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4.2 The Vector Space Rⁿ and Subspaces

In Section 4.1 we defined 3-dimensional space $R^{3}$ $R^{3}$ to be the set of all triples (x, y, z) of real numbers. This definition provides a mathematical model of the physical space in which we live, because geometric intuition and experience require that the location of every point be specified uniquely by three coordinates.

In science fiction, the fourth dimension often plays a rather exotic role. But there are common and ordinary situations where it is convenient to use four (or even more) coordinates rather than just two or three. For example, suppose we want to describe the motion of two points P and Q that are moving in the plane $R^{2}$ $R^{2}$ under the action of some given physical law. (See Fig. 4.2.1.) In order to tell where P and Q are at a given instant, we need to give two coordinates for P and two coordinates for Q. So let us write $P (x_{1}, x_{2})$ $P (x_{1}, x_{2})$ and $Q (x_{3}, x_{4})$ $Q (x_{3}, x_{4})$ to indicate these four coordinates. Then the two points P and Q determine a quadruple or 4-tuple $(x_{1}, x_{2}, x_{3}, x_{4})$ $(x_{1}, x_{2}, x_{3}, x_{4})$ of real numbers, and any such 4-tuple determines a possible pair of locations of P and Q. In this way the set of all pairs of points P and Q in the plane corresponds to the set of all 4-tuples of real numbers. By analogy with our definition of $R^{3}$ $R^{3}$ , we may define 4-dimensional space $R^{4}$ $R^{4}$ to be the set of all such 4-tuples $(x_{1}, x_{2}, x_{3}, x_{4})$ $(x_{1}, x_{2}, x_{3}, x_{4})$ . Then we can specify a pair of points P and Q in $R^{2}$ $R^{2}$ by specifying a single point $(x_{1}, x_{2}, x_{3}, x_{4})$ $(x_{1}, x_{2}, x_{3}, x_{4})$ in $R^{4}$ $R^{4}$ , and this viewpoint may simplify our analysis of the motions of the original points P and Q. For instance, it may turn out that their coordinates satisfy some single equation such as

3 x_{1} - 4 x_{2} + 2 x_{3} - 5 x_{4} = 0

$3 x_{1} - 4 x_{2} + 2 x_{3} - 5 x_{4} = 0$

that is better understood in terms of a single point in $R^{4}$ $R^{4}$ than in terms of the separate points P and Q in $R^{2}$ $R^{2}$ . Finally, note that in this example the fourth dimension is quite tangible—it refers simply to the second coordinate $x_{4}$ $x_{4}$ of the point Q.

FIGURE 4.2.1.

Two points $P (x_{1}, x_{2})$ $P (x_{1}, x_{2})$ and $Q (x_{3}, x_{4})$ $Q (x_{3}, x_{4})$ in $R^{2}$ $R^{2}$ .

But the number 4 is no more special in this context than the numbers 2 and 3. To describe similarly the motion of three points in $R^{3}$ $R^{3}$ , we would need 9 coordinates rather than 4. For instance, we might be studying the sun-earth-moon system in 3-dimensional space, with $(x_{1}, x_{2}, x_{3})$ $(x_{1}, x_{2}, x_{3})$ denoting the coordinates of the sun, $(x_{4}, x_{5}, x_{6})$ $(x_{4}, x_{5}, x_{6})$ the coordinates of the earth, and $(x_{7}, x_{8}, x_{9})$ $(x_{7}, x_{8}, x_{9})$ the coordinates of the moon at a given instant. Then the single list $(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9})$ $(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9})$ of 9 coordinates could be used to specify simultaneously the locations of the sun, earth, and moon. Application of Newton’s second law of motion to each of these three bodies would then lead to a system of differential equations involving the dependent variables $x_{i} (t), i = 1, 2, \dots, 9$ $x_{i} (t), i = 1, 2, \dots, 9$ as functions of time t. (We will discuss this very situation in Section 7.6.)

An efficient description of another situation might require fifty, a hundred, or even a thousand coordinates. For instance, an investor’s stock portfolio might contain 50 different high-tech stocks. If $x_{i}$ $x_{i}$ denotes the value of the ith stock on a particular day, then the state of the whole portfolio is described by the single list $(x_{1}, x_{2}, x_{3}, \dots, x_{49}, x_{50})$ $(x_{1}, x_{2}, x_{3}, \dots, x_{49}, x_{50})$ of 50 separate numbers. And the evolution of this portfolio with time could then be described by the motion of a single point in 50-dimensional space!

