Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

4.4 Bases and Dimension for Vector Spaces

An especially useful way of describing the solution space of a homogeneous linear system is to list explicitly a set S of solution vectors such that every solution vector is a unique linear combination of these particular ones. The following definition specifies the properties of such a set S of “basic” solution vectors, and the concept is equally important for vector spaces other than solution spaces.

In short, a basis for the vector space V is a linearly independent spanning set of vectors in V. Thus, if S={v1,v2,…,vn} $S = {v_{1}, v_{2}, \dots, v_{n}}$ is a basis for V, then any vector w in V can be expressed as a linear combination

w = c 1 v 1 + c 2 v 2 + \dots + c n v n

$w = c_{1} v_{1} + c_{2} v_{2} + \dots + c_{n} v_{n}$ (1)

of the vectors in S, and we saw in Section 4.3 that the linear independence of S implies that the coefficients c1,c2,…,cn $c_{1}, c_{2}, \dots, c_{n}$ in (1) are unique. That is, w cannot be expressed differently as a linear combination of the basis vectors v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ .

Example 1

The standard basis for Rn $R^{n}$ consists of the unit vectors

e 1 = (1, 0, 0, \dots, 0), e 2 = (0, 1, 0, \dots, 0), \dots, e n = (0, 0, 0, \dots, 1) .

$e_{1} = (1, 0, 0, \dots, 0), e_{2} = (0, 1, 0, \dots, 0), \dots, e_{n} = (0, 0, 0, \dots, 1) .$

If x=(x1,x2,…,xn) $x = (x_{1}, x_{2}, \dots, x_{n})$ is a vector in Rn $R^{n}$ , then

x = x 1 e 1 + x 2 e 2 + \dots + x n e n .

$x = x_{1} e_{1} + x_{2} e_{2} + \dots + x_{n} e_{n} .$

Thus the column vectors

e 1 = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 10 ⋮ 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥, e 2 = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 01 ⋮ 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥, \dots, e n = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 00 ⋮ 1 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥

$e_{1} = [\begin{array}{c} 1 \\ 0 \\ ⋮ \\ 0 \end{array}], e_{2} = [\begin{array}{c} 0 \\ 1 \\ ⋮ \\ 0 \end{array}], \dots, e_{n} = [\begin{array}{c} 0 \\ 0 \\ ⋮ \\ 1 \end{array}]$

of the n×n $n \times n$ identity matrix span Rn $R^{n}$ , and we noted in Example 4 of Section 4.3 that these standard unit vectors are linearly independent.

Example 2

Let v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ be n linearly independent vectors in Rn $R^{n}$ . We saw in Section 4.3 that any set of more than n vectors in Rn $R^{n}$ is linearly dependent. Hence, given a vector w in Rn $R^{n}$ , there exist scalars c,c1,c2,…,cn $c, c_{1}, c_{2}, \dots, c_{n}$ not all zero such that

c w + c 1 v 1 + c 2 v 2 + \dots + c n v n = 0 .

$c w + c_{1} v_{1} + c_{2} v_{2} + \dots + c_{n} v_{n} = 0 .$ (2)

If c were zero, then (2) would imply that the vectors v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ are linearly dependent. Hence c≠0 $c \neq 0$ , so Eq. (2) can be solved for w as a linear combination of v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ . Thus the linearly independent vectors v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ also span Rn $R^{n}$ and therefore constitute a basis for Rn $R^{n}$ .

Example 2 shows that any set of n linearly independent vectors in Rn $R^{n}$ is a basis for Rn $R^{n}$ . By Theorem 2 in Section 4.3, we can therefore determine whether n given vectors v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ form a basis for Rn $R^{n}$ by calculating the determinant of the n×n $n \times n$ matrix

A = [v 1 v 2 \dots v n]

$A = [v_{1} v_{2} \dots v_{n}]$

with these vectors as its column vectors. They constitute a basis for Rn $R^{n}$ if and only if detA≠0 $\det A \neq 0$ .

