Lecture 27 - Subspaces of \( \mathbb R^n \)

Learning Objectives

• Use set-builder notation to construct example elements of a given set
• Explain why a given set does (or does not) contain the zero vector
• Prove that a set is closed under vector addition, or give a counterexample to show that it is not
• Prove that a set is closed under scalar multiplication, or give a counterexample to show that it is not

Set-Builder Notation

In this lecture, we will be describing sets of vectors, which are subsets of \( \mathbb R^n \). We do this in two primary ways:

• Form Description. This describes the form that vectors in the set must take. It is easy to generate elements using this description, but harder to check whether a vector fits this form.
• Property Description. This describes a property that vectors must have to be in the set. It is easy to check whether a vector is in the set, but harder to generate specific elements.

Example 1 (Form Description). Let \( H = \left\{ \vectwo a {2a} : a \in \mathbb R \right\} \). List five different elements of \( H \).

This notation tells us that \( H \) is a subset of \( \mathbb R^2\) containing the vectors that have the form \( \vectwo a {2a} \), where \( a \) is any real number. To generate elements, we choose any real number for \( a \): \[ \vectwo 1 2, \vectwo 5 {10}, \vectwo {-2.5} {-5}, \vectwo {3\sqrt 7}{6\sqrt 7}, \vectwo \pi {2\pi}, \ldots \ \Box \]

Example 2 (Property Description). Let \( K = \left\{ \vecthree a b c : ab\gt c \right\} \). List three different elements of \( K \).

This notation tells us that \( K \) is a subset of \( \mathbb R^3 \) containing the vectors that have the property that the product of the first two entries is greater than the third entry. Due to the way \( K\) is defined, it's harder generate elements directly. Instead, we can just pick values for \( a, b, \) and \( c\) and check whether the property holds:

• \( \vecthree 1 2 3 \): Since \( 1\cdot 2 \not\gt 3 \), this vector is not in \( K \).
• \( \vecthree 2 4 6 \): Since \( 2\cdot 4 \gt 6 \), this vector is in \( K \).
• \( \vecthree 0 5 7 \): Since \( 0\cdot 5 \not\gt 7 \), this vector is not in \( K \).
• \( \vecthree {-1} {-2} {-3} \): Since \( (-1)\cdot (-2) \gt (-3) \), this vector is in \( K \).
• \( \vecthree 1 0 {-1} \): Since \( 1\cdot 0 \gt -1 \), this vector is in \( K \).

Subspaces

Certain sets of vectors form what are called subspaces. These sets are important to our study of linear algebra, and arise naturally from matrices, as we will see.

Definition. A subset \( H \) of \( \mathbb R^n \) is a subspace if it has these three properties:

1. Zero Vector. The zero vector \( \bbm 0 \) is in \( H \)
2. Closed Under Addition. If \( \bbm u, \bbm v \in H \), then \( \bbm u+\bbm v \in H \)
3. Closed Under Scalar Multiplication. If \( \bbm u \in H \) and \( c \in \mathbb R \), then \( c\ \bbm u \in H \)

Given a set of vectors, the way we demonstrate that this set is a subspace of \( \mathbb R^n \) is to write a short proof that the set satisfies each of these three properties.

Example 3. Let \( H = \left\{ \vectwo a {2a} : a \in \mathbb R \right\} \). Prove that \( H \) is a subspace of \( \mathbb R^2 \).

1. Zero Vector. If we let \( a = 0\), then \( \vectwo a {2a} = \vectwo 0 0 \in H \).
2. Closed Under Addition. Let \( \bbm u, \bbm v \in H \). By the definition of \( H \), we have \( \bbm u = \vectwo a {2a} \) and \( \bbm v = \vectwo b {2b} \) for some scalars \( a, b \in \mathbb R \). Now, \[ \bbm u+\bbm v = \vectwo a {2a} + \vectwo b {2b} = \vectwo {a+b} {2a+2b} = \vectwo {a+b} {2(a+b)} \in H. \]
3. Closed Under Scalar Multiplication. Let \( \bbm u \in H \) and let \( c\in \mathbb R \). Then \( \bbm u = \vectwo a {2a} \) for some scalar \( a \in \mathbb R \) and \[ c\ \bbm u = c\vectwo a {2a} = \vectwo {ca} {c(2a)} = \vectwo {ca} {2(ca)} \in H.\ \Box \]

If we want to demonstrate that a subset of \( \mathbb R^n \) is not a subspace, we need only find a single counterexample to any one of the three subspace properties.