An n-tuple of real numbers is an (ordered) list $(x_{1}, x_{2}, x_{3}, \dots, x_{n})$ $(x_{1}, x_{2}, x_{3}, \dots, x_{n})$ of n real numbers. Thus (1, 3, 4, 2) is a 4-tuple and $(0, - 3, 7, 5, 2, - 1)$ $(0, - 3, 7, 5, 2, - 1)$ is a 6-tuple. A 2-tuple is an ordered pair and a 3-tuple is an ordered triple of real numbers, so the following definition generalizes our earlier definitions of the plane $R^{2}$ $R^{2}$ and 3-space $R^{3}$ $R^{3}$ .

The elements of n-space $R^{n}$ $R^{n}$ are called points or vectors, and we ordinarily use boldface letters to denote vectors. The ith entry of the vector $x = (x_{1}, x_{2}, x_{3}, \dots, x_{n})$ $x = (x_{1}, x_{2}, x_{3}, \dots, x_{n})$ is called its ith coordinate or its ith component. For consistency with matrix operations, we agree that

x = (x_{1}, x_{2}, x_{3}, \dots, x_{n}) = [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{array}],

$x = (x_{1}, x_{2}, x_{3}, \dots, x_{n}) = [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{array}],$

so that the $n \times 1$ $n \times 1$ matrix, or column vector, is simply another notation for the same ordered list of n real numbers.

If $n > 3$ $n > 3$ , we cannot visualize vectors in $R^{n}$ $R^{n}$ in the concrete way that we can “see” vectors in $R^{2}$ $R^{2}$ and $R^{3}$ $R^{3}$ . Nevertheless, we saw in Section 4.1 that the geometric properties of $R^{2}$ $R^{2}$ and $R^{3}$ $R^{3}$ stem ultimately from algebraic operations with vectors, and these algebraic operations can be defined for vectors in $R^{n}$ $R^{n}$ by analogy with their definitions in dimensions 2 and 3.

If $u = (u_{1}, u_{2}, \dots, u_{n})$ $u = (u_{1}, u_{2}, \dots, u_{n})$ and $v = (v_{1}, v_{2}, \dots, v_{n})$ $v = (v_{1}, v_{2}, \dots, v_{n})$ are vectors in $R^{n}$ $R^{n}$ , then their sum is the vector $u + v$ $u + v$ given by

u + v = (u_{1} + v_{1}, u_{2} + v_{2}, \dots, u_{n} + v_{n}) .

$u + v = (u_{1} + v_{1}, u_{2} + v_{2}, \dots, u_{n} + v_{n}) .$ (1)

Thus addition of vectors in $R^{n}$ $R^{n}$ is defined in a componentwise manner, just as in $R^{2}$ $R^{2}$ and $R^{3}$ $R^{3}$ , and we can visualize $u + v$ $u + v$ as a diagonal vector in a parallelogram determined by u and v, as in Fig. 4.2.2. If c is a scalar—a real number—then the scalar multiple cu is also defined in componentwise fashion:

c u = (c u_{1}, c u_{2}, \dots, c u_{n}) .

$c u = (c u_{1}, c u_{2}, \dots, c u_{n}) .$ (2)

FIGURE 4.2.2.

The parallelogram law for addition of vectors, in $R^{n}$ $R^{n}$ just as in $R^{2}$ $R^{2}$ or $R^{3}$ $R^{3}$ .

We can visualize cu as a vector collinear with u, with only its length (and possibly its direction along the same line) altered by multiplication by the scalar c. See Fig. 4.2.3.

In Section 4.1 we saw that, for vectors in $R^{2}$ $R^{2}$ and $R^{3}$ $R^{3}$ , the operations in (1) and (2) satisfy a list of properties that constitute the definition of a vector space; the same is true for vectors in $R^{n}$ $R^{n}$ .

Definition of a Vector Space

Let V be a set of elements called vectors, in which the operations of addition of vectors and multiplication of vectors by scalars are defined. That is, given vectors u and v in V and a scalar c, the vectors $u + v$ $u + v$ and cu are also in V (so that V is closed under vector addition and multiplication by scalars). Then, with these operations, V is called a vector space provided that—given any vectors u, v, and w in V and any scalars a and b—the following properties hold true:

(a) $u + v = v + u$ $u + v = v + u$	(commutativity)
(b) $u + (v + w) = (u + v) + w$ $u + (v + w) = (u + v) + w$	(associativity)
(c) $u + 0 = 0 + u = u$ $u + 0 = 0 + u = u$	(zero element)
(d) $u + (- u) = (- u) + u = 0$ $u + (- u) = (- u) + u = 0$	(additive inverse)
(e) $a (u + v) = a u + a v$ $a (u + v) = a u + a v$	(distributivity)
(f) $(a + b) u = a u + b u$ $(a + b) u = a u + b u$
(g) $a (b u) = (a b) u$ $a (b u) = (a b) u$
(h) $(1) u = u$ $(1) u = u$

FIGURE 4.2.3.

