Vector Spaces and Subspaces

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Vector spaces abstract the arithmetic properties of addition and scalar multiplication of vectors. In $\R ^n$ we know how to add vectors and to multiply vectors by scalars. Indeed, it is straightforward to verify that each of the eight properties listed in Table ?? is valid for vectors in $V=\R ^n$ . Remarkably, sets that satisfy these eight properties have much in common with $\R ^n$ . So we define:

Let $V$ be a set having the two operations of addition and scalar multiplication. Then $V$ is a vector space if the eight properties listed in Table ?? hold. The elements of a vector space are called vectors.

The vector $0$ mentioned in (A3) in Table ?? is called the zero vector.

Table 1: Properties of Vector Spaces: suppose $u,v,w\in V$ and $r,s\in \R$ .


(A1)	Addition is commutative	$v+w=w+v$

(A2)	Addition is associative	$(u+v)+w = u+(v+w)$

(A3)	Additive identity $0$ exists	$v+0=v$

(A4)	Additive inverse $-v$ exists	$v+(-v) = 0$

(M1)	Multiplication is associative	$(rs)v = r(sv)$

(M2)	Multiplicative identity exists	$1v=v$

(D1)	Distributive law for scalars	$(r+s)v = rv+sv$

(D2)	Distributive law for vectors	$r(v+w) = rv+rw$

When we say that a vector space $V$ has the two operations of addition and scalar multiplication we mean that the sum of two vectors in $V$ is again a vector in $V$ and the scalar product of a vector with a number is again a vector in $V$ . These two properties are called closure under addition and closure under scalar multiplication.

In this discussion we focus on just two types of vector spaces: $\R ^n$ and function spaces. The reason that we make this choice is that solutions to linear equations are vectors in $\R ^n$ while solutions to linear systems of differential equations are vectors of functions.

An Example of a Function Space

For example, let $\cal F$ denote the set of all functions $f:\R \to \R$ . Note that functions like $f_1(t)=t^2-2t+7$ and $f_2(t)=\sin t$ are in $\cal F$ since they are defined for all real numbers $t$ , but that functions like $g_1(t)=\frac {1}{t}$ and $g_2(t)=\tan t$ are not in $\cal F$ since they are not defined for all $t$ .

We can add two functions $f$ and $g$ by defining the function $f+g$ to be: $(f+g)(t) = f(t) + g(t).$ We can also multiply a function $f$ by a scalar $c\in \R$ by defining the function $cf$ to be: $(cf)(t) = cf(t).$ With these operations of addition and scalar multiplication, $\cal F$ is a vector space; that is, $\cal F$ satisfies the eight vector space properties in Table ??. More precisely:

Define the zero function $\cal O$ by ${\cal O}(t) = 0 \quad \mbox {for all $t\in \R $.}$ For every $x$ in $\cal F$ the function $\cal O$ satisfies: $(x+{\cal O})(t) = x(t) + {\cal O}(t) = x(t) + 0 = x(t).$ Therefore, $x+{\cal O}=x$ and $\cal O$ is the additive identity in $\cal F$ .
Let $x$ be a function in $\cal F$ and define $y(t)=-x(t)$ . Then $y$ is also a function in $\cal F$ , and $(x+y)(t) = x(t) + y(t) = x(t) + (-x(t)) = 0 = {\cal O}(t).$ Thus, $x$ has the additive inverse $-x$ .

After these comments it is straightforward to verify that the remaining six properties in Table ?? are satisfied by functions in $\cal F$ .

Sets that are not Vector Spaces

It is worth considering how closure under vector addition and scalar multiplication can fail. Consider the following three examples.

(i): Let $V_1$ be the set that consists of just the $x$ and $y$ axes in the plane. Since $(1,0)$ and $(0,1)$ are in $V_1$ but $(1,0) + (0,1) = (1,1)$ is not in $V_1$ , we see that $V_1$ is not closed under vector addition. On the other hand, $V_1$ is closed under scalar multiplication.
(ii): Let $V_2$ be the set of all vectors $(k,\ell )\in \R ^2$ where $k$ and $\ell$ are integers. The set $V_2$ is closed under addition but not under scalar multiplication since $\half (1,0)=(\half ,0)$ is not in $V_2$ .
(iii): Let $V_3=[1,2]$ be the closed interval in $\R$ . The set $V_3$ is neither closed under addition ( $1+1.5=2.5\not \in V_3$ ) nor under scalar multiplication ( $4\cdot 1.5 = 6\not \in V_3$ ). Hence the set $V_3$ is not closed under vector addition and not closed under scalar multiplication.