Example 3

Let v1=(1,−1,−2,−3), v2=(1,−1,2,3), v3=(1,−1,−3,−2) $v_{1} = (1, - 1, - 2, - 3), v_{2} = (1, - 1, 2, 3), v_{3} = (1, - 1, - 3, - 2)$ , and v4=(0,3,−1,2) $v_{4} = (0, 3, - 1, 2)$ . Then we find that

∣ ∣ ∣ ∣ ∣ ∣ 1 - 1 - 2 - 3 1 - 1 23 1 - 1 - 3 - 2 03 - 1 2 ∣ ∣ ∣ ∣ ∣ ∣ = 30 \neq 0,

$| \begin{array}{r} 1 & 1 & 1 & 0 \\ - 1 & - 1 & - 1 & 3 \\ - 2 & 2 & - 3 & - 1 \\ - 3 & 3 & - 2 & 2 \end{array} | = 30 \neq 0,$

so it follows that {v1,v2,v3,v4} ${v_{1}, v_{2}, v_{3}, v_{4}}$ is a basis for R4 $R^{4}$ .

Theorem 1 has the following import: Just as in Rn $R^{n}$ , a basis for any vector space V contains the largest possible number of linearly independent vectors in V.

Proof:

We need to show that if m>n $m > n$ , then any set T={w1,w2,…,wm} $T = {w_{1}, w_{2}, \dots, w_{m}}$ of m vectors in V is linearly dependent. Because the basis S spans V, each vector wj $w_{j}$ in T can be expressed as a linear combination of v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ :

w 1 w 2 w m = = ⋮ = a 11 v 1 a 12 v 1 a 1 m v 1 + + + a 21 v 2 a 22 v 2 a 2 m v 2 + + + \dots \dots \dots + + + a n 1 v n a n 2 v n a n m v n .

$\begin{array}{rcccccccc} w_{1} & = & a_{11} v_{1} & + & a_{21} v_{2} & + & \dots & + & a_{n 1} v_{n} \\ w_{2} & = & a_{12} v_{1} & + & a_{22} v_{2} & + & \dots & + & a_{n 2} v_{n} \\ ⋮ \\ w_{m} & = & a_{1 m} v_{1} & + & a_{2 m} v_{2} & + & \dots & + & a_{n m} v_{n} . \end{array}$ (3)

Now we need to find scalars c1,c2,…,cm $c_{1}, c_{2}, \dots, c_{m}$ not all zero such that

c 1 w 1 + c 2 w 2 + \dots + c m w m = 0 .

$c_{1} w_{1} + c_{2} w_{2} + \dots + c_{m} w_{m} = 0 .$ (4)

Substituting in (4) the expressions in (3) for each wj $w_{j}$ , we get

+ + c 1 (a 11 v 1 + a 21 v 2 + \dots + a n 1 v n) c 2 (a 12 v 1 + a 22 v 2 + \dots + a n 2 v n) + \dots c m (a 1 m v 1 + a 2 m v 2 + \dots + a n m v n) = 0 .

$\begin{array}{l} c_{1} (a_{11} v_{1} + a_{21} v_{2} + \dots + a_{n 1} v_{n}) \\ + & c_{2} (a_{12} v_{1} + a_{22} v_{2} + \dots + a_{n 2} v_{n}) + \dots \\ + & c_{m} (a_{1 m} v_{1} + a_{2 m} v_{2} + \dots + a_{n m} v_{n)} = 0 . \end{array}$ (5)

Because the vectors v1,v2,…,vn $v_{1}, v_{2}, \dots, v_{n}$ are linearly independent, the coefficient in (5) of each vi $v_{i}$ must vanish:

a 11 c 1 a 21 c 1 a n 1 c 1 + + + a 12 c 2 a 22 c 2 a n 2 c 2 + + + \dots \dots \dots + + + a 1 m c m a 2 m c m a n m c m = = ⋮ = 00 0 .

$\begin{array}{rcrcrcrcl} a_{11} c_{1} & + & a_{12} c_{2} & + & \dots & + & a_{1 m} c_{m} & = & 0 \\ a_{21} c_{1} & + & a_{22} c_{2} & + & \dots & + & a_{2 m} c_{m} & = & 0 \\ ⋮ \\ a_{n 1} c_{1} & + & a_{n 2} c_{2} & + & \dots & + & a_{n m} c_{m} & = & 0 . \end{array}$ (6)

But because m>n $m > n$ , this is a homogeneous linear system with more unknowns than equations, and therefore has a nontrivial solution (c1,c2,…,cm) $(c_{1}, c_{2}, \dots, c_{m})$ . Then c1,c2,…, cm $c_{1}, c_{2}, \dots, c_{m}$ are scalars not all zero such that (4) holds, so we have shown that the set T of more than n vectors is linearly dependent.