Multiplication of the vector u by the scalar `c`, in $R^{n}$ $R^{n}$ just as in $R^{2}$ $R^{2}$ or $R^{3}$ $R^{3}$ .

In property (c), it is meant that there exists a zero vector 0 in V such that $u + 0 = u$ $u + 0 = u$ . The zero vector in $R^{n}$ $R^{n}$ is

0 = (0, 0, \dots, 0) .

$0 = (0, 0, \dots, 0) .$

Similarly, property (d) actually means that, given the vector u in V, there exists a vector $- u$ $- u$ in V such that $u + (- u) = 0$ $u + (- u) = 0$ . In $R^{n}$ $R^{n}$ , we clearly have

- u = (- u_{1}, - u_{2}, \dots, - u_{n}) .

$- u = (- u_{1}, - u_{2}, \dots, - u_{n}) .$

The fact that $R = R^{1}$ $R = R^{1}$ satisfies properties (a)–(h), with the real numbers playing the dual roles of scalars and vectors, means that the real line may be regarded as a vector space. If $n > 1$ $n > 1$ , then each of properties (a)–(h) may be readily verified for $R^{n}$ $R^{n}$ by working with components and applying the corresponding properties of real numbers. For example, to verify the commutativity of vector addition in (a), we begin with Eq. (1) and write

\begin{array}{rcl} u + v & = & (u_{1} + v_{1}, u_{2} + v_{2}, \dots, u_{n} + v_{n}) \\ = & (v_{1} + u_{1}, v_{2} + u_{2}, \dots, v_{n} + u_{n}) \\ = & v + u . \end{array}

$\begin{array}{rcl} u + v & = & (u_{1} + v_{1}, u_{2} + v_{2}, \dots, u_{n} + v_{n}) \\ = & (v_{1} + u_{1}, v_{2} + u_{2}, \dots, v_{n} + u_{n}) \\ = & v + u . \end{array}$

Thus n-space $R^{n}$ $R^{n}$ is a vector space with the operations defined in Eqs. (1) and (2). We will understand throughout that our scalars are the real numbers, though in more advanced treatments of linear algebra other scalars (such as the complex numbers) are sometimes used.

Observe that, in the definition of a vector space, nothing is said about the elements of V (the “vectors”) being n-tuples. They can be any objects for which addition and multiplication by scalars are defined and which satisfy properties (a)–(h). Although many of the vectors you will see in this text actually are n-tuples of real numbers, an example in which they are not may help to clarify the definition of a vector space.

Example 1

Let $F$ $F$ be the set of all real-valued functions defined on the real number line R. Then each vector in $F$ $F$ is a function f such that the real number f(x) is defined for all x in R. Given f and g in $F$ $F$ and a real number c, the functions $f + g$ $f + g$ and cf are defined in the natural way,

(f + g) (x) = f (x) + g (x)

$(f + g) (x) = f (x) + g (x)$

and

(c f) (x) = c (f (x)) .

$(c f) (x) = c (f (x)) .$

Then each of properties (a)–(h) of a vector space follows readily from the corresponding property of the real numbers. For instance, if a is a scalar, then

\begin{array}{rcl} [a (f + g)] (x) & = & a [(f + g) (x)] \\ = & a [f (x) + g (x)] \\ = & a f (x) + a g (x) \\ = & (a f + a g) (x) . \end{array}

$\begin{array}{rcl} [a (f + g)] (x) & = & a [(f + g) (x)] \\ = & a [f (x) + g (x)] \\ = & a f (x) + a g (x) \\ = & (a f + a g) (x) . \end{array}$

Thus $a (f + g) = a f + a g$ $a (f + g) = a f + a g$ , so $F$ $F$ enjoys property (e).

After verification of the other seven properties, we conclude that $F$ $F$ is, indeed, a vector space—one that differs in a fundamental way from each of the vector spaces $R^{n}$ $R^{n}$ . We naturally call $R^{n}$ $R^{n}$ an n-dimensional vector space, and in Section 4.4 we will define the word dimension in such a way that the dimension of $R^{n}$ $R^{n}$ actually is n. But the vector space $F$ $F$ of functions turns out not to have dimension n for any integer n; it is an example of an infinite-dimensional vector space. (See Section 4.4.)

