The Span and Basis of a Subspace

Section 4.2 The Span and Basis of a Subspace

Objectives

Define a subspace of a vector space.

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Show that a subset is a subspace.

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Define the span of a set of vectors in a vector space.

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Show that a set of vectors spans a subspace.

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Define the basis of a vector space.

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Find the representation of a vector with respect to a basis.

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In Section 2.5, we saw the span of vectors in $\mathbb{R}^n\text{.}$ We now extend this example to the span of any subspace. In addition, the notion of a basis of the subspace is introduced.

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Subsection 4.2.1 Linear Independence

In Section 2.6, the linear independence of vectors in $\mathbb{R}^n$ were defined and many examples shown. We extend this notion to vectors in a general vector space.

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Definition 4.2.1.

Let $S = \left\{\boldsymbol{v}_1, \boldsymbol{v}_2, \ldots \boldsymbol{v}_n\right\}$ be a set of vectors in a vectors space $V\text{.}$ The set $S$ is linear independent if for $c_1, c_2, \ldots, c_n \in \mathbb{R}\text{,}$ the equation $c_1 \boldsymbol{v}_1 + c_2 \boldsymbol{v}_2 + \cdots + c_n \boldsymbol{v}_n = 0$ implies that $c_1 = c_2 = \cdots = c_n = 0\text{.}$ If not, then the set is linearly dependent.

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Example 4.2.2.

Show that the set

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\begin{equation*} S = \left\{ 1, x, x^2 \right\} \end{equation*}

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is linearly independent in ${\cal P}_2\text{.}$

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Solution.

To show this, we need to show that if

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\begin{equation*} c_1 (1) + c_2 (x) + c_3 (x^2) = 0 \end{equation*}

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then $c_1 = c_2 = c_3 = 0\text{.}$ Note that the zero on the right hand side is the zero polynomial. Thus, equating coefficients, we get the system

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\begin{align*} c_1 \amp = 0 \\ c_2 \amp = 0 \\ c_3 \amp = 0 \end{align*}

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which has the only solution of $c_1 = c_2 = c_3 = 0\text{,}$ thus the set is linearly independent.

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Example 4.2.3.

Are the matrices in the set

\begin{equation*} \left\{ \begin{bmatrix} 1 \amp 1 \\ 1 \amp 1 \end{bmatrix}, \begin{bmatrix} 1 \amp -1 \\ 1 \amp -1 \end{bmatrix}, \begin{bmatrix} 2 \amp 2 \\ -2 \amp -2 \end{bmatrix} \right\} \end{equation*}

linearly independent or dependent in ${\cal M}_{2 \times 2}\text{?}$

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Solution.

To determine this, we need to solve

\begin{equation*} c_1 \begin{bmatrix} 1 \amp 1 \\ 1 \amp 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \amp -1 \\ 1 \amp -1 \end{bmatrix} + c_3 \begin{bmatrix} 2 \amp 2 \\ -2 \amp -2 \end{bmatrix} = \begin{bmatrix} 0 \amp 0 \\ 0 \amp 0 \end{bmatrix} \end{equation*}

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The solution of this is by equating terms or the system:

\begin{equation*} \begin{aligned} c_1 + c_2 + 2c_3 \amp = 0 \\ c_1 - c_2 + 2c_3 \amp = 0 \\ c_1 + c_2 - 2c_3 \amp = 0 \\ c_1 - c_2 - 2c_3 \amp = 0 \end{aligned} \end{equation*}

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This system has the unique solution $c_1 = c_2 = c_3 = 0\text{,}$ thus the set is linearly independent.

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Subsection 4.2.2 Subspaces

Recall that it is important to understand if a set of vectors is a span of a set or space, because we are able to take a linear combination of the vectors to get any other vector in the set. Since we have extended the notion of a vector space, the span will play the same role.

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Definition 4.2.4.

Let $U$ be a subset of a vector space $V\text{.}$ If $U$ is also a vector space, then $U$ is a subspace.

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The next three examples show that we have already seen subspaces because we know subsets of known vector spaces that are vector spaces themselves.

