Vector Spaces – Chapter 4

Doreen De LeonDepartment of Mathematics, California State University, Fresno

October 19, 2015

1 Why Study Vector Spaces?

Vector spaces have many applications. For one, they provide a framework to deal with an-alytical and geometrical problems, and are used in the Fourier transform. There are alsoapplications of vector spaces in optimization theory. The minimax theorem of game theorystating the existence of a unique payoff when all players play optimally can be formulatedand proven using vector space methods. Least squares estimation, which is used in amongother areas digital filter design, tracking (Kalman filters), control systems, etc. Represen-tation theory basically transforms problems in abstract algebra, particularly group theory,to problems in linear algebra. Such discussions are, however, beyond the scope of this class.Therefore, we limit ourselves to the above brief discussion, and launch directly into the topicat hand.

2 Introduction to Vector Spaces – Sections 4.1-4.2, 4.7

2.1 Vectors in Rn

We will use some familiar ideas about vectors to motivate the idea of a vector space, whichwe will introduce a bit later. First, a vector can be thought of as a directed line segment (orarrow) that has both magnitude and direction. In Calculus III, you probably defined vectoraddition in two ways:

• using the parallelogram law, where x + y is the diagonal of the parellelogram formedby x and y; and

1

• by adding corresponding elements of x and y to obtain x + y.

Then, to form the sum of three vectors, we simply add two vectors as above and then addthe third vector to the result. This can be performed equivalently by adding the first twovectors together, and then adding the third vector to the sum; or, by adding the last twovectors together, and then adding the first vector to the sum. So, we have that for all vectorsx, y, and z

x + y = y + x,

x + (y + z) = (x + y) + z.

The zero vector, denoted 0, is defined as the vector satisfying

x + 0 = x,

for all vectors x. We consider the zero vector as having zero magnitude and arbitrarydirection. Geometrically, we picture the zero vector as corresponding to a point in space.Let −x denote the vector that has the same magnitude as x but points in the oppositedirection. Then, if we add −x and x using either approach above, we obtain

x + (−x) = 0.

The vector −x so defined is the additive inverse of x. The above properties are thefundamental properties of vector addition.

We also need to define the operation of multiplication of a vector by a scalar, i.e., scalarmultiplication. Geometrically, if x is a vector and c is a scalar, then cx is defined by thevector whose magnitude is |c| times the magnitude of x and whose direction is the same as xif c > 0 and the opposite of x if c < 0. If c = 0, then cx = 0. Scalar multiplication has severalimportant properties, which can easily be verified, either geometrically or algebraically usingthe definitions of vector addition and scalar multiplication. For all vectors x and y and allscalars a, and b,

1x = x,

(ab)x = a(bx),

a(x + y) = ax + ay,

(a + b)x = ax + bx.

If you notice, we did not define “multiplication” of vectors. However, from Calculus III,you are familiar with the dot product and cross product of vectors. For the purpose oftalking about vector spaces, we will ignore these two operations on vectors, only concerningourselves with vector addition and scalar multiplication.

2

If we consider vectors in the plane, denoted R2, then each vector v = (v1, v2) can be iden-tified with a point in space with coordinates (v1, v2), and the direction and magnitude ofv are determined by considering the arrow with vertex at the origin, (0, 0). Then, for allvectors v = (v1, v2) and w = (w1, w2) and all scalars a, we define vector addition and scalarmultiplication as

v + w = (v1, v2) + (w1, w2) = (v1 + w1, v2 + w2) ,

av = a (v1, v2) = (av1, av2) .

We can easily verify the above listed properties for vectors in R2. For example, 0 = (0, 0)and since

(v1, v2) + (−v1,−v2) = (0, 0),

the additive inverse of v is −v = (−v1,−v2).

It is straightforward to extend these ideas to vectors in space, denoted R3 by simply addinga third component. Each vector v = (v1, v2, v3) can be identified with a point in space withcoordinates (v1, v2, v3), and the direction and magnitude of v are determined by consideringthe arrow with vertex at the origin, (0, 0, 0). Vector addition and scalar multiplication aredefined as follows. If v = (v1, v2, v3) and w = (w1, w2, w3) and c is a scalar,

v + w = (v1, v2, v3) + (w1, w2, w3) = (v1 + w1, v2 + w2, v3 + w3) ,

av = c (v1, v2, v3) = (cv1, cv2, cv3) .