Subspaces

Let $W = {0}$ $W = {0}$ be the subset of $R^{n}$ $R^{n}$ that contains only the zero vector 0. Then W satisfies properties (a)–(h) of a vector space, because each reduces trivially to the equation $0 = 0$ $0 = 0$ . Thus W is a subset of the vector space $V = R^{n}$ $V = R^{n}$ that is itself a vector space. According to the following definition, a subset of a vector space V that is itself a vector space is called a subspace of V.

In order for the subset W to be a subspace of the vector space V, it first must be closed under the operations of vector addition and multiplication by scalars. Then it must satisfy properties (a)–(h) of a vector space. But W “inherits” all these properties from V, because the vectors in W all are vectors in V, and the vectors in V all satisfy properties (a)–(h). The following “subspace criterion” says that, in order to determine whether the subset W is a vector space, we need only check the two closure conditions.

In Section 4.1 we saw that lines through the origin in $R^{2}$ $R^{2}$ form subspaces of $R^{2}$ $R^{2}$ and that lines and planes through the origin are subspaces of $R^{3}$ $R^{3}$ . The subspace W of the following example may be regarded as a higher-dimensional “plane” (or “hyperplane”) through the origin in $R^{n}$ $R^{n}$ .

Example 2

Let W be the subset of $R^{n}$ $R^{n}$ consisting of all those vectors $(x_{1}, x_{2}, \dots, x_{n})$ $(x_{1}, x_{2}, \dots, x_{n})$ whose coordinates satisfy the single homogeneous linear equation

a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = 0,

$a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = 0,$

where the given coefficients $a_{1}, a_{2}, \dots, a_{n}$ $a_{1}, a_{2}, \dots, a_{n}$ are not all zero. If $u = (u_{1}, u_{2}, \dots, u_{n})$ $u = (u_{1}, u_{2}, \dots, u_{n})$ and $v = (v_{1}, v_{2}, \dots, v_{n})$ $v = (v_{1}, v_{2}, \dots, v_{n})$ are vectors in W, then

\begin{array}{rcl} a_{1} (u_{1} & + & v_{1}) + a_{2} (u_{2} + v_{2}) + \dots + a_{n} (u_{n} + v_{n}) \\ = & (a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n}) + (a_{1} v_{1} + a_{2} + v_{2} + \dots + a_{n} v_{n}) \\ = & 0 + 0 = 0, \end{array}

$\begin{array}{rcl} a_{1} (u_{1} & + & v_{1}) + a_{2} (u_{2} + v_{2}) + \dots + a_{n} (u_{n} + v_{n}) \\ = & (a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n}) + (a_{1} v_{1} + a_{2} + v_{2} + \dots + a_{n} v_{n}) \\ = & 0 + 0 = 0, \end{array}$

so $u + v = (u_{1} + v_{1}, \dots, u_{n} + v_{n})$ $u + v = (u_{1} + v_{1}, \dots, u_{n} + v_{n})$ is also in W. If c is a scalar, then

\begin{array}{rcl} a_{1} (c u_{1}) + a_{2} (c u_{2}) + \dots + a_{n} (c u_{n}) & = & c (a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n}) \\ = & (c) (0) = 0, \end{array}

$\begin{array}{rcl} a_{1} (c u_{1}) + a_{2} (c u_{2}) + \dots + a_{n} (c u_{n}) & = & c (a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n}) \\ = & (c) (0) = 0, \end{array}$

so $c u = (c u_{1}, c u_{2}, \dots, c u_{n})$ $c u = (c u_{1}, c u_{2}, \dots, c u_{n})$ is in W. Thus we have shown that W satisfies conditions (i) and (ii) of Theorem 1 and is therefore a subspace of $R^{n}$ $R^{n}$ .

In order to apply Theorem 1 to show that W is a subspace of the vector space V, we must show that W satisfies both conditions in the theorem. But to apply Theorem 1 to show that W is not a subspace of V, we need only show either that condition (i) fails or that condition (ii) fails.

