1. Vector Spaces

In this section we state the axioms of a vector space; discuss the two major types of vector spaces that will be of interest to us, n-dimensional Euclidean space and function spaces; define what is meant by a subspace, and then examine some examples of subspaces.

The Axioms that Define a Vector space

A vector space V is a non-empty subset with elements u, v, w, .. , together with two operations, + and *, such that:

I. If v and w are elements of V then v + w is an element of V.

II. If c is any real number then c *v is an element of V.

In addition

III. If v and w are elements of V then v + w = w+v.

IV. If u, v and w are elements of V then u+ (v + w) = (u+w)+v.

V. There is a vector 0 in V such that v + 0 = v for all v in V; 0 is

called the 0 vector.

VI. Given a vector v in V, there is a vector w in V such that v + w = 0,

where 0 is the 0 vector. If v + w = 0, we write w = -v.

VII. If c and d are constants and v and w are vectors then

(c + d) *v = c*v + d*v, c*(v+w) = c*v + c*w,
(c d)*v = c*(d*v), 1*v = v.

Axiom I says that if v and w are vectors then the sum v + w is also a vector; this property is usually referred to as additive closure. Axiom II says that if c is a real number and v is a vector then c*v is also a vector; this property is called closure with respect to scalar multiplication.

The remaining axioms describe common features of mathematical systems:

III says that addition of vectors is commutative; IV says addition of vectors is associative; V describes 0 as the additive identity for vectors; VI says that any given vector v has an additive inverse w; VII describes various distributive laws for mixes of constants and vectors.

Example: 2-dimensional Euclidean Space

The two-dimensional plane

R(2) = { (x, y) : x real , y real}

is a familiar example of a vector space. Addition of vectors is defined by addition of components: If v = (x1, y1) and w = (x2, y2) then

v + w = (x1, y1) + (x2, y2) = (x1 + x2 , y1 + y2).

Thus, there is closure with respect to vector addition. If c is a constant and v =
(x, y), then c*v is defined by

c*v = c*(x, y) = (c x, c y),

so there is closure with respect to scalar multiplication.

Addition of vectors in R(2) is commutative. If v = (x1, y1) and w = (x2, y2) then,

since x1 + x2 = x2 + x1 for real numbers x1 and x2,

v + w = (x1, y1) + (x2, y2) = (x1 + x2 , y1 + y2) =

(x2 + x1 , y2 + y1) = (x2, y2) + (x1, y1) = w + v.

The commutativity of vectors in this space is said to hold "by inheritance". It holds,

as the above argument shows, because the addition of real numbers is commutative.

Associativity of vectors is also inherited from associativity of real numbers. Let u = (x1, y1), v = (x2, y2), w = (x3, y3). Then, since x1 + (x2 + x3) = (x1 + x2)+ x3 for

real numbers x1, x2 and x3, it follows that

u + (v + w) =

(x1, y1) + ((x2, y2) + (x3, y3)) =

(x1, y1) + (x2 + x3, y2 + y3) =

(x1 +( x2 + x3) , y1 + (y2 + y3)) =

(x1 +x2) + x3) , (y1 + y2)+ y3)) =

(x1 +x2, y1 + y2)+ (x3, y3) =

(u + v) + w.

The vector 0 = (0,0) is the additive identity: If v = (x, y) then

v + 0 = (x, y) + (0,0) = (x+0, y + 0) = (x, y).

Furthermore (-x, -y) is the additive inverse of the vector (x, y), for

(x, y) + (-x, -y) = (x +( -x), y +(-y)) = (0, 0) = 0.

The verification of the distributive laws of axiom 7 is based on repeated use of the definition of scalar multiplication and properties of the real numbers. For example, if v = (x, y) then

(c + d)*v =

(c + d)*(x, y) = ((c+d) x, (c+d) y) =

(c x + d x, c y + d y) = (c x, c y) + (d x, d y) =

c*(x, y) + d*(x, y) =

c*v + d*v.

The check that c*(v + w), (c d)*v = c*(d*v), and 1*v = v follows similar lines.

Thus the seven axioms for a vector space hold in R(2) when vector addition and scalar multiplication defined in the usual way.

Example: n-dimensional Euclidean Space

If R(3) is ordinary 3-dimensional space,

R(3) = { (x, y, z) : x real, y real, z real},

with addition of 3-tuples defined by

(x1, y1, z1) + (x2, y2, z2) = (x1 + x2, y1 + y2, z1 + z2)

and scalar multiplication of vectors defined by

c*(x, y, z) = (c x, c y, c z)

then the axioms of a vector space hold for R(3). The proof is almost identical to the one given for R(2); just replace 2-tuples by 3-tuples in the argument for R(2).

