Skip to main content

Matrix Theory From Eigenvalues to Data Insights

Yilan Tan

Section 4.2 Subspaces of \(\mathbb{R}^n\)

In this section, we introduce the concept of a subspace, which is a fundamental structure in linear algebra. A subspace is a subset of a vector space that is itself a vector space under the same operations.
🔗

Subsection 4.2.1 Definition and Basic Examples

Definition 4.2.1. Subspace.

A subset \(W\) of \(\mathbb{R}^n\) is called a subspace of \(\mathbb{R}^n\) if it satisfies the following three properties:
  1. The zero vector \(\mathbf{0}\) is in \(W\text{.}\)
    🔗
    🔗
  2. \(W\) is closed under vector addition: If \(\mathbf{u}, \mathbf{v} \in W\text{,}\) then \(\mathbf{u} + \mathbf{v} \in W\text{.}\)
    🔗
    🔗
  3. \(W\) is closed under scalar multiplication: If \(\mathbf{u} \in W\) and \(c \in \mathbb{R}\text{,}\) then \(c\mathbf{u} \in W\text{.}\)
    🔗
    🔗
🔗
🔗

Example 4.2.2. Trivial Subspaces.

For any \(n\text{,}\) there are always two trivial subspaces of \(\mathbb{R}^n\text{:}\)
  • The zero subspace \(\{\mathbf{0}\}\) containing only the zero vector.
    🔗
    🔗
  • The entire space \(\mathbb{R}^n\) itself.
    🔗
    🔗
🔗
🔗

Example 4.2.3. Lines Through the Origin.

Any line through the origin in \(\mathbb{R}^n\) is a subspace. For example, the set \(W = \{t\mathbf{v} : t \in \mathbb{R}\}\) where \(\mathbf{v}\) is a fixed nonzero vector in \(\mathbb{R}^n\) forms a subspace.
🔗
Solution.
We verify the three properties:
  1. When \(t = 0\text{,}\) we have \(\mathbf{0} = 0\mathbf{v} \in W\text{.}\)
    🔗
    🔗
  2. If \(t_1\mathbf{v}, t_2\mathbf{v} \in W\text{,}\) then \(t_1\mathbf{v} + t_2\mathbf{v} = (t_1 + t_2)\mathbf{v} \in W\text{.}\)
    🔗
    🔗
  3. If \(t\mathbf{v} \in W\) and \(c \in \mathbb{R}\text{,}\) then \(c(t\mathbf{v}) = (ct)\mathbf{v} \in W\text{.}\)
    🔗
    🔗
🔗
🔗
🔗
🔗

Subsection 4.2.2 The Null Space as a Subspace

Proof.

We need to verify the three properties of a subspace:
🔗
  1. The zero vector is in \(W\text{:}\) We have \(A\mathbf{0} = \mathbf{0}\text{,}\) so \(\mathbf{0} \in W\text{.}\)
    🔗
    🔗
  2. Closed under addition: Suppose \(\mathbf{u}, \mathbf{v} \in W\text{.}\) Then \(A\mathbf{u} = \mathbf{0}\) and \(A\mathbf{v} = \mathbf{0}\text{.}\) We need to show that \(\mathbf{u} + \mathbf{v} \in W\text{,}\) that is, \(A(\mathbf{u} + \mathbf{v}) = \mathbf{0}\text{.}\)
    \begin{align*} A(\mathbf{u} + \mathbf{v}) \amp = A\mathbf{u} + A\mathbf{v}\\ \amp = \mathbf{0} + \mathbf{0}\\ \amp = \mathbf{0} \end{align*}
    Therefore, \(\mathbf{u} + \mathbf{v} \in W\text{.}\)
    🔗
    🔗
  3. Closed under scalar multiplication: Suppose \(\mathbf{u} \in W\) and \(c \in \mathbb{R}\text{.}\) Then \(A\mathbf{u} = \mathbf{0}\text{.}\) We need to show that \(c\mathbf{u} \in W\text{,}\) that is, \(A(c\mathbf{u}) = \mathbf{0}\text{.}\)
    \begin{align*} A(c\mathbf{u}) \amp = c(A\mathbf{u})\\ \amp = c\mathbf{0}\\ \amp = \mathbf{0} \end{align*}
    Therefore, \(c\mathbf{u} \in W\text{.}\)
    🔗
    🔗
🔗
Since all three properties are satisfied, \(W\) is a subspace of \(\mathbb{R}^n\text{.}\)
🔗
🔗

Remark 4.2.5.

This theorem tells us that the solution set to any homogeneous linear system \(A\mathbf{x} = \mathbf{0}\) always forms a subspace. This is in contrast to non-homogeneous systems \(A\mathbf{x} = \mathbf{b}\) where \(\mathbf{b} \neq \mathbf{0}\text{,}\) whose solution sets (if non-empty) do not pass through the origin and therefore are not subspaces.
🔗
🔗

Example 4.2.6. Computing a Null Space.

Find the null space of the matrix
\begin{equation*} A = \begin{bmatrix} 1 \amp 2 \amp 3 \\ 2 \amp 4 \amp 6 \end{bmatrix} \end{equation*}
🔗
Solution.
We need to solve \(A\mathbf{x} = \mathbf{0}\text{.}\) Setting up the augmented matrix and row reducing:
\begin{equation*} \left[\begin{array}{ccc|c} 1 \amp 2 \amp 3 \amp 0 \\ 2 \amp 4 \amp 6 \amp 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 \amp 2 \amp 3 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \end{array}\right] \end{equation*}
From the reduced form, we have \(x_1 + 2x_2 + 3x_3 = 0\text{,}\) so \(x_1 = -2x_2 - 3x_3\text{.}\) The variables \(x_2\) and \(x_3\) are free. Setting \(x_2 = s\) and \(x_3 = t\text{,}\) we get:
\begin{equation*} \mathbf{x} = \begin{bmatrix} -2s - 3t \\ s \\ t \end{bmatrix} = s\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \end{equation*}
Therefore, the null space is
\begin{equation*} \operatorname{Null}(A) = \operatorname{Span}\left\{\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\right\} \end{equation*}
which is a 2-dimensional subspace (a plane through the origin) in \(\mathbb{R}^3\text{.}\)
🔗
🔗
🔗
🔗
🔗
PrevTopNext