Now let S={v1,v2,…,vn} $S = {v_{1}, v_{2}, \dots, v_{n}}$ and T={w1,w2,…,wm} $T = {w_{1}, w_{2}, \dots, w_{m}}$ be two different bases for the same vector space V. Because S is a basis and T is linearly independent, Theorem 1 implies that m≤n $m \leq n$ . Next, reverse the roles of S and T; the fact that T is a basis and S is linearly independent implies similarly that n≤m $n \leq m$ . Hence m=n $m = n$ , so we have proved the following theorem for any vector space with a finite basis.

A nonzero vector space V is called finite dimensional provided that there exists a basis for V consisting of a finite number of vectors from V. In this case the number n of vectors in each basis for V is called the dimension of V, denoted by n=dimV $n = \dim V$ . Then V is an n-dimensional vector space. The standard basis of Example 1 shows that Rn $R^{n}$ is, indeed, an n-dimensional vector space.

Note that the zero vector space {0} ${0}$ has no basis because it contains no linearly independent set of vectors. (Sometimes it is convenient to adopt the convention that the null set is a basis for {0} ${0}$ .) Here we define dim{0} $\dim {0}$ to be zero. A nonzero vector space that has no finite basis is called infinite dimensional. Infinite dimensional vector spaces are discussed in Section 4.5, but we include an illustrative example of one here.

Example 4

Let P $P$ be the set of all polynomials of the form

p (x) = a 0 + a 1 x + a 2 x 2 + \dots + a n x n,

$p (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n},$

where the largest exponent n≥0 $n \geq 0$ that appears is the degree of the polynomial p(x), and the coefficients a0,a1,a2,…,an $a_{0}, a_{1}, a_{2}, \dots, a_{n}$ are real numbers. We add polynomials in P $P$ and multiply them by scalars in the usual way—that is, by collecting coefficients of like powers of x. For instance, if

p (x) = 3 + 2 x + 5 x 3 and q (x) = 7 + 4 x + 3 x 2 + 9 x 4,

$p (x) = 3 + 2 x + 5 x^{3} and q (x) = 7 + 4 x + 3 x^{2} + 9 x^{4},$

then

(p + q) (x) = = (3 + 7) + (2 + 4) x + (0 + 3) x 2 + (5 + 0) x 3 + (0 + 9) x 4 10 + 6 x + 3 x 2 + 5 x 3 + 9 x 4

$\begin{array}{rcl} (p + q) (x) & = & (3 + 7) + (2 + 4) x + (0 + 3) x^{2} + (5 + 0) x^{3} + (0 + 9) x^{4} \\ = & 10 + 6 x + 3 x^{2} + 5 x^{3} + 9 x^{4} \end{array}$

and

(7 p) (x) = 7 (3 + 2 x + 5 x 3) = 21 + 14 x + 35 x 3 .

$(7 p) (x) = 7 (3 + 2 x + 5 x^{3}) = 21 + 14 x + 35 x^{3} .$

It is readily verified that, with these operations, P $P$ is a vector space. But P $P$ has no finite basis. For if p1,p2,…,pn $p_{1}, p_{2}, \dots, p_{n}$ are elements of P $P$ , then the degree of any linear combination of them is at most the maximum of their degrees. Hence no polynomial in P $P$ of higher degree lies in span{p1,p2,…,pn} $span {p_{1}, p_{2}, \dots, p_{n}}$ . Thus no finite subset of P $P$ spans P $P$ , and therefore P $P$ is an infinite dimensional vector space.

Here, our concern is with finite-dimensional vector spaces, and we note first that any proper subspace W of a finite-dimensional vector space V is itself finite dimensional, with dimW<dimV $\dim W < \dim V$ . For if dimV=n $\dim V = n$ , let k≤n $k \leq n$ be the largest integer such that W contains k linearly independent vectors v1,v2,…,vk $v_{1}, v_{2}, \dots, v_{k}$ . Then, by the same argument as in Example 2, we see that {v1,v2,…,vk} ${v_{1}, v_{2}, \dots, v_{k}}$ is a basis for W, so W is finite dimensional with dimW=k $\dim W = k$ , and k<n $k < n$ because W is a proper subspace.

