1. Vector Spaces 1 Vectors in Rn 2 Vector Spaces
3 Subspaces of Vector Spaces
4 Spanning Sets and Linear Independence
5 Basis and Dimension

2. 1 Vectors in Rn An ordered n-tuple : a sequence of n real numbers
Rn-space :
the set of all ordered n-tuples
R1-space = set of all real numbers
n = 1
(R1-space can be represented geometrically by the x-axis)
n = 2
R2-space = set of all ordered pair of real numbers
(R2-space can be represented geometrically by the xy-plane)
n = 3
R3-space = set of all ordered triple of real numbers
(R3-space can be represented geometrically by the xyz-space)
n = 4
R4-space = set of all ordered quadruple of real numbers

3. Notes:
(1) An n-tuple can be viewed as a point in Rn with the xi’s as its coordinates
(2) An n-tuple also can be viewed as a vector
in Rn with the xi’s as its components
Ex: or
a point
a vector

4. (two vectors in Rn)
Equality:
if and only if
Vector addition (the sum of u and v):
Scalar multiplication (the scalar multiple of u by c):
Notes: The sum of two vectors and the scalar multiple of a vector
in Rn are called the standard operations in Rn

5. Difference between u and v:
Zero vector :

6. Theorem 1: Properties of vector addition and scalar multiplication
Let u, v, and w be vectors in Rn, and let c and d be scalars
(1) u+v is a vector in Rn (closure under vector addition)
(2) u+v = v+u (commutative property of vector addition)
(3) (u+v)+w = u+(v+w) (associative property of vector addition)
(4) u+0 = u (additive identity property)
(5) u+(–u) = 0 (additive inverse property)
(6) cu is a vector in Rn (closure under scalar multiplication)
(7) c(u+v) = cu+cv (distributive property of scalar multiplication over vector addition)
(8) (c+d)u = cu+du (distributive property of scalar multiplication over real-number addition)
(9) c(du) = (cd)u (associative property of multiplication)
(10) 1(u) = u (multiplicative identity property)

7. Notes:
A vector in can be viewed as:
a 1×n row matrix (row vector): or
a n×1 column matrix (column vector):

8. Scalar multiplication
Vector addition
Scalar multiplication
Treated as 1×n row matrix
Treated as n×1 column matrix

9. 2 Vector Spaces Vector spaces :
Let V be a set on which two operations (vector addition and scalar multiplication) are defined. If the following ten axioms are satisfied for every u, v, and w in V and every scalar (real number) c and d, then V is called a vector space
Addition:
(1) u+v is in V
(2) u+v=v+u
(3) u+(v+w)=(u+v)+w
(4) V has a zero vector 0 such that for every u in V, u+0=u
(5) For every u in V, there is a vector in V denoted by –u
such that u+(–u)=0

10. Scalar multiplication: (6) is in V
(7)
(8)
(9)
(10)
※ Any set V that satisfies these ten properties (or axioms) is called a vector space, and the objects in the set are called vectors
※ Thus, we can conclude that Rn is of course a vector space

11. (the set of all m×n matrices with real-number entries)
Four examples of vector spaces are introduced as follows. (It is straightforward to show that these vector spaces satisfy the above ten axioms)
(1) n-tuple space: Rn
(standard vector addition)
(standard scalar multiplication for vectors)
(2) Matrix space :
(the set of all m×n matrices with real-number entries)
Ex: (m = n = 2)
(standard matrix addition)
(standard scalar multiplication for matrices)

12. (3) n-th degree or less polynomial space :
(the set of all real polynomials of degree n or less)
※ By the fact that the set of real numbers is closed under addition and multiplication, it is straightforward to show that Pn satisfies the ten axioms and thus is a vector space
(4) Continuous function space :
(the set of all real-valued continuous functions defined on the
entire real line)
※ By the fact that the sum of two continuous function is continuous and the product of a scalar and a continuous function is still a continuous function, is a vector space

13. Summary of important vector spaces
※ Each element in a vector space is called a vector, so a vector can be a real number, an n-tuple, a matrix, a polynomial, a continuous function, etc.