Example 3

Let W be the set of all those vectors $(x_{1}, x_{2}, x_{3}, x_{4})$ $(x_{1}, x_{2}, x_{3}, x_{4})$ in $R^{4}$ $R^{4}$ whose four coordinates are all nonnegative: $x_{i} \geq 0$ $x_{i} \geq 0$ for $i = 1$ $i = 1$ , 2, 3, 4. Then it should be clear that the sum of two vectors in W is also a vector in W, because the sum of two nonnegative numbers is nonnegative. Thus W satisfies condition (i) of Theorem 1. But if we take $u = (1, 1, 1, 1)$ $u = (1, 1, 1, 1)$ in W and $c = - 1$ $c = - 1$ , then we find that the scalar multiple

c u = (- 1) (1, 1, 1, 1) = (- 1, - 1, - 1, - 1)

$c u = (- 1) (1, 1, 1, 1) = (- 1, - 1, - 1, - 1)$

is not in W. Thus W fails to satisfy condition (ii) and therefore is not a subspace of $R^{4}$ $R^{4}$ .

Example 4

Let W be the set of all those vectors $(x_{1}, x_{2}, x_{3}, x_{4})$ $(x_{1}, x_{2}, x_{3}, x_{4})$ in $R^{4}$ $R^{4}$ such that $x_{1} x_{4} = 0$ $x_{1} x_{4} = 0$ . Now W satisfies condition (ii) of Theorem 1, because $x_{1} x_{4} = 0$ $x_{1} x_{4} = 0$ implies that $(c x_{1}) (c x_{4}) = 0$ $(c x_{1}) (c x_{4}) = 0$ for any scalar c. But if we take the vectors $u = (1, 1, 0, 0)$ $u = (1, 1, 0, 0)$ and $v = (0, 0, 1, 1)$ $v = (0, 0, 1, 1)$ in W, we see that their sum $u + v = (1, 1, 1, 1)$ $u + v = (1, 1, 1, 1)$ is not in W. Thus W does not satisfy condition (i) and therefore is not a subspace of $R^{4}$ $R^{4}$ .

Example 2 implies that the solution set of a homogeneous linear equation

a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = 0

$a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = 0$

in n variables is always a subspace of $R^{n}$ $R^{n}$ . Theorem 2 further implies that the same is true of a homogeneous system of linear equations. Recall that any such system of m equations in n unknowns can be written in the form $Ax = 0$ $Ax = 0$ , where A is the $m \times n$ $m \times n$ coefficient matrix and $x = (x_{1}, x_{2}, \dots, x_{n})$ $x = (x_{1}, x_{2}, \dots, x_{n})$ is regarded as a column vector. The solution set of $Ax = 0$ $Ax = 0$ is then the set of all vectors x in $R^{n}$ $R^{n}$ that satisfy this equation—that is, the set of all its solution vectors.

Note that, in order to conclude by Theorem 2 that the solution set of a linear system is a subspace, it is necessary for the system to be homogeneous. Indeed, the solution set of a nonhomogeneous linear system

Ax = b

$Ax = b$ (4)

with $b \neq 0$ $b \neq 0$ is never a subspace. For if u were a solution vector of the system in (4), then

A (2 u) = 2 (A u) = 2 b \neq b

$A (2 u) = 2 (A u) = 2 b \neq b$

because $b \neq 0$ $b \neq 0$ . Thus the scalar multiple 2u of the solution vector u is not a solution vector. Therefore, Theorem 1 implies that the solution set of (4) is not a subspace.

Because the solution set of a homogeneous linear system is a subspace, we often call it the solution space of the system. The subspaces of $R^{n}$ $R^{n}$ are the possible solution spaces of a homogeneous linear system with n unknowns, and this is one of the principal reasons for our interest in subspaces of $R^{n}$ $R^{n}$ .

At opposite extremes as subspaces of $R^{n}$ $R^{n}$ lie the zero subspace ${0}$ ${0}$ and $R^{n}$ $R^{n}$ itself. Every other subspace of $R^{n}$ $R^{n}$ , each one that is neither ${0}$ ${0}$ nor $R^{n}$ $R^{n}$ , is called a proper subspace of $R^{n}$ $R^{n}$ . The proper subspaces of $R^{n}$ $R^{n}$ play the same role in $R^{n}$ $R^{n}$ that lines and planes through the origin play in $R^{3}$ $R^{3}$ . In the following two sections we develop the tools that are needed to analyze the structure of a given proper subspace of $R^{n}$ $R^{n}$ . In particular, given the homogeneous system $Ax = 0$ $Ax = 0$ , we ask how we can describe its solution space in a concise and illuminating way, beyond the mere statement that it is the set of all solution vectors of the system. Example 5 illustrates one possible way of doing this.