Moreover, an n-dimensional vector space V contains proper subspaces of each dimension k=1,2,…,n−1 $k = 1, 2, \dots, n - 1$ . For instance, if {v1,v2,…,vn} ${v_{1}, v_{2}, \dots, v_{n}}$ is a basis for V and k<n $k < n$ , then W=span{v1,v2,…,vk} $W = span {v_{1}, v_{2}, \dots, v_{k}}$ is a subspace of dimension k. Thus R4 $R^{4}$ contains proper subspaces of dimensions 1, 2, and 3; R5 $R^{5}$ contains proper subspaces of dimensions 1, 2, 3, and 4; and so on. The proper subspaces of Rn $R^{n}$ of dimensions 1,2,…,n−1 $1, 2, \dots, n - 1$ are the higher-dimensional analogues of lines and planes through the origin in R3 $R^{3}$ .

Suppose that V is an n-dimensional vector space and that S={v1,v2,…,vn} $S = {v_{1}, v_{2}, \dots, v_{n}}$ is a set of n vectors in V. Then, in order to show that S is a basis for V, it is not necessary to prove both that S is linearly independent and that S spans V, because it turns out (for n vectors in an n-dimensional vector space) that either property of S implies the other. For instance, if S={v1,v2,…,vn} $S = {v_{1}, v_{2}, \dots, v_{n}}$ is a set of n linearly independent vectors in V, then Theorem 1 and the argument of Example 2 (once again) imply that S spans V and, hence, is a basis for V. This proves part (a) of Theorem 3. The remaining parts are left to the problems. (See Problems 27–32.)

Part (c) of Theorem 3 is often applied in the following form: If W is a k-dimensional subspace of the n-dimensional vector space V, then any basis {v1,v2,…,vk} ${v_{1}, v_{2}, \dots, v_{k}}$ for W can be “extended” to a basis {v1,v2,…,vn} ${v_{1}, v_{2}, \dots, v_{n}}$ for V, consisting of the original basis vectors for W together with n−k $n - k$ additional vectors vk+1,vk+2,…,vn $v_{k + 1}, v_{k + 2}, \dots, v_{n}$ .

Bases for Solution Spaces

We consider now the homogeneous linear system

A x = 0,

$Ax = 0,$ (7)

in which A is an m×n $m \times n$ matrix, so the system consists of m equations in the n variables x1,x2,…,xn $x_{1}, x_{2}, \dots, x_{n}$ . Its solution space W is then a subspace of Rn $R^{n}$ . We want to determine the dimension of W and, moreover, to find an explicit basis for W. Thus we seek a maximal set of linearly independent solution vectors of (7).

Recall the Gaussian elimination method of Section 3.2. We use elementary row operations to reduce the coefficient matrix A to an echelon matrix E and note the (nonzero) leading entries in the rows of E. The leading variables are those that correspond to the columns of E containing the leading entries. The remaining variables (if any) are the free variables. If there are no free variables, then our system has only the trivial solution, so W={0} $W = {0}$ . If there are free variables, we set each of them (separately) equal to a parameter and solve (by back substitution) for the leading variables as linear combinations of these parameters. The solution space W is then the set of all solution vectors obtained in this manner (for all possible values of the parameters).

To illustrate the general situation, let us suppose that the leading variables are the first r variables x1,x2,…,xr $x_{1}, x_{2}, \dots, x_{r}$ , so the k=n−r $k = n - r$ variables xr+1,xr+2,…,xn $x_{r + 1}, x_{r + 2}, \dots, x_{n}$ are free variables. The reduced system Ex=0 $E x = 0$ then takes the form

b 11 x 1 + b 12 x 2 + \dots + b 1 r x r + \dots + b 1 n x n b 22 x 2 + \dots + b 2 r x r + \dots + b 2 n x n b r r x r + \dots + b r n x n 00 = = ⋮ = = ⋮ = 00000