In general, if R(n) is defined to be the set of all real n-tuples

R(n) = { (x1, x2, .. , xn) : xi real, i = 1, 2, .. , n},

addition of elements of R(n) is defined by

(x1, x2, .. , xn) + (y1, y2, .. , yn) = (x1 + y1, x2 + y2, .. , xn + yn),

and scalar multiplication is defined by

c*(x1, x2, .. , xn) = (c x1, c x2, .. , c xn)

then the axioms for a vector space hold for R(n). The verification is, again, almost identical to the one given for R(2); just replace 2-tuples by n-tuples in the argument for R(2).

There is one big difference between R(2) or R(3) and the vector spaces R(n) where n >= 4. The first two (n = 2 or n = 3) can be thought of as geometrical structures. The 2-dimensional plane or 3-dimensional space can be pictured in our minds. The larger spaces R(n), n >= 4, do not have an intuitive description. Nevertheless, they are vector spaces, for they satisfy the axioms of a vector space.

Function Spaces

Let F be the set of all functions defined on some fixed domain D. The domain D can be nearly anything. It could be an open interval (7, 9), a closed interval
[3, 12], or the real line (-, ). The important feature is that it is fixed.

We wish to show that F is a vector space.

To that end recall, first, the difference between f and f(x): f is the name of the rule that defines a function; f(x) is the value the function has at the point x. A function f is defined by specifying the value f(x) it has at each x in the domain D.

This distinction between the name of the function, f, and its value at x, f(x), is usually not made in earlier classes; there is no need to. However, it is useful in more abstract settings.

So, suppose that f and g are functions in F. Define the their sum f+g by

(I) (f+g)(x) = f(x) + g(x) for all x in D.

This defines f+g for we are specifying the value it has at each x in D. As for scalar multiplication by: if c is a real number and f is in F, then c* f is defined by

(II) (c*f)(x) = c f(x) for all x in D.

Then (I) is the operation of addition of vector pairs and (II) is scalar multiplication.

Verifying axioms III through VII is straightforward. Commutativity, for example, follows from commutativity of the real numbers:

(f+g)(x) = f(x) + g(x) for all x in D

(f + g)(x) = (g+f)(x) for all x in D f+g = g + f.

(g+f)(x) = g(x) + f(x) for all x in D

Thus, commutativity is inherited from the reals. Associativity is proved in a similar fashion, using the assoiativity of the real numbers.

The function z defined by z(x) = 0 for all x in D is the 0 vector in F. The inverse of f is -f, and the distributive laws of Axiom 7 follow from the definitions of addition and scalar multiplication, and the distributive laws for real numbers.

It may be strange, at first, to think of a function f as a "vector"; it is quite different from a line with an arrow at one end. But the set of all functions all functions f in F, with vector addition defined by I and scalar multiplication defined by II does satisfy all the axioms of a vector space. So, F is a vector space and its elements are called vectors.

Subspaces

Definition Let V be a vector space and W be a non-empty subset of V that is a vector space, relative to the operations of vector addition and scalar multiplication defined for V. Then W is called a subspace of V.

Theorem 1. 1 Let V be a vector space and W be a non-empty subset of V such that

(I) If v and w are in W then v + w is in W.

(II) If v is in W and c is a real number then c*v is in W.

Then W is a subspace of V.

This theorem says that you only need check the first two axioms for a vector space in order to verify that W is a subspace.

The validity of the remaining axioms follows from the fact that V is a vector space or from II. For example, if v and w are in W then the commutativity axiom,
v + w = w + v, holds in V, so holds for pairs v and w in W. The validity of the associative and distributive laws in W is also inherited from the space V.

The fact that the 0-vector 0 is in W follows from the fact that 0*v = 0 for any vector v in V. For, by the distributive laws and the fact that 0 + 0 = 0,

(()+0)*v= 0*v

(0+0)*v = 0*v + 0*v = 2*(0*v).

Thus,

2*(0*v) = 0*v 0*v = 0,

so 0*v = 0, as claimed. Taking c = 0 in (II), we get 0*v = 0 is in W.

from (II) of Theorem 1.1, with c = 0, and the fact that if 0 is the real number zero, 0 is the 0-vector, and v is any vector in V then 0*v = 0. (See problem _)

Finally, if v is in W then the additive inverse -v is also in W; to see this, take c = - 1 in (II).

Example 1.1 Let V = R(4) = { (x, y, z, u) : x, y, z, u real} and W be the set of all 4-tuples (x, y, z, u) that satisfy the two equations

3 x + 4 y + 5 z + 6 u = 0

2 x + 3 y + 6 z + 5 u = 0.

Then W is a subspace of V.