Example 5

In Example 4 of Section 3.4 we considered the homogeneous system

\begin{array}{rcrcl} x_{1} + 3 x_{2} - 15 x_{3} & + & 7 x_{4} & = & 0 \\ x_{1} + 4 x_{2} - 19 x_{3} & + & 10 x_{4} & = & 0 \\ 2 x_{1} + 5 x_{2} - 26 x_{3} & + & 11 x_{4} & = & 0 . \end{array}

$\begin{array}{rcrcl} x_{1} + 3 x_{2} - 15 x_{3} & + & 7 x_{4} & = & 0 \\ x_{1} + 4 x_{2} - 19 x_{3} & + & 10 x_{4} & = & 0 \\ 2 x_{1} + 5 x_{2} - 26 x_{3} & + & 11 x_{4} & = & 0 . \end{array}$ (5)

The reduced echelon form of the coefficient matrix of this system is

[\begin{array}{r} 1 & 0 & - 3 & - 2 \\ 0 & 1 & - 4 & 3 \\ 0 & 0 & 0 & 0 \end{array}] .

$[\begin{array}{r} 1 & 0 & - 3 & - 2 \\ 0 & 1 & - 4 & 3 \\ 0 & 0 & 0 & 0 \end{array}] .$

Hence $x_{1}$ $x_{1}$ and $x_{2}$ $x_{2}$ are the leading variables and $x_{3}$ $x_{3}$ and $x_{4}$ $x_{4}$ are free variables. Back substitution yields the general solution

x_{3} = s, x_{4} = t, x_{2} = 4 s - 3 t, x_{1} = 3 s + 2 t

$x_{3} = s, x_{4} = t, x_{2} = 4 s - 3 t, x_{1} = 3 s + 2 t$

in terms of arbitrary parameters s and t. Thus a typical solution vector of the system in (5) has the form

x = [\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} 3 s + 2 t \\ 4 s - 3 t \\ s \\ t \end{array}] = s [\begin{array}{l} 3 \\ 4 \\ 1 \\ 0 \end{array}] + t [\begin{array}{r} 2 \\ - 3 \\ 0 \\ 1 \end{array}] .

$x = [\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} 3 s + 2 t \\ 4 s - 3 t \\ s \\ t \end{array}] = s [\begin{array}{l} 3 \\ 4 \\ 1 \\ 0 \end{array}] + t [\begin{array}{r} 2 \\ - 3 \\ 0 \\ 1 \end{array}] .$

It follows that the solution space of the system in (5) can be described as the set of all linear combinations of the form

x = s u + t v,

$x = s u + t v,$ (6)

where $u = (3, 4, 1, 0)$ $u = (3, 4, 1, 0)$ and $v = (2, - 3, 0, 1)$ $v = (2, - 3, 0, 1)$ . Thus we have found two particular solution vectors u and v of our system that completely determine its solution space [by the formula in (6)].

4.2 Problems

In Problems 1–14, a subset W of some n-space $R^{n}$ $R^{n}$ is defined by means of a given condition imposed on the typical vector $(x_{1}, x_{2}, \dots, x_{n})$ $(x_{1}, x_{2}, \dots, x_{n})$ . Apply Theorem 1 to determine whether or not W is a subspace of $R^{n}$ $R^{n}$ .