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Example 4.2.5.

We showed in Example 4.1.3 that the set of all lines in $\mathbb{R}^2$ that pass through the origin is a vector space. Since the set is a subset of $\mathbb{R}^2\text{,}$ it is a subspace of $\mathbb{R}^2$ as well.

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Example 4.2.6.

Show that $\mathbb{R}^2$ is a subspace of $\mathbb{R}^3\text{.}$

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Solution.

Since $\mathbb{R}^2$ is itself a vector space and a subset of $\mathbb{R}^3\text{,}$ then $\mathbb{R}^2$ is a subspace.

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Example 4.2.7.

Recall that the set

\begin{equation*} {\cal P}_2=\{ a_0 + a_1 x + a_2 x^2\; | \; a_0, a_1, a_2 \in \mathbb{R} \} \end{equation*}

is the set of all quadratic functions.

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The set ${\cal P}_1 = \{a_0 + a_1 x \; | \; a_0, a_1 \in \mathbb{R} \}$ of all linear functions is itself a vector space as well as a subset of ${\cal P}_2\text{,}$ therefore ${\cal P}_1$ is a subspace of ${\cal P}_2\text{.}$

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In addition, the set $\{ a x^2\; | \; a \in \mathbb{R}\}$ is a vector space as well as a subset of ${\cal P}_2\text{,}$ therefore it is a subspace.

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The above examples show that there are many already known subspaces. There are many cases though that aren’t evident or to show it is a subspace, we would need to prove all 10 properties that it is a vector space. The next lemma, however, shows that isn’t the case.

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Lemma 4.2.8.

Let $S$ be a nonempty subset $S$ of a vector space $V\text{,}$ under the inherited operations of vector addition and scalar multiplication. If for all $\boldsymbol{s}_1, \boldsymbol{s}_2 \in S$ and $r_1, r_2 \in \mathbb{R}\text{,}$

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\begin{equation*} r_1 \boldsymbol{s}_1 + r_2 \boldsymbol{s}_2 \in S \end{equation*}

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then $S$ is a subspace.

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Proof.

This means that if $S$ is a subset of $V\text{,}$ a vector space, to prove that $S$ is a subspace, we only need to check if $r_1 \boldsymbol{s}_1 + r_2 \boldsymbol{s}_2 \in S\text{.}$

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Since $S$ is a subspace of $V\text{,}$ properties (2), (3), (5) and (7)-(10) of Definition 4.1.1 hold for $S\text{.}$ Thus we only need to prove closure under addition and scalar multiplication as well as the existence of the identity element.

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Property 1: Because $r_1 \boldsymbol{s}_1 + r_2 \boldsymbol{s}_2 \in S\text{,}$ let $r_1=r_2=1\text{,}$ thus $\boldsymbol{s}_1+\boldsymbol{s}_2 \in S\text{.}$

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Property 4: If $r_1=0$ and $r_2=0\text{,}$ then this shows that $\boldsymbol{0} \in S\text{,}$ so there is an identity element.

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Property 6: Because $r_1 \boldsymbol{s}_1 + r_2 \boldsymbol{s}_2 \in S\text{,}$ let $r_2=0\text{,}$ thus $r_1 \boldsymbol{s}_1 \in S\text{.}$

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We will use this definition to prove that certain sets are subspaces.

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Example 4.2.9.

Show that

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\begin{equation*} V = \left\{ \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \; | \; v_2 = k v_1 \right\} \end{equation*}

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(that is, all vectors on a line of slope $k$) is a subspace of $\mathbb{R}^2\text{.}$

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Solution.