And again, we can show that the properties given above for vector addition and scalarmultiplication are satisfied by R3, with 0 = (0, 0, 0) and −v = (−v1,−v2,−v3).

We can generalize this notion of vectors that we can visualize to vectors with any number ofcomponents. The set of all vectors with n real-valued components is denoted by Rn. Vectoraddition and scalar multiplication in Rn are defined componentwise. Given any two vectorsu and v in Rn, defined by

u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) ,

and any scalar c, the standard vector addition and scalar multiplication are defined as follows:

u + v = (u1 + v1, u2 + v2, . . . , un + vn)

cu = (cu1, cu2, . . . , cun)

We can show that for any positive integer n, Rn possesses all of the algebraic propertiesfor vector addition and scalar multiplication as R2 and R3. These properties give us theframework to define a general notion, that of vector spaces.

3

2.2 Vector Spaces

We will start with a set of elements V , which we will call vectors. They may be vectors inthe sense of vectors discussed previously, or they may be some other mathematical construct,such as functions, matrices, or polynomials.

Associated with this set of vectors are two operations, vector addition and scalar multipli-cation.

• Vector addition can be thought of as a rule for combining two vectors in V . To simplifythings notationally, we will use the + sign to denote vector addition. So, the result ofadding two vectors u and v will be denoted u + v.

• Scalar multiplication can be thought of as a rule for combining each vector in V withany scalar, and we will use the usual notation av to denote the result of multiplyingthe vector v by the scalar a.

So, what is a vector space?

Definition: Let V be a set of elements called vectors, in which the operations of additionof vectors and multiplication of vectors by scalars are defined. Then, given any vectorsu,v,w ∈ V and any scalars a and b, V is a vector space if all of the following propertiesare satisfied.

(1) u + v ∈ V (V is closed under vector addition).

(2) au ∈ V (V is closed under scalar multiplication).

(3) u + v = v + u (commutativity).

(4) u + (v + w) = (u + v) + w (associativity).

(5) There exists a zero element in V , 0 ∈ V , such that u + 0 = 0 + u = u.

(6) There exists an element in V , −u, called the additive inverse, such that

u + (−u) = −u + u = 0.

(7) a(u + v) = au + av.

(8) (a + b)u = au + bu.

(9) a(bu) = (ab)u.

(10) (1)u = u.

4

2.3 Subspaces

Definition: Let W be a nonempty subset of the vector space V . Then W is a subspaceof V provided that W is itself a vector space with the operations of addition and scalarmultiplication as defined in V .

Theorem 1. A subset W of a vector space V is a subspace of V if and only if it satisfies

(1) W contains the zero vector for V ;

(2) W is closed under vector addition (i.e., for every u,v ∈ W , then u + v ∈ W ); and

(3) W is closed under scalar multiplication (i.e., for every u ∈ W and c is a scalar,then cu ∈ W ).

Examples:

1) W = {(x, y, z) : z = 0}. Is W a subspace of R3? Why or why not?

• zero check0 = (0, 0, 0) ∈ W. X

• closure under vector addition

Let u = (u1, u2, 0) ∈ W,

v = (v1, v2, 0) ∈ W

u + v = (u1 + v1, u2 + v2, 0) ∈ W. X

• closure under scalar multiplication

Let c be a scalar and let u = (u1, u2, 0) ∈ W .

cu = (cu1, cu2, 0) ∈ W. X

Since W ⊂ R3 contains the zero vector and is closed under vector addition and scalarmultiplication, W is a subspace of R3.

2) W = {(x, y, z) : y = 1} Is W a subspace of R3?

(0, 0, 0) /∈ W , so W is not a subspace of R3.

Note that in addition, we can show that this set is not closed under vector addition orscalar multiplication, but it suffices to show that only one property is not satisfied.