W is the set of all vectors in $R^{3}$ $R^{3}$ such that $x_{3} = 0$ $x_{3} = 0$ .
W is the set of all vectors in $R^{3}$ $R^{3}$ such that $x_{1} = 5 x_{2}$ $x_{1} = 5 x_{2}$ .
W is the set of all vectors in $R^{3}$ $R^{3}$ such that $x_{2} = 1$ $x_{2} = 1$ .
W is the set of all vectors in $R^{3}$ $R^{3}$ such that $x_{1} + x_{2} + x_{3} = 1$ $x_{1} + x_{2} + x_{3} = 1$ .
W is the set of all vectors in $R^{4}$ $R^{4}$ such that $x_{1} + 2 x_{2} + 3 x_{3} + 4 x_{4} = 0$ $x_{1} + 2 x_{2} + 3 x_{3} + 4 x_{4} = 0$ .
W is the set of all vectors in $R^{4}$ $R^{4}$ such that $x_{1} = 3 x_{3}$ $x_{1} = 3 x_{3}$ and $x_{2} = 4 x_{4}$ $x_{2} = 4 x_{4}$ .
W is the set of all vectors in $R^{2}$ $R^{2}$ such that $| x_{1} | = | x_{2} |$ $| x_{1} | = | x_{2} |$ .
W is the set of all vectors in $R^{2}$ $R^{2}$ such that ${(x_{1})}^{2} + {(x_{2})}^{2} = 0$ ${(x_{1})}^{2} + {(x_{2})}^{2} = 0$ .
W is the set of all vectors in $R^{2}$ $R^{2}$ such that ${(x_{1})}^{2} + {(x_{2})}^{2} = 1$ ${(x_{1})}^{2} + {(x_{2})}^{2} = 1$ .
W is the set of all vectors in $R^{2}$ $R^{2}$ such that $| x_{1} | + | x_{2} | = 1$ $| x_{1} | + | x_{2} | = 1$ .
W is the set of all vectors in $R^{4}$ $R^{4}$ such that $x_{1} + x_{2} = x_{3} + x_{4}$ $x_{1} + x_{2} = x_{3} + x_{4}$ .
W is the set of all vectors in $R^{4}$ $R^{4}$ such that $x_{1} x_{2} = x_{3} x_{4}$ $x_{1} x_{2} = x_{3} x_{4}$ .
W is the set of all vectors in $R^{4}$ $R^{4}$ such that $x_{1} x_{2} x_{3} x_{4} = 0$ $x_{1} x_{2} x_{3} x_{4} = 0$ .
W is the set of all those vectors in $R^{4}$ $R^{4}$ whose components are all nonzero.

In Problems 15–18, apply the method of Example 5 to find two solution vectors u and v such that the solution space is the set of all linear combinations of the form $s u + t v$ $s u + t v$ .

$\begin{array}{r} x_{1} & - & 4 x_{2} & + & x_{3} & - & 4 x_{4} & = & 0 \\ x_{1} & + & 2 x_{2} & + & x_{3} & + & 8 x_{4} & = & 0 \\ x_{1} & + & x_{2} & + & x_{3} & + & 6 x_{4} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & - & 4 x_{2} & + & x_{3} & - & 4 x_{4} & = & 0 \\ x_{1} & + & 2 x_{2} & + & x_{3} & + & 8 x_{4} & = & 0 \\ x_{1} & + & x_{2} & + & x_{3} & + & 6 x_{4} & = & 0 \end{array}$
$\begin{array}{r} x_{1} & - & 4 x_{2} & - & 3 x_{3} & - & 7 x_{4} & = & 0 \\ 2 x_{1} & + & x_{2} & + & x_{3} & + & 7 x_{4} & = & 0 \\ x_{1} & + & 2 x_{2} & + & 3 x_{3} & + & 11 x_{4} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & - & 4 x_{2} & - & 3 x_{3} & - & 7 x_{4} & = & 0 \\ 2 x_{1} & + & x_{2} & + & x_{3} & + & 7 x_{4} & = & 0 \\ x_{1} & + & 2 x_{2} & + & 3 x_{3} & + & 11 x_{4} & = & 0 \end{array}$
$\begin{array}{r} x_{1} & + & 3 x_{2} & + & 8 x_{3} & - & x_{4} & = & 0 \\ x_{1} & - & 3 x_{2} & - & 10 x_{3} & + & 5 x_{4} & = & 0 \\ x_{1} & + & 4 x_{2} & + & 11 x_{3} & - & 2 x_{4} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & + & 3 x_{2} & + & 8 x_{3} & - & x_{4} & = & 0 \\ x_{1} & - & 3 x_{2} & - & 10 x_{3} & + & 5 x_{4} & = & 0 \\ x_{1} & + & 4 x_{2} & + & 11 x_{3} & - & 2 x_{4} & = & 0 \end{array}$
$\begin{array}{r} x_{1} & + & 3 x_{2} & + & 2 x_{3} & + & 5 x_{4} & - & x_{5} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & 4 x_{3} & + & 11 x_{4} & + & 2 x_{5} & = & 0 \\ 2 x_{1} & + & 6 x_{2} & + & 5 x_{3} & + & 12 x_{4} & - & 7 x_{5} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & + & 3 x_{2} & + & 2 x_{3} & + & 5 x_{4} & - & x_{5} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & 4 x_{3} & + & 11 x_{4} & + & 2 x_{5} & = & 0 \\ 2 x_{1} & + & 6 x_{2} & + & 5 x_{3} & + & 12 x_{4} & - & 7 x_{5} & = & 0 \end{array}$

In Problems 19–22, reduce the given system to echelon form to find a single solution vector u such that the solution space is the set of all scalar multiples of u.