Subspaces

Let $V$ be a vector space. A nonempty subset $W\subset V$ is a subspace if $W$ is a vector space using the operations of addition and scalar multiplication defined on $V$ .

Note that in order for a subset $W$ of a vector space $V$ to be a subspace it must be closed under addition and closed under scalar multiplication. That is, suppose $w_1,w_2\in W$ and $r\in \R$ . Then

$w_1+w_2 \in W$ , and
$rw_1\in W$ .

The $x$ -axis and the $xz$ -plane are examples of subsets of $\R ^3$ that are closed under addition and closed under scalar multiplication. Every vector on the $x$ -axis has the form $(a,0,0)\in \R ^3$ . The sum of two vectors $(a,0,0)$ and $(b,0,0)$ on the $x$ -axis is $(a+b,0,0)$ which is also on the $x$ -axis. The $x$ -axis is also closed under scalar multiplication as $r(a,0,0)=(ra,0,0)$ , and the $x$ -axis is a subspace of $\R ^3$ . Similarly, every vector in the $xz$ -plane in $\R ^3$ has the form $(a_1,0,a_3)$ . As in the case of the $x$ -axis, it is easy to verify that this set of vectors is closed under addition and scalar multiplication. Thus, the $xz$ -plane is also a subspace of $\R ^3$ .

In Theorem ?? we show that every subset of a vector space that is closed under addition and scalar multiplication is a subspace. To verify this statement, we need the following lemma in which some special notation is used. Typically, we use the same notation $0$ to denote the real number zero and the zero vector. In the following lemma it is convenient to distinguish the two different uses of $0$ , and we write the zero vector in boldface.

Let $V$ be a vector space, and let ${\bf 0}\in V$ be the zero vector. Then $0v={\bf 0} \quad \mbox {and}\quad (-1)v=-v$ for every vector in $v\in V$ .

Proof

Let $v$ be a vector in $V$ and use (D1) to compute $0v + 0v = (0+0)v = 0v.$ By (A4) the vector $0v$ has an additive inverse $-0v$ . Adding $-0v$ to both sides yields $(0v + 0v) + (-0v) = 0v + (-0v) = {\bf 0}.$ Associativity of addition (A2) now implies $0v + (0v + (-0v)) = {\bf 0}.$ A second application of (A4) implies that $0v + {\bf 0} = {\bf 0}$ and (A3) implies that $0v={\bf 0}$ .

Next, we show that the additive inverse $-v$ of a vector $v$ is unique. That is, if $v + a = {\bf 0}$ , then $a=-v$ .

Before beginning the proof, note that commutativity of addition (A1) together with (A3) implies that ${\bf 0}+v=v$ . Similarly, (A1) and (A4) imply that $-v+v={\bf 0}$ .

To prove uniqueness of additive inverses, add $-v$ to both sides of the equation $v + a = {\bf 0}$ yielding $-v + (v + a) = -v + {\bf 0}.$ Properties (A2) and (A3) imply $(-v + v) + a = -v.$ But $(-v + v) + a = {\bf 0} + a = a.$ Therefore $a = -v$ , as claimed.

To verify that $(-1)v = -v$ , we show that $(-1)v$ is the additive inverse of $v$ . Using (M1), (D1), and the fact that $0v = {\bf 0}$ , calculate $v + (-1)v = 1v + (-1)v =(1-1)v = 0v = {\bf 0}.$ Thus, $(-1)v$ is the additive inverse of $v$ and must equal $-v$ , as claimed. $\blacksquare$

Let $W$ be a subset of the vector space $V$ . If $W$ is closed under addition and closed under scalar multiplication, then $W$ is a subspace.