$\begin{array}{rcl} b_{11} x_{1} + b_{12} x_{2} + \dots + b_{1 r} x_{r} + \dots + b_{1 n} x_{n} & = & 0 \\ b_{22} x_{2} + \dots + b_{2 r} x_{r} + \dots + b_{2 n} x_{n} & = & 0 \\ ⋮ \\ b_{r r} x_{r} + \dots + b_{r n} x_{n} & = & 0 \\ 0 & = & 0 \\ ⋮ \\ 0 & = & 0 \end{array}$ (8)

with the last m−r $m - r$ equations being “trivial.” We set

x r + 1 = t 1, x r + 2 = t 2, \dots, x n = t k

$x_{r + 1} = t_{1}, x_{r + 2} = t_{2}, \dots, x_{n} = t_{k}$ (9)

and then solve (by back substitution) the equations in (8) for the leading variables

x 1 x 2 x r = = ⋮ = c 11 t 1 + c 12 t 2 + \dots + c 1 k t k c 21 t 1 + c 22 t 2 + \dots + c 2 k t k c r 1 t 1 + c r 2 t 2 + \dots + c r k t k .

$\begin{array}{rcl} x_{1} & = & c_{11} t_{1} + c_{12} t_{2} + \dots + c_{1 k} t_{k} \\ x_{2} & = & c_{21} t_{1} + c_{22} t_{2} + \dots + c_{2 k} t_{k} \\ ⋮ \\ x_{r} & = & c_{r 1} t_{1} + c_{r 2} t_{2} + \dots + c_{r k} t_{k} . \end{array}$ (10)

The typical solution vector (x1,x2,…,xn) $(x_{1}, x_{2}, \dots, x_{n})$ is given in terms of the k parameters t1,t2,…,tk $t_{1}, t_{2}, \dots, t_{k}$ by the equations in (9) and (10).

We now choose k particular solution vectors v1,v2,…,vk $v_{1}, v_{2}, \dots, v_{k}$ as follows: To get vj $v_{j}$ we set the jth parameter tj $t_{j}$ equal to 1 and set all other parameters equal to zero. Then

v j = (c 1 j, c 2 j, \dots, c r j, 0, \dots, 1, \dots, 0),

$v_{j} = (c_{1 j}, c_{2 j}, \dots, c_{r j}, 0, \dots, 1, \dots, 0),$ (11)

with the 1 appearing as the (r+j)th $(r + j) th$ entry. The vectors v1,v2,…,vk $v_{1}, v_{2}, \dots, v_{k}$ are the column vectors of the n×k $n \times k$ matrix

⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ c 11 c 21 ⋮ c r 1 10 ⋮ 0 c 12 c 22 ⋮ c r 2 01 ⋮ 0 \dots \dots ⋱ \dots \dots \dots ⋱ \dots c 1 k c 2 k ⋮ c r k 00 ⋮ 1 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ .

$[\begin{array}{c} c_{11} & c_{12} & \dots & c_{1 k} \\ c_{21} & c_{22} & \dots & c_{2 k} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ c_{r 1} & c_{r 2} & \dots & c_{r k} \\ 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & 1 \end{array}] .$ (12)

Because of the presence of the lower k×k $k \times k$ identity matrix, it is clear that the vectors v1,v2,…,vk $v_{1}, v_{2}, \dots, v_{k}$ are linearly independent. (See Problem 36.) But Eqs. (9) and (10) show that the typical solution vector x is a linear combination

x = t 1 v 1 + t 2 v 2 + \dots + t k v k .

$x = t_{1} v_{1} + t_{2} v_{2} + \dots + t_{k} v_{k} .$ (13)

Therefore, the vectors v1,v2,…,vk $v_{1}, v_{2}, \dots, v_{k}$ defined in (11) form a basis for the solution space W of the original system in (7).

The following algorithm summarizes the steps in this procedure.

Example 5

Find a basis for the solution space of the homogeneous linear system

3 x 1 2 x 1 3 x 1 + + + 6 x 2 4 x 2 6 x 2 - - - x 3 x 3 2 x 3 - - - 5 x 4 3 x 4 4 x 4 + + + 5 x 5 2 x 5 x 5 = = = 00 0 .

