Kuzee
Vector Algebra and Analytic Geometry of Space

Subspaces

The Definition of a Subspace

Subspace

Let VV be a vector space and let WW be a nonempty subset of VV. If WW also forms a vector space with respect to the addition and scalar multiplication of VV, then WW is called a subspace of VV.

An Equivalent Definition

An equivalent definition of a subspace

WW is a subspace of VV if and only if WW satisfies:

  1. Closure under addition: for any a⃗,b⃗∈W\vec{a}, \vec{b} \in W, we have a⃗+b⃗∈W\vec{a} + \vec{b} \in W
  2. Closure under scalar multiplication: for any a⃗∈W\vec{a} \in W and k∈Rk \in \mathbb{R}, we have ka⃗∈Wk\vec{a} \in W

Examples of Subspaces

The Trivial Subspaces

  1. The zero subspace: {0⃗}\{ \vec{0} \}, containing only the zero vector
  2. The whole space: VV itself

Geometric Subspaces

In R3\mathbb{R}^3:

  • a straight line: one passing through the origin
  • a plane: one passing through the origin
  • the origin: the zero subspace

Algebraic Subspaces

A subspace of the matrix space

In Mn×nM_{n \times n}, the set of symmetric matrices:

S={A∈Mn×n∣AT=A}S = \{ A \in M_{n \times n} \mid A^T = A \}

The Spanning Subspace

Spanning subspace

Let SS be a subset of the vector space VV. The set of all the linear combinations containing SS is called the subspace spanned by SS, written ⟨S⟩\langle S \rangle or span(S)\mathrm{span}(S).

Properties of the Spanning Subspace

The form of a spanning subspace

⟨S⟩={∑i=1kcia⃗i∣ci∈R,a⃗i∈S}\langle S \rangle = \left\{ \sum_{i=1}^k c_i \vec{a}_i \mid c_i \in \mathbb{R}, \vec{a}_i \in S \right\}

The Dimension of a Subspace

Theorem1

If WW is a subspace of VV, then:

  1. dim⁡W≤dim⁡V\dim W \leq \dim V
  2. if dim⁡W=dim⁡V\dim W = \dim V, then W=VW = V

The Sum of Two Subspaces

The sum of two subspaces

Let W1,W2W_1, W_2 be subspaces of VV. Then:

W1+W2={w⃗1+w⃗2∣w⃗1∈W1,w⃗2∈W2}W_1 + W_2 = \{ \vec{w}_1 + \vec{w}_2 \mid \vec{w}_1 \in W_1, \vec{w}_2 \in W_2 \}

The dimension formula for a sum of subspaces

dim⁡(W1+W2)=dim⁡W1+dim⁡W2−dim⁡(W1∩W2)\dim(W_1 + W_2) = \dim W_1 + \dim W_2 - \dim(W_1 \cap W_2)

Exercises

Exercise 1

Prove that the set W={(x,y,z)∈R3∣x+y+z=0}W = \{ (x, y, z) \in \mathbb{R}^3 \mid x + y + z = 0 \} is a subspace of R3\mathbb{R}^3.

subspaceverifying closurea linear equation

Approach:

Verify closure under addition and under scalar multiplication.

Steps:

  1. Closure under addition: let a⃗=(x1,y1,z1),b⃗=(x2,y2,z2)∈W\vec{a} = (x_1, y_1, z_1), \vec{b} = (x_2, y_2, z_2) \in W Then x1+y1+z1=0x_1 + y_1 + z_1 = 0 and x2+y2+z2=0x_2 + y_2 + z_2 = 0 a⃗+b⃗=(x1+x2,y1+y2,z1+z2)\vec{a} + \vec{b} = (x_1+x_2, y_1+y_2, z_1+z_2) (x1+x2)+(y1+y2)+(z1+z2)=0+0=0(x_1+x_2) + (y_1+y_2) + (z_1+z_2) = 0 + 0 = 0 so a⃗+b⃗∈W\vec{a} + \vec{b} \in W

  2. Closure under scalar multiplication: let a⃗=(x,y,z)∈W\vec{a} = (x, y, z) \in W and k∈Rk \in \mathbb{R} Then x+y+z=0x + y + z = 0 ka⃗=(kx,ky,kz)k\vec{a} = (kx, ky, kz) kx+ky+kz=k(x+y+z)=k⋅0=0kx + ky + kz = k(x + y + z) = k \cdot 0 = 0 so ka⃗∈Wk\vec{a} \in W

Answer: WW is a subspace of R3\mathbb{R}^3

Exercise 2

Find the subspace spanned by the set of vectors {(1,0,1),(0,1,1)}\{(1,0,1), (0,1,1)\}.

a spanning subspacelinear combinationsa parametric form

Approach:

Find the form of all the linear combinations.

Steps:

  1. Put v⃗=x(1,0,1)+y(0,1,1)=(x,y,x+y)\vec{v} = x(1,0,1) + y(0,1,1) = (x, y, x + y)

  2. The general form of the vector: (x,y,x+y)(x, y, x + y), where x,y∈Rx, y \in \mathbb{R}

  3. This can be written as: (x,y,z)(x, y, z), where z=x+yz = x + y

  4. That is, all the vectors satisfying z=x+yz = x + y

Answer: the spanning subspace is {(x,y,x+y)∣x,y∈R}\{(x, y, x+y) \mid x, y \in \mathbb{R}\}

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
WWSubspacesubspace WA subset of a vector space
⟨S⟩\langle S \rangleSpanning subspacespan of SThe subspace spanned by the set SS
W1+W2W_1 + W_2Sum of subspacessum of subspacesThe sum of two subspaces
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