Example 4. Let \( K = \left\{ \vecthree a b c : ab\gt c \right\} \). Prove that \( K \) is not a subspace of \( \mathbb R^3 \).

We only need a single counterexample, so any one of the following would suffice to answer this question:

1. Zero Vector. The zero vector has \( a=b=c=0 \), and \( 0\cdot 0 \not\gt 0 \), so \( \bbm 0 \notin K \). \( \Box \)
2. Closed Under Addition. Let \( \bbm u = \vecthree 2 4 6 \) and \( \bbm v = \vecthree {-1} {-2} {-3} \). Since \( 2\cdot 4 \gt 6 \) and \( (-1)(-2)\gt -3 \), we have \( \bbm u, \bbm v \in K \). However, \( \bbm u + \bbm v = \vecthree 1 2 3 \) which is not in \( K \) since \( 1\cdot 2 \not \gt 3 \). Thus, \( K \) is not closed under addition. \( \Box \)
3. Closed Under Scalar Multiplication. Let \( \bbm u = \vecthree 1 2 {-3} \). Since \( 1\cdot 2 \gt -3 \), we have \( \bbm u \in K \). Let \( c = -1\). Now, \( c\ \bbm u = \vecthree {-1} {-2} 3 \), and this vector is not in \( K \) since \( (-1)(-2) \not\gt 3 \). Thus, \( K \) is not closed under scalar multiplication. \( \Box \)

A Subspace Spanned By a Set

Recall from Lecture 8 that the span of a set of vectors is the set of all linear combinations of those vectors. Specifically, if \( \bbm v_1, \bbm v_2, \ldots, \bbm v_p \in \mathbb R^n \), then \[ {\rm Span} \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} = \{ c_1 \bbm v_1 + c_2 \bbm v_2 + \cdots + c_p \bbm v_p : c_1, c_2, \ldots, c_p \in \mathbb R \}. \]

We can now revisit the span of a set of vectors and say that the span of any set of vectors in \( \mathbb R^n \) is a subspace of \( \mathbb R^n \).

Theorem (Subspace Spanned By a Set). Let \( \bbm v_1, \bbm v_2, \ldots, \bbm v_p \in \mathbb R^n \). Then \( {\rm Span} \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \) is a subspace of \( \mathbb R^n \).

Proof. Let \( S = {\rm Span} \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_p \} \). We prove that \( S \) satisfies the three subspace properties.

1. Zero Vector. If we set \( c_1=c_2=\cdots=c_p = 0\), we see that \( \bbm 0 \in S \).
2. Closed Under Addition. Let \( \bbm u, \bbm v \in S \). Then \( \bbm u = c_1 \bbm v_1 + c_2 \bbm v_2 + \cdots + c_p \bbm v_p \) and \( \bbm v = d_1 \bbm v_1 + d_2 \bbm v_2 + \cdots + d_p \bbm v_p \) for some scalars \( c_i, d_i \in \mathbb R \). Now, \[ \bbm u + \bbm v = (c_1+d_1)\bbm v_1 + (c_2+d_2)\bbm v_2 + \cdots + (c_p+d_p)\bbm v_p \in S. \]
3. Closed Under Scalar Multiplication. Let \( \bbm u \in S \) and let \( r \in \mathbb R \) be a scalar. Then \( \bbm u = c_1 \bbm v_1 + c_2 \bbm v_2 + \cdots + c_p \bbm v_p \) for some scalars \( c_i \in \mathbb R \). Now, \[ r \bbm u = (rc_1)\bbm v_1 + (rc_2)\bbm v_2 + \cdots + (rc_p)\bbm v_p \in S.\ \Box \]

Important (But Boring) Subspaces

There are two more important subspaces of \( \mathbb R^n \) to discuss: the set \( \{ \bbm 0 \} \) containing only the zero vector, and the set \( \mathbb R^n \) containing every vector in \( \mathbb R^n \). These two sets satisfy the three subspace properties (check for yourself!) and can be considered "extreme" cases. The subspace \( \{ \bbm 0 \} \) is the smallest possible subspace of \( \mathbb R^n \), and is contained in every other subspace (since every subspace contains the zero vector). The subspace \( \mathbb R^n \) is the largest possible subspace and contains all other subspaces (since it contains every vector).

In Lecture 28 and Lecture 29, we will study two subspaces associated with a matrix, and we will be particularly interested in the cases where these subspaces are \( \{ \bbm 0 \} \) or \( \mathbb R^n \).