5

3) W = {(x, y, z) : x2 + y2 − z2 = 0}. Is W a subspace of R3?

Here, we see that W contains the zero vector, so at first glance it might seem that Wis a subspace of R3. We must check if this is true.

• zero check0 = (0, 0, 0) ∈ W since 02 + 02 − 02 = 0.

• closure under vector addition

Let u = (u1, u2, u3) ∈ W =⇒ u21 + u2

2 − u23 = 0,

v = (v1, v2, v3) ∈ W =⇒ v21 + v22 − v23 = 0

u + v = (u1 + v1, u2 + v2, u3 + v3) .

We know that u + v ∈ W if (u1 + v1)2 + (u2 + v2)

2 − (u3 + v3)2 = 0.

(u1 + v1)2 + (u2 + v2)

2 − (u3 + v3)2 = u2

1 + 2u1v1 + v21 + u22 + 2u2v2 + v22 − u2

3 − 2u3v3 − v23

=(u21 + u2

2 − u23

)+(v21 + v22 − v23

)+ 2u1v1 + 2u2v2 − 2u3v3

= 0 + 0 + 2 (u1v1 + u2v2 − u3v3)

= 2 (u1v1 + u2v2 − u3v3) 6= 0 for arbitrary vectors u and v.

If we have an idea that a set is not a subspace of a vector space, then we can simplyfind an example where closure under either vector addition or scalar multiplicationfails. In this case, we choose

u = (1, 0, 1) ∈ W (since 12 + 02 − 12 = 0)

v = (0, 1, 1) ∈ W (since 01 + 12 − 12 = 0).

We see that u + v = (1, 1, 2) /∈ W (since 12 + 12 − 22 = −2 6= 0). So, W is not closedunder vector addition, and therefore, W is not a subspace of R3.

2.4 Solution Subspaces

Theorem 2. If A is a constant m× n matrix, then the solution set of the system

Ax = 0 (1)

is a subspace of Rn, called the solution space of the system.

Proof. Let W be the set of all solutions to (1). Verify the conditions for a subspace.

• zero check: 0 ∈ W since the trivial solution is always a solution to a homogeneoussystem of equations.

6

• closure under addition: If u,v ∈ W then

Au = 0 and Av = 0

Is u + v ∈ W?A(u + v) = Au + Av = 0 + 0 = 0.

Since A(u + v) = 0, u + v is a solution to (1) and u + v ∈ W .

• closure under scalar multiplication: If u ∈ W and c is a scalar, then

A(cu) = c(Au) = c0 = 0.

Since A(cu) = 0, cu is a solution to (1) and cu ∈ W .

For all of the above reasons, W is a subspace of Rn.

2.5 General Vector Spaces

The term “vector” in vector space can be interpreted in a more general sense.

Examples:

(1) Given m and n positive integers, define Mmn as the set of all m×n matrices with realentries. Then Mmn is a vector space.

→ Matrices play the role of vectors, with

• matrix addition defining “vector” addition

• multiplication of a matrix by a scalar defining scalar multiplication

It is easily verified using these definitions of vector addition and scalar multiplicationthat Mmn satisfies properties (1)-(10) of the definition of a vector space.

(2) F is the set of all real-valued functions defined on R.

→ Functions play the role of vectors, with the following definitions for vector additionand scalar multiplication.

• For f, g ∈ F , (f + g)(x) = f(x) + g(x) ← vector addition

• For f ∈ F and c ∈ R, (cf)(x) = cf(x) ← scalar multiplication.

It is easily verified using these definitions of vector addition and scalar multiplicationthat F is a vector space, with the zero element being the function f(x) = 0 (i.e., thefunction whose value is zero for all x).

7

(3) Cn[a, b] is the set of all continuous functions with n continuous derivatives on [a, b].

Cn[a, b] is a vector space with the same properties as F .

(4) P is the set of all polynomials.

P is a vector space with the polynomials playing the role of vectors.

Note: P is also a subspace of F since P ⊂ F .

(5) Pn is the set of all polynomials of degree at most n.

Pn is a vector space with the same properties as P .

In fact, Pn is a subspace of P .