$\begin{array}{rcrcrcrcrcl} 3_{x 1} & + & 6_{x 2} & - & x_{3} & - & 5 x_{4} & + & 5 x_{5} & = & 0 \\ 2_{x 1} & + & 4_{x 2} & - & x_{3} & - & 3 x_{4} & + & 2 x_{5} & = & 0 \\ 3_{x 1} & + & 6_{x 2} & - & 2 x_{3} & - & 4 x_{4} & + & x_{5} & = & 0 . \end{array}$ (14)

Solution

We readily reduce the coefficient matrix A to the echelon form

E = ⎡ ⎣ ⎢ 100200010 - 2 - 1 0 340 ⎤ ⎦ ⎥ .

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

The leading entries are in the first and third columns, so the leading variables are x1 $x_{1}$ and x3 $x_{3}$ ; the free variables are x2, x4 $x_{2}, x_{4}$ , and x5 $x_{5}$ . To avoid subscripts, we use r, s, and t rather than t1, t2 $t_{1}, t_{2}$ , and t3 $t_{3}$ to denote the three parameters. Thus we set

x 2 = r, x 4 = s, and x 5 = t .

$x_{2} = r, x_{4} = s, and x_{5} = t .$ (15)

Then back substitution in the reduced system

x 1 + 2 x 2 x 3 - - 2 x 4 x 4 + + 3 x 5 4 x 5 = = 00

$\begin{array}{r} x_{1} & + & 2 x_{2} & - & 2 x_{4} & + & 3 x_{5} & = & 0 \\ x_{3} & - & x_{4} & + & 4 x_{5} & = & 0 \end{array}$

yields

x 1 = - 2 r + 2 s - 3 t and x 3 = s - 4 t .

$x_{1} = - 2 r + 2 s - 3 t and x_{3} = s - 4 t .$ (16)

The equations in (15) and (16) give the typical solution vector (x1,x2,x3,x4,x5) $(x_{1}, x_{2}, x_{3}, x_{4}, x_{5})$ in terms of the parameters r, s, and t.

With r=1 $r = 1$ and s=t=0 $s = t = 0$ ,

we obtain v 1 = (- 2, 1, 0, 0, 0) .

$we obtain v_{1} = (- 2, 1, 0, 0, 0) .$

With s=1 $s = 1$ and r=t=0 $r = t = 0$ ,

we obtain v 2 = (2, 0, 1, 1, 0) .

$we obtain v_{2} = (2, 0, 1, 1, 0) .$

With t=1 $t = 1$ and r=s=0 $r = s = 0$ ,

we obtain v 3 = (- 3, 0, - 4, 0, 1) .

$we obtain v_{3} = (- 3, 0, - 4, 0, 1) .$

Thus the solution space of the system in (14) is a 3-dimensional subspace of R5 $R^{5}$ with basis {v1,v2,v3} ${v_{1}, v_{2}, v_{3}}$ .

4.4 Problems

In Problems 1–8, determine whether or not the given vectors in Rn $R^{n}$ form a basis for Rn $R^{n}$ .

v1=(4,7), v2=(5,6) $v_{1} = (4, 7), v_{2} = (5, 6)$
v1=(3,−1,2), v2=(6,−2,4), v3=(5,3,−1) $v_{1} = (3, - 1, 2), v_{2} = (6, - 2, 4), v_{3} = (5, 3, - 1)$
v1=(1,7,−3), v2=(2,1,4), v3=(6,5,1), v4=(0,7,13) $v_{1} = (1, 7, - 3), v_{2} = (2, 1, 4), v_{3} = (6, 5, 1), v_{4} = (0, 7, 13)$
v1=(3,−7,5,2), v2=(1,−1,3,4), v3=(7,11,3,13) $v_{1} = (3, - 7, 5, 2), v_{2} = (1, - 1, 3, 4), v_{3} = (7, 11, 3, 13)$
v1=(0,7,−3), v2=(0,5,4), v3=(0,5,10) $v_{1} = (0, 7, - 3), v_{2} = (0, 5, 4), v_{3} = (0, 5, 10)$
v1=(0,0,1), v2=(0,1,2), v3=(1,2,3) $v_{1} = (0, 0, 1), v_{2} = (0, 1, 2), v_{3} = (1, 2, 3)$
v1=(0,0,1), v2=(7,4,11), v3=(5,3,13) $v_{1} = (0, 0, 1), v_{2} = (7, 4, 11), v_{3} = (5, 3, 13)$
v1=(2,0,0,0), v2=(0,3,0,0), v3=(0,0,7,6), v4=(0,0,4,5) $v_{1} = (2, 0, 0, 0), v_{2} = (0, 3, 0, 0), v_{3} = (0, 0, 7, 6), v_{4} = (0, 0, 4, 5)$

In Problems 9–11, find a basis for the indicated subspace of R3 $R^{3}$ .