To show that W is a subspace we must first show two things:

(I) If v = (x1, y1, z1, u1) is in W and w = (x2, y2, z2, u2) then

v + w = (x1 + x2, y1 + y2, z1 + z2, u1+ u2) is in W.

(II) If v = (x, y, z, u) is in W and c is a real number then

c*v = (c x, c y, c z, c u) is in W.

We start by showing (I), W is closed with respect to vector addition:



1
v = (x1, y1, z1, u1) is in W

w = (x2, y2, z2, u2) is in W


2		3 x1 + 4 y1 + 5 z1 + 6 u1 = 0 2 x1 + 3 y1 + 6 z1 + 5 u1 = 0

3 x2 + 4 y2 + 5 z2 + 6 u2 = 0 2 x2 + 3 y2 + 6 z2 + 5 u2 = 0


3		3 (x1 +x2)+ 4 (y1 + y2) + 5 (z1 +z2)+ 6 (u1 + u2) = 0

2 (x1 +x2)+ 3 (y1 + y2) + 6 (z1 +z2)+ 5 (u1 + u2) = 0

v+ w = (x1 +x2, y1 +y2, z1 +z2, u1 +u2) is in W

In words:

1. Assume that v = (x1, y1, z1, u1) and w = (x2, y2, z2, u2) are in W.

2. This means that each 4-tuple satisfies both of the given equations

3. Adding the two equations in 2 we find that the 4-tuple

(x1 + x2, y1 + y2, z1 + z2, u1+ u2) satisfies both equations.

4. This says that v + w is in W; W is closed with respect to

vector addition.

Next, we must show that W is closed with respect to scalar multiplication:



1
v = (x, y, z, u) is in W

c is a real number


2		3 x + 4 y + 5 z + 6 u = 0

2 x + 3 y + 6 z + 5 u = 0


3		c(3 x + 4 y + 5 z + 6 u)= c 0 = 0

c(2 x + 3 y + 6 z + 5 u) = c 0 = 0


4		3 c x + 4 c y + 5 c z + 6 c u = 0

2 c x + 3 c y + 6 c z + 5 c u = 0

c*v = (c x, c y, c z, c u) is in W

In words:

1. Assume that v = (x, y, z, u) is in W and c is a real number.

2. Then the 4-tuple satisfies the two, given equations.

3. Furthermore, the equations both hold if we multiply them by c.

4. The constant c can be distributed inwards. This shows that the

4-tuple (c x, c y, c z, c u) satisfies the given pair of equations.

5. Thus c*v = (c x, c y, c z, c u) is in W; W is closed with

respect to scalar multiplication.

At this point we have shown that W is closed with respect to vector addition and with respect to scalar multiplication. Thus, by Theorem 1.1, it is a subspace of V.

Example 1.2 Let M be the vector space of all 2 x 2 matrices with addition defined by matrix addition,


and scalar multiplication defined by


A 2 x 2 matrix is said to be symmetric if the entry in its first row, second column is equal to the entry in its second row, first column. That is, A has the form

Let S be the subset of all symmetric 2 x 2 matrices. Thus


Then M is a vector space and S is a subspace of M.

Proof: First, M is a vector space : There is closure with respect to vector addition and scalar multiplication. The 2 x 2 0-matrix is the additive identity, 0, in M. The other axioms hold by the definition of matrix addition, scalar multiplication and properties of the real numbers.

Next, S is a subspace of M. It is closed with respect to vector addition:

If is symmetric and is symmetric

then is symmetric.

In addition, S is closed with respect to vector addition:

If c is a constant and is symmetric

then is symmetric.

So, by Theorem 1.1, S is a subspace of M.

Example 1.3 Let F be the vector space of all functions defined on the domain D = [-1, 1]. Let C be the subset of all functions in F that are continuous on D. Then C is a subspace of V.

Proof: By elementary calculus:

If f and g are continuous functions on D

then f + g is a continuous function on D,

so C is closed with respect to addition. In addition, by calculus,

If c is a real number and f is a continuous function on D

then c*f is a continuous function on D,

so C is closed with respect to scalar multiplication. Thus, by Theorem 1.1, C is a subspace of F.

Example 1.4. A function f, defined on the domain D, is said to be bounded on D if there is a constant M such that |f(x)| <= M for all x in D. The exact value of M is not important; any M such that |f(x)| <= M for all x in D will do.

Let F be the vector space of all functions defined on the domain D = [-1, 1]. Let B be the set of all functions in F that are bounded on D. Then B is a subspace of V.

Speaking in geometric terms, the condition |f(x)| <= M for all x in D means that the graph of f on D lies between -M and M. The function graphed below on
[-1, 1] lies between -M and M where M is about 1.8.