2.6 Subspaces of General Vector Spaces

In order to determine if a subset of a vector space is a subspace, we must apply Theorem 1.In other words, we must determine if closure under vector addition and scalar multiplicationare satisfied.

Examples:

(1) W is the set of diagonal 2× 2 matrices. Is W a subspace of M22? Why or why not?

• zero check

0 =

(0 00 0

)∈ W. X

• closure under vector addition

First, we need to choose two arbitrary vectors in W . Let

A =

(a1 00 a2

)∈ W,B =

(b1 00 b2

)∈ W.

Then,

A + B =

(a1 00 a2

)+

(b1 00 b2

)=

(a1 + b1 0

0 a2 + b2

)∈ W. X

8

• closure under scalar multiplication

Let c be a scalar and let A ∈ W . Then

cA = c

(a1 00 a2

)=

(ca1 00 ca2

)∈ W. X

Since W satisfies all of the above properties, W is a subspace of M22.

(2) W is the set of all solutions of the differential equation

y′ + p(x)y = 0.

Is W a subspace of C1(I)? Why or why not?

• zero check

The zero element is in W , since the function y = 0 is a solution to the differentialequation.

• closure under vector addition

First, we need to choose two arbitrary vectors in W . Let

f, g ∈ W =⇒ f ′ + p(x)f = 0g′ + p(x)g = 0

Then, we need to determine if f + g ∈ W . In other words, we need to check iff + g is a solution to the differential equation:

(f + g)′ + p(x)(f + g) = f ′ + g′ + p(x)f + p(x)g

= (f ′ + p(x)f) + (g′ + p(x)g)

= 0 + 0 (since f, g ∈ W )

= 0

=⇒ f + g ∈ W X.

• closure under scalar multiplication

Let c be a scalar and let f ∈ W .

We need to determine if cf ∈ W . In other words, we need to check if cf is asolution to the differential equation:

(cf)′ + p(x)(cf) = cf ′ + cp(x)f

= c(f ′ + p(x)f)

= c(0) (since f ∈ W )

= 0

=⇒ cf ∈ W X.

9

Since W satisfies the above properties, W is a subspace of F .

(3) W is the set of all solutions of the differential equation

y′ + p(x)y = x.

Is W a subspace of C1(I)? Why or why not?

W does not contain the zero element (the function y = 0 is not a solution, because0′ + p(x) · 0 = 0 6= x). Therefore, W is not a subspace of F .

(4) W is the set of all polynomials of the form a0 + a1x + a2x2 such that a0 = 2a2. Is W

a subspace of P2? Why or why not?

• zero check

The zero element, 0 = 0 + 0x + 0x2 ∈ W . X

• closure under vector addition

First, we need to choose two arbitrary polynomials in W . Let

p, q ∈ W =⇒ p(x) = a0 + a1x + a2x2, where a0 = 2a2, or p(x) = 2a2 + a1x + a2x

2

q(x) = b0 + b1x + b2x2, where b0 = 2b2, or q(x) = 2b2 + a1x + b2x

2.

Then, we need to determine if p + q ∈ W .

(p + q)(x) = p(x) + q(x)

=(2a2 + a1x + a2x

2)

+(2b2 + b1x + b2x

2)

= 2(a2 + b2) + (a1 + b1)x + (a2 + b2)x2.

So, p + q ∈ W . X

• closure under scalar multiplication

Let c be a scalar and let p ∈ W .

We need to determine if cp ∈ W .

(cp)(x) = cp(x)

= c(2a2 + a1x + a2x2)

= 2(ca2) + ca1x + (ca2)x2.

So, cp ∈ W . X

Since W satisfies all of the above properties, W is a subspace of P2

10

(5) W is the set of all polynomials in P2 whose coefficients are odd integers. Is W asubspace of P2? Why or why not?

0 = 0 + 0x + 0x2 /∈ W , because its coefficients are all 0 and 0 is not an odd integer.Therefore, W is not a subspace of P2.