$\begin{array}{r} x_{1} & - & 3 x_{2} & - & 5 x_{3} & - & 6 x_{4} & = & 0 \\ 2 x_{1} & + & x_{2} & + & 4 x_{3} & + & 4 x_{4} & = & 0 \\ x_{1} & + & 3 x_{2} & + & 7 x_{3} & + & x_{4} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & - & 3 x_{2} & - & 5 x_{3} & - & 6 x_{4} & = & 0 \\ 2 x_{1} & + & x_{2} & + & 4 x_{3} & + & 4 x_{4} & = & 0 \\ x_{1} & + & 3 x_{2} & + & 7 x_{3} & + & x_{4} & = & 0 \end{array}$
$\begin{array}{r} x_{1} & + & 5 x_{2} & + & x_{3} & - & 8 x_{4} & = & 0 \\ 2 x_{1} & + & 5 x_{2} & - & 5 x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & x_{3} & - & 9 x_{4} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & + & 5 x_{2} & + & x_{3} & - & 8 x_{4} & = & 0 \\ 2 x_{1} & + & 5 x_{2} & - & 5 x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & x_{3} & - & 9 x_{4} & = & 0 \end{array}$
$\begin{array}{r} x_{1} & + & 7 x_{2} & + & 2 x_{3} & - & 3 x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & x_{3} & - & 4 x_{4} & = & 0 \\ 3 x_{1} & + & 5 x_{2} & - & x_{3} & - & 5 x_{4} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & + & 7 x_{2} & + & 2 x_{3} & - & 3 x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & x_{3} & - & 4 x_{4} & = & 0 \\ 3 x_{1} & + & 5 x_{2} & - & x_{3} & - & 5 x_{4} & = & 0 \end{array}$
$\begin{array}{r} x_{1} & + & 3 x_{2} & + & 3 x_{3} & + & 3 x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & 5 x_{3} & - & x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & 4 x_{3} & - & 4 x_{4} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & + & 3 x_{2} & + & 3 x_{3} & + & 3 x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & 5 x_{3} & - & x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & 4 x_{3} & - & 4 x_{4} & = & 0 \end{array}$
Show that every subspace W of a vector space V contains the zero vector 0.
Apply the properties of a vector space V to show each of the following.
1. $0 u = 0$ $0 u = 0$ for every u in V.
2. $c 0 = 0$ $c 0 = 0$ for every scalar c.
3. $(- 1) u = - u$ $(- 1) u = - u$ for every u in V.
Do not assume that the vectors in V are n-tuples of real numbers.
Show that the nonempty subset W of a vector space V is a subspace of V if and only if for every pair of vectors u and v in W and every pair of scalars a and b, $a u + b v$ $a u + b v$ is also in W.
Prove: If u is a (fixed) vector in the vector space V, then the set W of all scalar multiples cu of u is a subspace of V.
Let u and v be (fixed) vectors in the vector space V. Show that the set W of all linear combinations $a u + b v$ $a u + b v$ of u and v is a subspace of V.
Suppose that A is an $n \times n$ $n \times n$ matrix and that k is a (constant) scalar. Show that the set of all vectors x such that $Ax = k x$ $Ax = k x$ is a subspace of $R^{n}$ $R^{n}$ .
Let A be an $n \times n$ $n \times n$ matrix, b be a nonzero vector, and $x_{0}$ $x_{0}$ be a solution vector of the system $Ax = b$ $Ax = b$ . Show that x is a solution of the nonhomogeneous system $Ax = b$ $Ax = b$ if and only if $y = x - x_{0}$ $y = x - x_{0}$ is a solution of the homogeneous system $A y = 0$ $A y = 0$ .
Let U and V be subspaces of the vector space W. Their intersection $U \cap V$ $U \cap V$ is the set of all vectors that are both in U and in V. Show that $U \cap V$ $U \cap V$ is a subspace of W. If U and V are two planes through the origin in $R^{3}$ $R^{3}$ , what is $U \cap V$ $U \cap V$ ?
Let U and V be subspaces of the vector space W. Their sum $U + V$ $U + V$ is the set of all vectors w of the form

$w = u + v,$ $w = u + v,$

where u is in U and v is in V. Show that $U + V$ $U + V$ is a subspace of W. If U and V are lines through the origin in $R^{3}$ $R^{3}$ , what is $U + V$ $U + V$ ?

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Table of Contents for 4.2 The Vector Space Rn and Subspaces

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Table of Contents for
4.2 The Vector Space Rn and Subspaces