Proof: We have to show that $W$ is a vector space using the operations of addition and scalar multiplication defined on $V$ . That is, we need to verify that the eight properties listed in Table ?? are satisfied. Note that properties (A1), (A2), (M1), (M2), (D1), and (D2) are valid for vectors in $W$ since they are valid for vectors in $V$ .
It remains to verify (A3) and (A4). Let $w\in W$ be any vector. Since $W$ is closed under scalar multiplication, it follows that $0w$ and $(-1)w$ are in $W$ . Lemma ?? states that $0w=0$ and $(-1)w=-w$ ; it follows that $0$ and $-w$ are in $W$ . Hence, properties (A3) and (A4) are valid for vectors in $W$ , since they are valid for vectors in $V$ . $\blacksquare$

Examples of Subspaces of $\R ^n$

Let $V$ be a vector space. Then the subsets $V$ and $\{0\}$ are always subspaces of $V$ . A subspace $W\subset V$ is proper if $W\neq 0$ and $W\neq V$ .
Lines through the origin are subspaces of $\R ^n$ . Let $w\in \R ^n$ be a nonzero vector and let $W=\{rw:r\in \R \}$ . The set $W$ is closed under addition and scalar multiplication and is a subspace of $\R ^n$ by Theorem ??. The subspace $W$ is just a line through the origin in $\R ^n$ , since the vector $rw$ points in the same direction as $w$ when $r>0$ and the exact opposite direction when $r<0$ .
Planes containing the origin are subspaces of $\R ^3$ . To verify this point, let $P$ be a plane through the origin and let $N$ be a vector perpendicular to $P$ . Then $P$ consists of all vectors $v\in \R ^3$ perpendicular to $N$ ; using the dot-product (see Chapter ??, (??)) we recall that such vectors satisfy the linear equation $N\cdot v = 0$ . By superposition, the set of all solutions to this equation is closed under addition and scalar multiplication and is therefore a subspace by Theorem ??.

In a sense that will be made precise all subspaces of $\R ^n$ can be written as the span of a finite number of vectors generalizing Example ??(b) or as solutions to a system of linear equations generalizing Example ??(c).

Examples of Subspaces of the Function Space $\cal F$

Let $\cal P$ be the set of all polynomials in $\cal F$ . The sum of two polynomials is a polynomial and the scalar multiple of a polynomial is a polynomial. Thus, $\cal P$ is closed under addition and scalar multiplication, and $\cal P$ is a subspace of $\cal F$ .

As a second example of a subspace of $\cal F$ , let $\CCone$ be the set of all continuously differentiable functions $u:\R \to \R$ . A function $u$ is in $\CCone$ if $u$ and $u'$ exist and are continuous for all $t\in \R$ . Examples of functions in $\CCone$ are:

(i): Every polynomial $p(t)=a_mt^m+a_{m-1}t^{m-1}+\cdots +a_1t+a_0$ is in $\CCone$ .
(ii): The function $u(t)=e^{\lambda t}$ is in $\CCone$ for each constant $\lambda \in \R$ .
(iii): The trigonometric functions $u(t)=\sin (\lambda t)$ and $v(t)=\cos (\lambda t)$ are in $\CCone$ for each constant $\lambda \in \R$ .
(iv): $u(t)=t^{7/3}$ is twice differentiable everywhere and is in $\CCone$ .

Equally there are many commonly used functions that are not in $\CCone$ . Examples include:

(i): $u(t)=\frac {1}{t-5}$ is neither defined nor continuous at $t=5$ .
(ii): $u(t)=|t|$ is not differentiable (at $t=0$ ).
(iii): $u(t)=\csc (t)$ is neither defined nor continuous at $t=k\pi$ for any integer $k$ .

The subset $\CCone \subset {\cal F}$ is a subspace and hence a vector space. The reason is simple. If $x(t)$ and $y(t)$ are continuously differentiable, then $\frac {d}{dt}(x+y) = \frac {dx}{dt}+\frac {dy}{dt}.$ Hence $x+y$ is differentiable and is in $\CCone$ and $\CCone$ is closed under addition. Similarly, $\CCone$ is closed under scalar multiplication. Let $r\in \R$ and let $x\in \CCone$ . Then $\frac {d}{dt}(rx)(t) = r\frac {dx}{dt}(t).$ Hence $rx$ is differentiable and is in $\CCone$ .