3 Linear Combinations and Independence of Vectors –

Section 4.3, 4.7

Definition: A linear combination of vectors v1,v2, . . . ,vn is defined as

c1v1 + c2v2 + · · ·+ cnvn

for some constants c1, c2, ...cn.

Example: v1 = (1, 0, 0), v2 = (0, 1, 0), v3 = (0, 0, 1). Then

(1, 2, 3) = 1(1, 0, 0) + 2(0, 1, 0) + 3(0, 0, 1)

is a linear combination of v1,v2,v3.

Definition: Suppose that v1,v2, . . . ,vn are vectors in a vector space V . Then v1,v2, . . . ,vn

span V if every vector v ∈ V can be written as a linear combination of v1,v2, . . . ,vn.

Examples:

(1) Do v1 = (1, 0, 0), v2 = (0, 1, 0), v3 = (0, 0, 1) span R3?

Yes, becuase every vector w = (x1, x2, x3) ∈ R3 can be written as

w = x1v1 + x2v2 + x3v3.

(2) Do v1 = (1, 0), v2 = (1, 1), v3 = (2, 1) span R2?

We must see if we can find constants c1, c2, c3 that solve c1v1 + c2v2 + c3v3 = w,where w = (a, b) is an arbitrary vector in R2, or

c1(1, 0) + c2(1, 1) + c3(2, 1) = (a, b).

This yields the system of equations:

c1 + c2 + 2c3 = a

c2 + c3 = b

11

Solving: (1 1 2 | a0 1 1 | b

)is already in row echelon form. We see that c3 is arbitrary, so we know that there areinfinitely many solutions to this system of equations =⇒ the vectors span R2.

(3) Do v1 = (1, 1, 0), v2 = (0, 1, 1) span R3?

We must see if we can find constants c1, c2 that solve c1v1 + c2v2 = w, where w =(a, b, c) is an arbitrary vector in R3, or

c1(1, 1, 0) + c2(0, 1, 1) = (a, b, c).

This yields the system of equations:

c1 = a

c1 + c2 = b.

c2 = c

Solving:1 0 | a1 1 | b0 1 | c

r2→r2−r1−−−−−−→

1 0 | a0 1 | b− a0 1 | c

r3→r3−r2−−−−−−→

1 0 | a0 1 | b− a0 0 | c− b + a

The last equation gives us 0 = c − b + a. This is not true for every a, b, and c.Therefore, the vectors do not span R3.

Theorem 3. Let v1,v2, . . . ,vk be vectors in a vector space V . The set W of all linearcombinations of v1,v2, . . . ,vk is a subspace of V .

The set W defined in the theorem is denoted by W = span{v1,v2, . . . ,vk}.

Example: Is v = (1, 2, 1) in span{(1, 0, 0), (1, 1, 1), (2,−1, 0)}?

We must determine if there are constants c1, c2, c3 that solve

(1, 2, 1) = c1(1, 0, 0) + c2(1, 1, 1) + c3(2,−1, 0).

Simplifying, we obtain(1, 2, 1) = (c1 + c2 + 2c3, c2 − c3, c2) .

This gives the following system of equations:

c1 + c2 + 2c3 = 1

c2 − c3 = 2.

c2 = 1

12

Solving, c2 = 1 =⇒ c3 = −1. Then, plugging into the first equation, we obtain c1 = 2.Since we can solve for c1, c2, and c3, v is in the span of the three vectors.

Definition: The vectors v1,v2, . . . ,vn in a vector space V are linearly independent if

c1v1 + c2v2 + · · ·+ cnvn = 0

is only true for c1 = c2 = · · · = cn = 0. Otherwise, the vectors are linearly dependent.

The idea is that if one or more of the c1, c2, . . . , cn are nonzero, we can write one vector interms of the others (so, in some sense, the value of one vector depends on the value of theothers), so the vectors are linearly dependent.

Examples:

(1) Are v1 = (1, 0, 1), v2 = (1, 1, 0), and v3 = (0, 1, 1) linearly independent? Why or whynot?

We must determine the constants c1, c2, c3 that solve c1v1 + c2v2 + c3v3 = 0, or

c1(1, 0, 1) + c2(1, 1, 0) + c3(0, 1, 1) = (0, 0, 0).