14. Theorem 2: Properties of scalar multiplication
Let v be any element of a vector space V, and let c be any
scalar. Then the following properties are true
(the additive inverse of v equals ((–1)v)

15. Notes: To show that a set is not a vector space, you need
only find one axiom that is not satisfied
(it is not closed under scalar multiplication)
scalar
Pf:
Ex 1: The set of all integers is not a vector space
integer
noninteger
Ex 2: The set of all (exact) second-degree polynomial functions is
not a vector space
Pf: Let and
(it is not closed under vector addition)

16. V=R2=the set of all ordered pairs of real numbers
Ex 3:
V=R2=the set of all ordered pairs of real numbers
vector addition:
scalar multiplication:
(nonstandard definition)
Verify V is not a vector space
Sol:
This kind of setting can satisfy the first nine axioms of the definition of a vector space (you can try to show that), but it violates the tenth axiom
the set (together with the two given operations) is
not a vector space

17. 3 Subspaces of Vector Spaces
a vector space
a nonempty subset
The nonempty subset W is called a subspace if W is a vector space under the operations of addition and scalar multiplication defined in V
Trivial subspace :
Every vector space V has at least two subspaces
(1) Zero vector space {0} is a subspace of V
(It satisfies the ten axioms)
(2) V is a subspace of V
※ Any subspaces other than these two are call proper (or nontrivial) subspaces

18. 4.4 Spanning Sets and Linear Independence
Linear combination :

19. Ex : Finding a linear combination
Sol:

20. (this system has infinitely many solutions)

22. The span of a set: span(S)
If S={v1, v2,…, vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S,
Definition of a spanning set of a vector space:
If every vector in a given vector space V can be written as a linear combination of vectors in a set S, then S is called a spanning set of the vector space V

23. Note: The above statement can be expressed as follows

24. Ex 5: A spanning set for R3
Sol:

26. Theorem 5: span(S) is a subspace of V
If S={v1, v2,…, vk} is a set of vectors in a vector space V, then
span(S) is a subspace of V
span(S) is the smallest subspace of V that contains S, i.e., every other subspace of V containing S must contain span(S)

27. Pf:
(a)

28. (b)
(because U is closed under vector addition and scalar multiplication)
※ For example, V = R5, S = {(1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0)} and thus span(S) = R3, and U = R4, U contains S and contains span(S) as well

29. Definitions of Linear Independence (L.I.) and Linear Dependence (L.D.) :

30. Ex 8: Testing for linear independence
Determine whether the following set of vectors in R3 is L.I. or L.D.
Sol:

31. Ex 9: Testing for linear independence
Determine whether the following set of vectors in P2 is L.I. or L.D.
Sol:
c1v1+c2v2+c3v3 = 0
i.e.,
c1(1+x – 2x2) + c2(2+5x – x2) + c3(x+x2) = 0+0x+0x2
c1+2c = 0
c1+5c2+c3 = 0
–2c1– c2+c3 = 0
This system has infinitely many solutions
(i.e., this system has nontrivial solutions, e.g., c1=2, c2= – 1, c3=3)
S is (or v1, v2, v3 are) linearly dependent

32. Ex 10: Testing for linear independence
Determine whether the following set of vectors in the 2×2
matrix space is L.I. or L.D.
Sol:
c1v1+c2v2+c3v3 = 0

33. 2c1+3c2+ c3 = 0
c = 0
2c2+2c3 = 0
c1+ c = 0
(This system has only the trivial solution)
c1 = c2 = c3= 0
S is linearly independent

34. Theorem 6: A property of linearly dependent sets
A set S = {v1,v2,…,vk}, k2, is linearly dependent if and
only if at least one of the vectors vi in S can be written as
a linear combination of the other vectors in S
Pf:
()
c1v1+c2v2+…+ckvk = 0
 ci  0 for some i

35. vi = d1v1+…+di-1vi-1+di+1vi+1+…+dkvk
Let
vi = d1v1+…+di-1vi-1+di+1vi+1+…+dkvk
 d1v1+…+di-1vi-1–vi +di+1vi+1+…+dkvk = 0
(there exits at least this nontrivial solution)
 c1=d1 , c2=d2 ,…, ci=–1 ,…, ck=dk
S is linearly dependent
Corollary to Theorem 4.8:
Two vectors u and v in a vector space V are linearly dependent (for k = 2 in Theorem 6) if and only if one is a scalar multiple of the other