The Vector Space $(\CCone )^n$

Another example of a vector space that combines the features of both $\R ^n$ and $\CCone$ is $(\CCone )^n$ . Vectors $u\in (\CCone )^n$ have the form $u(t) = (u_1(t),\ldots ,u_n(t)),$ where each coordinate function $u_j(t)\in \CCone$ . Addition and scalar multiplication in $(\CCone )^n$ are defined coordinatewise — just like addition and scalar multiplication in $\R ^n$ . That is, let $u,v$ be in $(\CCone )^n$ and let $r$ be in $\R$ , then

$\begin{eqnarray*} (u+v)(t) & = & (u_1(t)+v_1(t),\ldots,u_n(t)+v_n(t)) \\ (ru)(t) & = & (ru_1(t),\ldots,ru_n(t)). \end{eqnarray*}$

The set $(\CCone )^n$ satisfies the eight properties of vector spaces and is a vector space. Solutions to systems of $n$ linear ordinary differential equations are vectors in $(\CCone )^n$ .

Exercises

Verify that the set $V_1$ consisting of all scalar multiples of $(1,-1,-2)$ is a subspace of $\R ^3$ .

Let $V_2$ be the set of all $2\times 3$ matrices. Verify that $V_2$ is a vector space.

Let $A=\left (\begin {array}{rrr} 1 & 1 & 0\\ 1 & -1 & 1 \end {array} \right ).$ Let $V_3$ be the set of vectors $x\in \R ^3$ such that $Ax=0$ . Verify that $V_3$ is a subspace of $\R ^3$ . Compare $V_1$ with $V_3$ .

In Exercises ?? – ?? you are given a vector space $V$ and a subset $W$ . For each pair, decide whether or not $W$ is a subspace of $V$ .

$V=\R ^3$ and $W$ consists of vectors in $\R ^3$ that have a $0$ in their first component.

$V=\R ^3$ and $W$ consists of vectors in $\R ^3$ that have a $1$ in their first component.

$V=\R ^2$ and $W$ consists of vectors in $\R ^2$ for which the sum of the components is $1$ .

$V=\R ^2$ and $W$ consists of vectors in $\R ^2$ for which the sum of the components is $0$ .

$V=\CCone$ and $W$ consists of functions $x(t)\in \CCone$ satisfying $\int _{-2}^4x(t)dt =0$ .

$V=\CCone$ and $W$ consists of functions $x(t)\in \CCone$ satisfying $x(1)=0$ .

$V=\CCone$ and $W$ consists of functions $x(t)\in \CCone$ satisfying $x(1)=1$ .

In Exercises ?? – ?? which of the sets $S$ are subspaces?

$S = \{(a,b,c)\in \R ^3 : a\ge 0, \; b\ge 0,\; c\ge 0\}$ .

$S = \{(x_1,x_2,x_3)\in \R ^3 : a_1x_1+a_2x_2+a_3x_3=0 \mbox { where } a_1,a_2,a_3\in \R \mbox { are fixed}\}$ .

$S = \{(x,y)\in \R ^2: (x,y) \mbox { is on the line through } (1,1) \mbox { with slope } 1\}$ .

$S = \{x\in \R ^2: Ax=0\}$ where $A$ is a $3\times 2$ matrix.

$S = \{x\in \R ^2: Ax=b\}$ where $A$ is a $3\times 2$ matrix and $b\in \R ^3$ is a fixed nonzero vector.

Let $V$ be a vector space and let $W_1$ and $W_2$ be subspaces. Show that the intersection $W_1\cap W_2$ is also a subspace of $V$ .

For which scalars $a,b,c$ do the solutions to the equation $ax+by = c$ form a subspace of $\R ^2$ ?

For which scalars $a,b,c,d$ do the solutions to the equation $ax+by+cz = d$ form a subspace of $\R ^3$ ?

Show that the set of all solutions to the differential equation $\dot {x}=2x$ is a subspace of $\CCone$ .

Recall from equation (??) of Section ?? that solutions to the system of differential equations $\frac {dX}{dt} = \mattwo {-1}{3}{3}{-1} X$ are $X(t) = \alpha e^{2t}\vectwo {1}{1} + \beta e^{-4t}\vectwo {1}{-1}.$ Use this formula for solutions to show that the set of solutions to this system of differential equations is a vector subspace of $(\CCone )^2$ .