This yields the system of equations:

c1 + c2 = 0

c2 + c3 = 0.

c1 + c3 = 0

Solving:1 1 0 | 00 1 1 | 01 0 1 | 0

r2→r2−r1−−−−−−→

1 1 0 | 00 1 1 | 00 −1 1 | 0

r3→r3+r2−−−−−−→

1 1 0 | 00 1 1 | 00 0 2 | 0

Solving, we see that c1 = c2 = c3 = 0, and the vectors are linearly independent.

(2) Are v1 = (1, 1, 1), v2 = (1, 0, 1), and v3 = (0, 1, 0) linearly independent? Why or whynot?

By inspection,

(1, 1, 1) = (1, 0, 1) + (0, 1, 0)

=⇒ v1 = v2 + v3

=⇒ v1 − v2 − v3 = 0

Therefore, the vectors are not linearly independent.

13

If we could not see this by inspection, we would try to determine the constants c1, c2, c3that solve c1v1 + c2v2 + c3v3 = 0, or

c1(1, 1, 1) + c2(1, 0, 1) + c3(0, 1, 0) = (0, 0, 0).

This yields the system of equations:

c1 + c2 = 0

c1 + c3 = 0.

c1 + c2 = 0

Solving: 1 1 0 | 01 0 1 | 01 1 0 | 0

r2→r2−r1−−−−−−→r3→r3−r1

1 1 0 | 00 −1 1 | 00 0 0 | 0

From this, we see that c3 is arbitrary. Therefore, the trivial solution is not the onlysolution, and the vectors are not linearly independent.

(3) Are 1 + x, 1− x, 1− x2 linearly independent? Why or why not?

We need to determine what constants c1, c2, c3 solve

c1(1 + x) + c2(1− x) + c3(1− x2) = 0.

Collecting terms, we obtain

−c3x2 + (c1 − c2)x + (c1 + c2 + c3) = 0.

Equate coefficients of like terms:

c1 + c2 + c3 = 0

c1 − c2 = 0.

−c3 = 0

Solving, we see that c1 = c2 = c3 = 0, and the functions are linearly independent.

For functions in C(n−1)[a, b], there is a different approach for determining linear independence.Let f1, f2, . . . , fn ∈ C(n−1)[a, b]. We will start our search instead by looking for a direct wayto determine if the functions are linearly dependent. If the functions are linearly dependent,then there exist constants c1, c2, . . . , cn not all zero such that

c1f1(t) + c2f2(t) + · · · cnfn(t) = 0 (2)

for each t ∈ [a, b].

14

Differentiating both sides of (2) with respect to t gives

c1f′1(t) + c2f

′2(t) + · · · cnf ′n(t) = 0.

We may continue differentiating to obtain a system of n equations in n unknowns (the ci).

c1f1(t) + c2f2(t) + · · · + cnfn(t) = 0c1f′1(t) + c2f

′2(t) + · · · + cnf

′n(t) = 0

......

c1f(n−1)1 (t) + c2f

(n−1)2 (t) + · · · + cnf

(n−1)n (t) = 0.

(3)

Writing (3) in matrix form gives:f1(t) f2(t) · · · fn(t)f ′1(t) f ′2(t) · · · f ′n(t)

......

...

f(n−1)1 (t) f

(n−1)2 (t) · · · f

(n−1)n (t)

c = 0, (4)

where c = (c1, c2, . . . , cn). So, if the functions are linearly dependent in C(n−1)[a, b], for eacht ∈ [a, b], the coefficient matrix in (4) is singular. In other words, the determinant of thecoefficient matix in (4) is 0 for each t ∈ [a, b]. This leads to the following definition anduseful theorem.

Definition: Let f1, f2, . . . , fn ∈ C(n−1)[a, b]. The Wronskian of f1, f2, . . . , fn on [a, b],denoted W [f1, f2, . . . , fn](t), is the determinant of the coefficient matrix in (4).

