \text{trace}(A) + \text{trace}(B) = 0 + 0 = 0 . Thus, A + B \in W .
  3. Closed under scalar multiplication: If A \in W and c \in \mathbb{R} , then \text{trace}(cA) = c \cdot \text{trace}(A) = c \cdot 0 = 0 . Thus, cA \in W .

Since all conditions are satisfied, W is a subspace of M_{2 \times 2} .

Example 4: Trivial Subspaces

  1. The zero subspace {0} is always a subspace of any vector space because it satisfies all subspace properties trivially.
  2. The entire vector space V is also a subspace of V .

Example 5: Subspace of Polynomials

Let V be the space of all polynomials of degree at most 3:
V = \{p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 \mid a_0, a_1, a_2, a_3 \in \mathbb{R}\}.

Define a subset W as:
W = \{p(x) \in V \mid p(1) = 0\}.

Verifying W is a subspace:

  1. Contains the zero polynomial: The zero polynomial p(x) = 0 satisfies p(1) = 0 , so p(x) \in W .
  2. Closed under addition: If p(x), q(x) \in W , then p(1) = 0 and q(1) = 0 . For (p + q)(x) = p(x) + q(x) :
    (p + q)(1) = p(1) + q(1) = 0 + 0 = 0.
    Thus, p(x) + q(x) \in W .
  3. Closed under scalar multiplication: If p(x) \in W and c \in \mathbb{R} , then (cp)(1) = c \cdot p(1) = c \cdot 0 = 0 . Thus, cp(x) \in W .

Since all conditions are satisfied, W is a subspace of V .

Non-Examples:

A plane in \mathbb{R}^3 that does not pass through the origin is not a subspace.
A set of vectors that does not include the zero vector is not a subspace because it fails to meet one of the fundamental requirements for being a subspace.
For example: Consider a vector space \mathbb{R}^2 . Let’s say we have a set W consisting of all vectors of the form (x, y) where x and y are real numbers, but the zero vector (0, 0) is excluded.

  • If \mathbf{u} = (1, 2) is in W , then -\mathbf{u} = (-1, -2) must also be in W , and their sum \mathbf{u} + (-\mathbf{u}) = (0, 0) must also be in W .
  • Since (0, 0) is not in W , W fails to be a subspace.

Without the zero vector, the set cannot satisfy these necessary conditions, thereby disqualifying it as a subspace.

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