The plane with equation x−2y+5z=0 $x - 2 y + 5 z = 0$ .
The plane with equation y=z $y = z$ .
The line of intersection of the planes described in Problems 9 and 10.

In Problems 12–14, find a basis for the indicated subspace of R4 $R^{4}$ .

The set of all vectors of the form (a, b, c, d) for which a=b+c+d $a = b + c + d$ .
The set of all vectors of the form (a, b, c, d) such that a=3c $a = 3 c$ and b=4d $b = 4 d$ .
The set of all vectors of the form (a, b, c, d) for which a+2b=c+3d=0 $a + 2 b = c + 3 d = 0$ .

In Problems 15–26, find a basis for the solution space of the given homogeneous linear system.

x12x1−−2x23x2+−3x3x3==00 $\begin{array}{r} x_{1} & - & 2 x_{2} & + & 3 x_{3} & = & 0 \\ 2 x_{1} & - & 3 x_{2} & - & x_{3} & = & 0 \end{array}$
x13x1++3x28x2++4x37x3==00 $\begin{array}{r} x_{1} & + & 3 x_{2} & + & 4 x_{3} & = & 0 \\ 3 x_{1} & + & 8 x_{2} & + & 7 x_{3} & = & 0 \end{array}$
x12x1−−3x25x2++2x37x3−−4x43x4==00 $\begin{array}{r} x_{1} & - & 3 x_{2} & + & 2 x_{3} & - & 4 x_{4} & = & 0 \\ 2 x_{1} & - & 5 x_{2} & + & 7 x_{3} & - & 3 x_{4} & = & 0 \end{array}$
x12x1++3x26x2++4x39x3++5x45x4==00 $\begin{array}{r} x_{1} & + & 3 x_{2} & + & 4 x_{3} & + & 5 x_{4} & = & 0 \\ 2 x_{1} & + & 6 x_{2} & + & 9 x_{3} & + & 5 x_{4} & = & 0 \end{array}$
x12x1x1−++3x2x23x2−−+9x34x33x3−++5x411x413x4===000 $\begin{array}{r} x_{1} & - & 3 x_{2} & - & 9 x_{3} & - & 5 x_{4} & = & 0 \\ 2 x_{1} & + & x_{2} & - & 4 x_{3} & + & 11 x_{4} & = & 0 \\ x_{1} & + & 3 x_{2} & + & 3 x_{3} & + & 13 x_{4} & = & 0 \end{array}$
x1x1x1−++3x24x23x2−++10x311x38x3+−−5x42x4x4===000 $\begin{array}{r} x_{1} & - & 3 x_{2} & - & 10 x_{3} & + & 5 x_{4} & = & 0 \\ x_{1} & + & 4 x_{2} & + & 11 x_{3} & - & 2 x_{4} & = & 0 \\ x_{1} & + & 3 x_{2} & + & 8 x_{3} & - & x_{4} & = & 0 \end{array}$
x12x1x1−−+4x2x22x2−++3x3x33x3−++7x47x411x4===000 $\begin{array}{r} x_{1} & - & 4 x_{2} & - & 3 x_{3} & - & 7 x_{4} & = & 0 \\ 2 x_{1} & - & x_{2} & + & x_{3} & + & 7 x_{4} & = & 0 \\ x_{1} & + & 2 x_{2} & + & 3 x_{3} & + & 11 x_{4} & = & 0 \end{array}$
x12x1x1−−−2x24x22x2−++3x3x33x3−++16x417x426x4===000 $\begin{array}{r} x_{1} & - & 2 x_{2} & - & 3 x_{3} & - & 16 x_{4} & = & 0 \\ 2 x_{1} & - & 4 x_{2} & + & x_{3} & + & 17 x_{4} & = & 0 \\ x_{1} & - & 2 x_{2} & + & 3 x_{3} & + & 26 x_{4} & = & 0 \end{array}$
x12x12x1+++5x25x27x2+++13x311x317x3+++14x412x419x4===000 $\begin{array}{r} x_{1} & + & 5 x_{2} & + & 13 x_{3} & + & 14 x_{4} & = & 0 \\ 2 x_{1} & + & 5 x_{2} & + & 11 x_{3} & + & 12 x_{4} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & + & 17 x_{3} & + & 19 x_{4} & = & 0 \end{array}$
x1x12x1++3x27x2−+−4x32x310x3−+−8x4x419x4+++6x53x513x5===000 $\begin{array}{r} x_{1} & + & 3 x_{2} & - & 4 x_{3} & - & 8 x_{4} & + & 6 x_{5} & = & 0 \\ x_{1} & + & 2 x_{3} & + & x_{4} & + & 3 x_{5} & = & 0 \\ 2 x_{1} & + & 7 x_{2} & - & 10 x_{3} & - & 19 x_{4} & + & 13 x_{5} & = & 0 \end{array}$
x12x13x1+++2x24x26x2+++7x37x35x3−−−9x411x411x4+++31x534x529x5===000 $\begin{array}{r} x_{1} & + & 2 x_{2} & + & 7 x_{3} & - & 9 x_{4} & + & 31 x_{5} & = & 0 \\ 2 x_{1} & + & 4 x_{2} & + & 7 x_{3} & - & 11 x_{4} & + & 34 x_{5} & = & 0 \\ 3 x_{1} & + & 6 x_{2} & + & 5 x_{3} & - & 11 x_{4} & + & 29 x_{5} & = & 0 \end{array}$
3x15x12x1+++x28x25x2−+3x32x3+−−11x42x4x4+++10x57x514x5===000 $\begin{array}{r} 3 x_{1} & + & x_{2} & - & 3 x_{3} & + & 11 x_{4} & + & 10 x_{5} & = & 0 \\ 5 x_{1} & + & 8 x_{2} & + & 2 x_{3} & - & 2 x_{4} & + & 7 x_{5} & = & 0 \\ 2 x_{1} & + & 5 x_{2} & - & x_{4} & + & 14 x_{5} & = & 0 \end{array}$