Theorem 4. Let f1, f2, . . . , fn ∈ C(n−1)[a, b]. If there exists a point t0 ∈ [a, b] such thatW [f1, f2, . . . , fn](t0) 6= 0 then f1, f2, . . . , fn are linearly independent.

Example: Are the functions f1(x) = tan x, f2(x) = sec x linearly independent?

W (x) =

∣∣∣∣ f1(x) f2(x)f ′1(x) f ′2(x)

∣∣∣∣=

∣∣∣∣ tanx secxsec2 x secx tanx

∣∣∣∣= secx tan2 x− secx sec2 x

= secx(tan2 x− sec2 x)

= − secx

If x = 0, then W (x) = −1 6= 0. Therefore, the functions are linearly independent.

What is the point of all of this work?

15

4 Basis and Dimension for Vector Spaces – Sections

4.4, 4.7

Idea: A basis represents the “building blocks” of a vector space.

Example: i = (1, 0) and j = (0, 1) form a basis of R2.

Definition: A set of vectors S in a vector space V is a basis of V if

(i) the vectors in S are linearly independent, and

(ii) the vectors in S span V .

Example: R2

(a, b)ab

(a, b) = (a, 0) + (0, b)

= a(1, 0) + b(0, 1)

→ We can write any vector in R2 as a linear combination of (1, 0) and (0, 1)=⇒ (1, 0), (0, 1) span R2.

(1, 0) and (0, 1) are linearly independent:

c1(1, 0) + c2(0, 1) = (0, 0) =⇒ c1 = 0 = c2.

So, sincespan

+linear independence

= basis,

16

{(1, 0), (0, 1)} is a basis of R2.

Is {(1, 0), (0, 1)} the only basis of R2? No. There are infinitely many bases of R2.

Example: {(1, 1), (0, 1)} is a basis of R2. Why?

• {(1, 1), (0, 1)} spans R2

(a, b) = a(1, 1) + (b− a)(0, 1)

• (1, 1) and (0, 1) are linearly independent:

c1(1, 1) + c2(0, 1) = (0, 0)

=⇒ c1 = 0

c1 + c2 = 0 =⇒ c2 = 0.

Theorem 5. Let S = {v1, v2, . . . , vn} be a basis of vector space V . Then any set of morethan n vectors is linearly dependent.

Example: v1 = (1, 2), v2 = (1, 0), v3 = (0, 1) are linearly dependent.(We can also see this since (1, 2) = 1(1, 0) + 2(0, 1)).

Theorem 6. Any two bases of a vector space contain the same number of vectors.

Definitions:

• A nonzero vector space V is called finite-dimensional provided that there exists abasis for V consisting of a finite number of vectors in V .

• The number of vectors in each basis of a finite-dimensional vector space V is thedimension of V , denoted dim(V ).

Examples:

(1) dim(R2) = 2.

(2) dim(Rn) = n.

(3) M22 =

(a bc d

)is a 4-dimensional vector space. Why? A basis of M22 is{(

1 00 0

),

(0 10 0

),

(0 01 0

),

(0 00 1

)}.

17

(4) A basis of Pn is {1, x, x2, . . . , xn} =⇒ dim(Pn) = n + 1.

Note: dim({0}) = 0.

Idea so far: Every basis of a (finite-dimensional) vector space has the same number of vectors.The number of vectors in a basis of a vector space V is called its dimension, denoted dim(V ).

Why do we care?

Theorem 7. Let V be an n-dimensional vector space and let S be a subset of V . Then

(a) If S is linearly independent and consists of n vectors, S is a basis of V .

(b) If S spans V and consists of n vectors, then S is a basis of V .

(c) If S is linearly independent, then S is contained in a basis of V .

(d) If S spans V , then S contains a basis of V .

Idea: If we know the dimension of a vector space, then determining if a set of vectors is abasis becomes easier:

correct # of vectors+

linear independence

= basis, orcorrect # of vectors

+span

= basis.

If either of the two conditions fails to hold, the set of vectors is not a basis.

Examples:

(1) Is {(1, 0), (2, 1), (−1, 2)} a basis of R2? Why or why not?

No. A basis of R2 must have two vectors (dim(R2) = 2) and our set has three vectors.