We will use Lemma 4.2.8. Let

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\begin{align*} \boldsymbol{u} \amp = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \amp \boldsymbol{v} \amp = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \end{align*}

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be elements of $V\text{.}$ That is $v_2 = k v_1$ and $u_2 = ku_1\text{.}$ Then

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\begin{align*} r_1 \boldsymbol{u} + r_2 \boldsymbol{v} \amp = r_1 \begin{bmatrix} u_1 \\ k u_1 \end{bmatrix} + r_2 \begin{bmatrix} v_1 \\ k v_1 \end{bmatrix} = \begin{bmatrix} r_1 u_1 + r_2 v_1 \\ r_1 k u_1 + r_2 k v_1 \end{bmatrix} \\ \amp = \begin{bmatrix} r_1 u_1 + r_2 v_1 \\ k (r_1 u_1 + r_2 v_1) \end{bmatrix} \end{align*}

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which is an element of $V$ because the second component is $k$ times the first one. Thus $V$ is a subspace of $\mathbb{R}^2\text{.}$

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Example 4.2.10.

Show using Lemma 4.2.8 that

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\begin{equation*} V =\left\{ \begin{bmatrix} a \amp 0 \\ 0 \amp b \end{bmatrix} \; | \; a,b \in \mathbb{R} \right\} \end{equation*}

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the set of all diagonal matrices is a subspace of ${\cal M}_{2 \times 2}\text{,}$ the vector space of all $2 \times 2$ matrices.

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Solution.

In this case, if we show that for any two matrices

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\begin{align*} A \amp = \begin{bmatrix} a_1 \amp 0 \\ 0 \amp b_1 \end{bmatrix}, \amp B \amp = \begin{bmatrix} a_2 \amp 0 \\ 0 \amp b_2 \end{bmatrix} \end{align*}

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and scalars $r_1, r_2 \in \mathbb{R}$ that $r_1 A + r_2 B$ is in the set.

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\begin{align*} r_1 A + r_2 B \amp = r_1 \begin{bmatrix} a_1 \amp 0 \\ 0 \amp b_1 \end{bmatrix}+ r_2 \begin{bmatrix} a_2 \amp 0 \\ 0 \amp b_2 \end{bmatrix} \\ \amp = \begin{bmatrix} r_1 a_1 + r_2 a_2 \amp 0 \\ 0 \amp r_1 b_1 + r_2 b_2 \end{bmatrix} \end{align*}

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which is a diagonal matrix, therefore in $V\text{,}$ thus this is a subspace.

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In Section 2.7, we explored the null space of a matrix. In the next lemma, we show that any null space is a subspace.

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Lemma 4.2.11.

Let $A$ be an $m$ by $n$ matrix. The null space of $A$ is a subspace of $\mathbb{R}^n\text{.}$

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Proof.

We will use Lemma 4.2.8 to solve this. Let both $\boldsymbol{x}$ and $\boldsymbol{y}$ be in the null space of $A\text{.}$ This means that $A\boldsymbol{x}=\boldsymbol{0}$ and $A\boldsymbol{y}=\boldsymbol{0}\text{.}$ We need to show that $r_1 \boldsymbol{x} + r_2 \boldsymbol{y}$ is in the null space of $A\text{.}$

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\begin{align*} A(r_1 \boldsymbol{x} + r_2 \boldsymbol{y}) \amp = r_1 A\boldsymbol{x} + r_2 A\boldsymbol{y} \\ \amp = r_1 (\boldsymbol{0}) + r_2 (\boldsymbol{0}) = \boldsymbol{0} \end{align*}

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Vectors in the null space are vectors of length $n\text{,}$ so the null space is a subset of $\mathbb{R}^n$ and since $r_1 \boldsymbol{x} + r_2 \boldsymbol{y}$ is in the null space of $A\text{,}$ then the null space is a subspace of $\mathbb{R}^n\text{.}$

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This is an important result that we will see in eigenvalues in the next chapter.

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Subsection 4.2.3 The Span of a set of vectors

We saw in Section 2.5 the span of a set of vectors in $\mathbb{R}^n\text{.}$ We now generalize this to any vector space.

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Definition 4.2.12.

The span of a nonempty subset $S=\{\boldsymbol{s}_1, \boldsymbol{s}_2, \ldots, \boldsymbol{s}_n\}$ of a vector space is the set of all linear combinations of the vectors in $S\text{.}$ That is,

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\begin{equation*} \text{span}(S) = \{ c_1 \boldsymbol{s}_1 + c_2 \boldsymbol{s}_2 + \cdots + c_n \boldsymbol{s}_n\; | \; \text{$c_1, c_2, \ldots, c_n \in \mathbb{R}, \boldsymbol{s}_1, \boldsymbol{s}_2, \ldots, \boldsymbol{s}_n \in S$} \}. \end{equation*}

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To show that a subset of vectors span a subspace $S\text{,}$ we need to show that any vector in $S$ can be written as a linear combination of the spanning vectors.