So suppose that f and g are in B. Then there are constants M and N such that |f(x)| <= M and |g(x)| <= N for all x in D. The two graphs would look like this:







Thus the graph of f + g will lie between -(M + N) and (M+N):

The graphs show that if f is bounded and g is bounded then f + g is bounded. Thus, the set B of bounded functions is closed with respect to vector addition.

Next, if c is a real number and f is a bounded function, with |f(x)| <= M, then the graph of c*f lies between -|c| M and c |M|, so c*f is a bounded function.


This shows that the set of bounded functions is closed with respect to scalar multiplication, and concludes the proof that B is a subspace of F.

Summary on Proving That a Subset is a Subspace

In order to prove that a given subset is a subspace you should start by reading, rereading, and then reading once again the terms that define the subspace.

Do this until you understand the definition.

In Examples 1.1 through 1.4 the terms that defined the subspace were:

1. "all 4-tuples (x, y, z, u) that satisfy the two equations

3 x + 4 y + 5 z + 6 u = 0

2 x + 3 y + 6 z + 5 u = 0"

2. "the subset of all symmetric 2 x 2 matrices"

3. "subset of all functions in F that are continuous on D."

4. "the set of all functions in F that are bounded on D."

Once you understand the defining conditions for the subspace you can go on to proving the two parts of Theorem 1:

If v and w are two vectors that satisfy the defining conditions

then the sum v + w satisfies them.

If c is a constant c is real and v is a vector satisfying the defining

conditions then the scalar product c*v satisfies them.

The emphasis here is on understanding the definitions. You must take the time to integrate them into your thinking if you are to carry out these two proofs on closure with respect to vector addition and scalar multiplication.

Exercises

1. Let W be the set of 4-tuples (x, y, z, u) in R(4) that satisfy the system of equations

2 x - 5 y + 3 z + 4 u = 0

2 x + 5 y - 3 z + 4 u = 0

2 x + 5 y + 3 z - 4 u = 0.

Prove that W is a subspace of R(4).

2. Let W be the set of 4-tuples (x, y, z, u) in R(4) that satisfy the system of equations

2 x - 5 y + 3 z + 4 u = 0

2 x + 5 y - 3 z + 4 u = 0

2 x + 5 y + 3 z - 4 u = 1.

Is W a subspace of R(4) ? Why or why not?

3. Let R(2) be the 2-dimensional plane. Let W be the first quadrant of the plane, so

W = { (x, y) : x >= 0, y >= 0}.

Is W a subspace of the vector space R(2) ? Why or why not?

4. Let S be the set of all 4-tuples (x,y,z,u) in R(4) that are solutions of the system of equations


where, as usual, |y| is the absolute value of y. Is S a subspace of V? If it is, prove it. If not, give a specific example illustrating why it isn't.

5. Let V = R(5). Let S be the set of all 5-tuples of the form (x, y, y, x, x + y). Prove that S is a subspace of V.

6. Let V = R(5). Let S be the set of all 5-tuples of the form (x, y, x y, y, x). Is S a subspace of V? If it is, prove it. If not, give a specific example illustrating why it isn't.

7. Let V = R(5). Let

That is, S is the set of 5-tuples whose first component is greater than or equal to its second component, whose second component is greater than its third component, etc.

Is S a subspace of V? If it is, prove it. If not, give a specific example illustrating why it isn't.

8. Let M be the vector space of all 2 x 2 matrices. Let be A be a given, fixed 2 x 2 matrix. Let W be the set of all 2 x 2 matrices that commute with A, so

W = { B : B A = A B}.

Prove that W is a subspace of M.

9. Let M be the vector space of all 2 x 2 matrices. Let W be the set of all non-singular matrices in M. Is W a subspace of M ? Why or why not? Is the set of all singular matrices in M a subspace?

10. Let M be the vector space of all 4 x 4 matrices. Let (x, y, z, u) be a given, fixed
4-tuple and let W be the set of all 4 x 4 matrices A such that

(x, y, z, u)A = (0, 0, 0, 0).

prove that W is a subspace of V.

11. Let F be the space of all functions defined on D = [-1, 1]. Let W be the subset of all functions f in F that are differentiable. Prove that W is a subspace of F.

12. Let F be the space of all functions defined on D = [-1, 1]. Let W be the subset of all functions f in F that have, at most, a finite number of discontinuities in D. Thus, f(x) is continuous at all x in D except, perhaps, at finite set of numbers x1 , .. , xm in D. Prove that W is a subspace of F.

13. Let F be the space of all functions defined on D = [-1, 1]. Let W be the subset of all functions f in F that have the property f(-x) = -f(x) for all x in D. Prove that W is a subset of F.

14. Recall that if S and T are sets then the intersection of S and T, ST, is the set of all elements that are in both S and T. Let V be a vector space and S and T be subspaces of V. Prove that ST is a subspace of V.