Can we find a basis of R2 that is a subset of these vectors?

Yes! {(1, 0), (2, 1)} is a basis for R2. (Verify this.)

(2) Is {(1, 0, 1), (0, 1, 0), (1,−1, 1)} a basis of R3? Why or why not?

No. (1, 0, 1) + (−1)(0, 1, 0) = (1,−1, 1) =⇒ the vectors are linearly dependent.

(3) Is {(1, 1, 1), (0, 1, 0)} a basis for R3? Why or why not?

No. dim(R3) = 3 =⇒ need three vectors for a basis, and our set has only two vectors.

18

Can we find a basis of R3 that contains these vectors?

Yes! The vectors are linearly independent, so we just need to find a third vec-tor to add to the set that will make the set linearly independent. For example,{(1, 1, 1), (0, 1, 0), (0, 0, 1)} is a basis for R3. (Verify this.)

(4) Is {1− x, x2, 1 + 2x} a basis of P2? Why or why not?

dim(P2) = 3, so we need only verify that the functions are linearly independent.

c1(1− x) + c2(x2) + c3(1 + 2x) = 0

c2x2 + (−c1 + 2c3)x + c1 + c3 = 0

Equating coefficients of like terms gives:

c2 = 0

−c1 + 2c3 = 0 =⇒ c1 = 2c3

c1 + c3 = 0 =⇒ 2c3 + c3 = 0 =⇒ c3 = 0

=⇒ c1 = 0

Since c1 = c2 = c3 = 0 the functions are linearly independent =⇒ the set is a basisof P2.

Note: We could have used the Wronskian to determine if the polynomials are linearlyindependent.

Example: Find a basis of the subspace S of M33 defined by the set of 3 × 3 symmetricmatrices.

The first step is to write an arbitrary element of the set S,

A =

a b cb d ec d f

.

Then, use this to determine a set of matrices that spans:a b cb d ec e f

=

a 0 00 0 00 0 0

+

0 b 0b 0 00 0 0

+

0 0 c0 0 0c 0 0

+

0 0 00 d 00 0 0

+

0 0 00 0 e0 e 0

+

0 0 00 0 00 0 f

= a

1 0 00 0 00 0 0

+ b

0 1 01 0 00 0 0

+ c

0 0 10 0 01 0 0

+ d

0 0 00 1 00 0 0

+ e

0 0 00 0 10 1 0

+ f

0 0 00 0 00 0 1

.

19

Therefore, a basis for S is1 0 0

0 0 00 0 0

,

0 1 01 0 00 0 0

,

0 0 10 0 01 0 0

,

0 0 00 1 00 0 0

,

0 0 00 0 10 1 0

,

0 0 00 0 00 0 1

,

since this set of matrices spans S and is linearly independent (easily verified).

4.1 Basis for Solution Spaces

The procedure for determining the basis of a solution space will be illustrated by example.

Examples:

(1) Find a basis for the solution space of

3x1 + 4x2 + 5x3 = 0

2x1 + 3x2 + 4x3 = 0.(3 4 5 | 02 3 4 | 0

)r2→r2− 2

3r1−−−−−−→(

3 4 5 | 00 1

323| 0

)r2→3r2−−−−→

(3 4 5 | 00 1 2 | 0

)So,

x2 + 2x3 = 0

=⇒ x2 = −2x3

3x1 + 4x2 + 5x3 = 0

=⇒ x1 =−5x3 − 4x2

3

x3 is arbitrary, so let x3 = r. Then

x2 = −2r

x1 =−5r − 4(−2r)

3= r

So, x = (r,−2r, r) = r(1,−2, 1)=⇒ {(1,−2, 1)} is a basis for the solution space.

(2) Find a basis for the plane defined by

2x + 3y + z = 0.

Solving for x gives

x =−3y − z

2.

20

y and z are arbitrary, so let y = t and z = w.

=⇒ x = −3t

2− w

2.

So,

x =

t(−3

2

)+ w

(−1

2

)tw

= t

−32

10

+ w

−12

01

.

Therefore, −3

2

10

,

−12

01

for the plane.

21