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Example 4.2.13.

Show that the set $\{2+x,1,x+x^2\}$ spans $\mathcal{P}_2\text{.}$

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Solution.

In this case, we need show that a general polynomial in $\mathcal{P}_2$ can be written as a linear combination of elements of the given set. That is

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\begin{equation*} c_1 (2+x) + c_2 (1) + c_3 (x+x^2) = a_0 + a_1 x + a_2 x^2 \end{equation*}

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and if there is a solution for the $c$’s, then that shows the the set spans $\mathcal{P}_2\text{.}$ To find the solution, use the technique of equating coefficients. Write down the coefficients for the constant terms, $x$ terms and $x^2$ terms respectively.

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\begin{align*} 2 c_1 + c_2 \amp = a_0 \\ c_1 + c_3 \amp = a_1 \\ c_3 \amp = a_2 \end{align*}

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This has a solution $c_3=a_2, c_1 = a_1-a_2$ and $c_2 = a_0 - 2(a_1-a_2)\text{,}$ which means that a linear combination of the three “vectors” can form any quadratic function, thus the given set spans $\mathcal{P}_2\text{.}$

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Lemma 4.2.14.

The span of any subset of a vector space is a subspace.

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Proof.

Let $S$ be the subset and $\boldsymbol{s}_1, \boldsymbol{s}_2, \ldots, \boldsymbol{s}_n$ be the elements of $S\text{.}$ Using Lemma 4.2.8, we need to check that $\text{span}(S)$ is closed under linear combinations. Let

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\begin{align*} \boldsymbol{v} \amp = c_1 \boldsymbol{s}_1 + c_2 \boldsymbol{s}_2 + \cdots + c_n \boldsymbol{s}_n, \\ \boldsymbol{w} \amp = k_1 \boldsymbol{s}_1 + k_2 \boldsymbol{s}_2 + \cdots + k_n \boldsymbol{s}_n \end{align*}

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then

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\begin{align*} r_1 \boldsymbol{v} + r_2 \boldsymbol{w} \amp = r_1 (c_1 \boldsymbol{s}_1 + c_2 \boldsymbol{s}_2 + \cdots + c_n \boldsymbol{s}_n) + r_2 (k_1 \boldsymbol{s}_1 + k_2 \boldsymbol{s}_2 + \cdots + k_n \boldsymbol{s}_n) \\ \amp = (r_1 c_1 + r_2 k_1) \boldsymbol{s}_1 + (r_1c_2 + r_2 k_2) \boldsymbol{s}_2 + \cdots + (r_1 c_n + r_2 k_n) \boldsymbol{s}_n \end{align*}

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Since this shows that $r_1 \boldsymbol{v} + r_2 \boldsymbol{w}$ is in $S\text{,}$ then $S$ is a subspace.

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This lemma allows us to talk about a vector space in terms of the vectors that span it. For example, instead of thinking of $\mathcal{P}_2\text{,}$we think of the span of $\{2+x,1,x+x^2\}$ (in this case, it may not be more helpful, but other cases it is).

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Example 4.2.15.

Show that the following vectors span $\mathbb{R}^3\text{:}$

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\begin{align*} \boldsymbol{e}_1 \amp = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \amp \boldsymbol{e}_2 \amp = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \amp \boldsymbol{e}_3 \amp = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \end{align*}

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Solution.

Because the vector $\boldsymbol{x}=[x_1,x_2,x_3]^{\intercal}$ can be written $\boldsymbol{x}=x_1 \boldsymbol{e}_1 + x_2 \boldsymbol{e}_2 + x_3 \boldsymbol{e}_3\text{,}$ then these vectors span $\mathbb{R}^3\text{.}$

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