36. 4.5 Basis and Dimension Basis : V: a vector space S ={v1, v2, …, vn}V
Spanning
Sets
Bases
Linearly
Independent
V: a vector space
S ={v1, v2, …, vn}V V
S spans V (i.e., span(S) = V)
S is linearly independent
 S is called a basis for V
Notes:
A basis S must have enough vectors to span V, but not so many vectors that one of them could be written as a linear combination of the other vectors in S

37. Notes:
(1) the standard basis for R3:
{i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)
(2) the standard basis for Rn :
{e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0),…, en=(0,0,…,1)
Ex: For R4,
{(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}

38. (3) the standard basis matrix space:
Ex: matrix space:
(4) the standard basis for Pn(x):
{1, x, x2, …, xn}
Ex: P3(x)
{1, x, x2, x3}

39. Ex 2: The nonstandard basis for R2
Because the coefficient matrix of this system has a nonzero determinant, the system has a unique solution for each u. Thus you can conclude that S spans R2
Because the coefficient matrix of this system has a nonzero determinant, you know that the system has only the trivial solution. Thus you can conclude that S is linearly independent
According to the above two arguments, we can conclude that S is a (nonstandard) basis for R2

40. Theorem 7: Uniqueness of basis representation for any vectors
If is a basis for a vector space V, then every
vector in V can be written in one and only one way as a linear
combination of vectors in S
Pf:
(1) span(S) = V
(2) S is linearly independent
span(S) = V
Let
v = c1v1+c2v2+…+cnvn
v = b1v1+b2v2+…+bnvn
 v+(–1)v = 0 = (c1–b1)v1+(c2 – b2)v2+…+(cn – bn)vn
 c1= b1 , c2= b2 ,…, cn= bn
(i.e., unique basis representation)

41. Theorem 8: Bases and linear dependence
If is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent (In other words, every linearly independent set contains at most n vectors)
Pf:
Let
S1 = {u1, u2, …, um} , m > n
uiV

42. Consider
k1u1+k2u2+…+kmum= 0
(if ki’s are not all zero, S1 is linearly dependent)
 d1v1+d2v2+…+dnvn= 0
(di = ci1k1+ci2k2+…+cimkm)
 di=0 i
i.e.,
Theorem : If the homogeneous system has fewer equations (n equations) than variables (k1, k2, …, km), then it must have infinitely many solution
m > n  k1u1+k2u2+…+kmum = 0 has nontrivial (nonzero) solution
 S1 is linearly dependent

43. Theorem 9: Number of vectors in a basis
If a vector space V has one basis with n vectors, then every
basis for V has n vectors
Pf:
※ According to Thm.8, every linearly independent set contains at most n vectors in a vector space if there is a set of n vectors spanning that vector space
S ={v1, v2, …, vn}
are two bases with different sizes for V
S'={u1, u2, …, um}
※ For R3, since the standard basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} can span this vector space, you can infer any basis that can span R3 must have exactly 3 vectors
※ For example, S={(1, 2, 3), (0, 1, 2), (–2, 0, 1)} is another basis of R3 (because S can span R3 and S is linearly independent), and S has 3 vectors

44. Infinite dimensional:
The dimension of a vector space V is defined to be the number of vectors in a basis for V
V: a vector space
S: a basis for V
 dim(V) = #(S)
(the number of vectors in a basis S)
Finite dimensional:
A vector space V is finite dimensional if it has a basis consisting of a finite number of elements
Infinite dimensional:
If a vector space V is not finite dimensional, then it is called infinite dimensional

45. Ex 9: Finding the dimension of a subspace of R3
(a) W={(d, c–d, c): c and d are real numbers}
(b) W={(2b, b, 0): b is a real number}
(Hint: find a set of L.I. vectors that spans the subspace, i.e., find a basis for the subspace.)
Sol:
(a)
(d, c– d, c) = c(0, 1, 1) + d(1, – 1, 0)
 S = {(0, 1, 1) , (1, – 1, 0)}
(S is L.I. and S spans W)
 S is a basis for W
 dim(W) = #(S) = 2
(b)
 S = {(2, 1, 0)} spans W and S is L.I.
 S is a basis for W
 dim(W) = #(S) = 1

46. Ex 11: Finding the dimension of a subspace of M22
Let W be the subspace of all symmetric matrices in M22.
What is the dimension of W?
Sol:
spans W and S is L.I.
 S is a basis for W
 dim(W) = #(S) = 3