Problems 27 through 36 further explore independent sets, spanning sets, and bases.

Suppose that S is a set of n linearly independent vectors in the n-dimensional vector space V. Prove that S is a basis for V.
Suppose that S is a set of n vectors that span the n-dimensional vector space V. Prove that S is a basis for V.
Let {v1,v2,…,vk} ${v_{1}, v_{2}, \dots, v_{k}}$ be a basis for the proper subspace W of the vector space V, and suppose that the vector v of V is not in W. Show that the vectors v1,v2,…,vk,v $v_{1}, v_{2}, \dots, v_{k}, v$ are linearly independent.
Use the result of Problem 29 to prove that every linearly independent set of vectors in a finite-dimensional vector space V is contained in a basis for V.
Suppose that the vectors v1,v2,…,vk,vk+1 $v_{1}, v_{2}, \dots, v_{k}, v_{k + 1}$ span the vector space V and that vk+1 $v_{k + 1}$ is a linear combination of v1,v2,…,vk $v_{1}, v_{2}, \dots, v_{k}$ . Show that the vectors v1,v2,…,vk $v_{1}, v_{2}, \dots, v_{k}$ span V.
Use the result of Problem 31 to prove that every spanning set for a finite-dimensional vector space V contains a basis for V.
Let S be a linearly independent set of vectors in the finite-dimensional vector space V. Then S is called a maximal linearly independent set provided that if any other vector is adjoined to S, then the resulting set is linearly dependent. Prove that every maximal linearly independent set in V is a basis for V.
Let S be a finite set of vectors that span the vector space V. Then S is called a minimal spanning set provided that no proper subset of S spans V. Prove that every minimal spanning set in V is a basis for V.
Let S be a finite set of vectors that span the vector space V. Then S is called a uniquely spanning set provided that each vector in V can be expressed in one and only one way as a linear combination of the vectors in S. Prove that every uniquely spanning set in V is a basis for V.
Apply the definition of linear independence to show directly that the column vectors of the matrix in (12) are linearly independent.

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for 4.4 Bases and Dimension for Vector Spaces

Create new playlist

Sign In

Sign Up

Table of Contents for
4.4 Bases and Dimension for Vector Spaces