Definition of a Vector Space

A collection of vectors: V , scalars for scaling : R;A vector addition: +, A scalar multiplication: ·

for vector addition:

(i) (closure of vector addition) u+ v is always in V .

(ii) (commutative law) u+ v = v + u.

(iii) (associative law) (u+ v) +w = u+ (v +w).

(iv) (existence of zero vector) has 0 s.t. u+ 0 = u.

(v) (existence of negative vector) for each u ∈ V , there is avector −u s.t. u+ (−u) = 0.

1

...continued for scalar multiplication:

(vi) (closure of scalar multiplication) cu is always in V .

(vii) (distributive law) c(u+ v) = cu+ cv.

(viii) (distributive law) (c+ d)u = cu+ du.

(ix) (compatibility) c(du) = (cd)u.

(x) (normalization) 1u = u.

Def: A non-empty set of vectors V with vector addition“+” and scalar multiplication “·” satisfying all the aboveproperties is called a vector space over R.

2

Facts: (i) The zero vector 0 so defined is unique.

0 = 0+w = w

(ii) The negative vector for each u is unique.

−u = −u+ 0 = −u+ (u+w)

= (−u+ u) +w = 0+w = w

We define vector subtraction u− v := u+ (−v).

(iii) c0 = 0.

(iv) 0u = 0.

(v) −u = (−1)u.

3

Examples of Common Vector Spaces

• {0}, zero vector space.

• Rn: ordered n-tuples of real numbers with entry-wisevector addition and scalar multiplication.

u+ v =

u1 + v1...

un + vn

, cu =

cu1...

cun

Blue-print for vector space.

4

• S, the doubly infinite sequences of numbers:

{yk} = (. . . , y−2, y−1, y0, y1, y2, . . .)

with component-wise addition and scalar multiplication.

{yk}+ {zk} = (. . . , y−2, y−1, y0, y1, y2, . . .)

+ (. . . , z−2, z−1, z0, z1, z2, . . .)

:= (. . . , y−1 + z−1, y0 + z0, y1 + z1, . . .)

c{yk} := (. . . , cy−2, cy−1, cy0, cy1, cy2, . . .)

zero vector: {0}, sequence of zeros.

5

• Pn: collection of polynomials with degree at most equalto n and coefficients chosen from R:

p(t) = p0 + p1t+ . . .+ pntn,

with polynomial operations.→ “vector addition”: “p(t) + q(t)” is defined as:

p(t) + q(t) = (p0 + q0) + (p1 + q1)t+ . . .+ (pn + qn)tn;

→ “scalar multiplication”: “cp(t)” is defined as:

cp(t) = (cp0) + (cp1)t+ . . .+ (cpn)tn.

→ “zero vector”: the zero polynomial 0(t) ≡ 0.• P: all polynomials.

6

• The collection of all real-valued functions on a set D, i.e.f : D → R. (D usually is an interval.)

→ “vector addition”: the function “f + g” is defined as:

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

→ “scalar multiplication”: the function “cf” is definedas:

(cf)(x) = cf(x) for all x in D.

→ “zero vector”: the zero function 0(x) which sendsevery x in D to 0, i.e.

0(x) = 0 for all x in D.

7

• Mm×n: collection of all m × n matrices with entries inR with matrix addition and scalar multiplication.

A+B =(aij + bij

), cA =

(caij

).

zero vector: m× n zero matrix Om×n.

. . . and many more.

8

Subspace: A subspace H of V is a non-empty subset of Vsuch that H itself forms a vector space under the same vectoraddition and scalar multiplication induced from V .

Checking: Following conditions are automatically valid:

(ii) u+ v = v + u.

(iii) (u+ v) +w = u+ (v +w).

(vi) c(u+ v) = cu+ cv.

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

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

(x) 1u = u.

9

Need to verify the remainings:

(i) the sum u+ v is always in H.

(iv) there is a zero vector 0 in H.

(v) for each u ∈ H, there is a negative vector −u ∈ H.

(vi) the scalar multiple cu is always in H.

[Same zero vector, negative vector for H.]

.... but we know ...

• −u = (−1)u for any u.

10

Alternative Definition: H is a subspace of V if:

1. 0 of V is in H.

2. sum of two vectors in H is again in H.

3. the scalar multiple of a vector in H is again in H.

In fact, (2) & (3) can be combined together as a single check-ing condition:

Thm: H is a subspace of V iff H contains 0 and:

u,v ∈ H ⇒ au+ bv ∈ H for all scalars a, b.

11

Examples of Subspaces

1. V is a subspace of itself.

2. {0} is a subspace of V , called the zero subspace.

3. R2 is not a subspace of R3, strictly speaking.

3′. The following subset of R3 is a subspace of R3:

H =

st0

: s, t in R

4. Pn is a subspace of P;

Pn is a subspace of Pm if n ≤ m.

12

5. The collection of n× n symmetric matrices H is a sub-space of Mn×n.

5′. The collection of n× n skew-symmetric matrices is alsoa subspace of Mn×n.

6. The collection Ve of even functions from R to R:

f(−x) = f(x) for all x ∈ R

and the collection Vo of odd functions:

f(−x) = −f(x) for all x ∈ R

are both subspaces of the vector space of functions.

13

Linear Combination and Span

Similar to vectors in Rn, we define:

Def: Let S = {v1, . . . ,vk} be a collection of vectors in Vand let c1, . . . , ck be scalars. The following y is called alinear combination (l.c.) of vectors in S:

y = c1v1 + . . .+ ckvk.

Note: When S contains infinitely many vectors, a l.c. ofvectors in S means a l.c. of a finite subset of vectors in S.

Rmk: Sum of infinite number of vectors is not defined (yet).

[We need a concept of “limit” to do this.]

14

Example: In function space, let S = {sin2 x, cos2 x}. Thenthe constant function 1(x) ≡ 1 is a l.c. of “vectors” in S:

1(x) = 1 · sin2 x+ 1 · cos2 x, for all x in D.

Example: In polynomial space, let S = {1, t, t2, t3, . . .}.Then any polynomial p(t) is a l.c. of “vectors” in S.

p1(t) = 1 + t+ t3, S1 = {1, t, t3}p2(t) = 3t− 5t100, S2 = {t, t100}

p3(t) = 0, S3 = {1}

15

Exercise: Consider the function space.

Is sin3 x a l.c. of S = {sinx, sin 2x, sin 3x}?

Exercise: Consider the function space.

Is cosx a l.c. of S = {sinx, sin 2x, sin 3x}?

***

16

Span of a Collection of Vectors

Def: Let S = {v1, . . . ,vk}. The collection of all possible l.c.of vectors in S is called the span of S:

SpanS := {c1v1 + . . .+ ckvk | c1, . . . , ck are scalars.}.

Note: When S contains infinitely many vectors, SpanS isthe collection of all possible l.c. of any finite subset of vectorsin S.Example: In P, consider S = {1, t2, t4, . . .}. Then

SpanS = collection of polynomials with only even powers.

17

Thm: Let v1, v2 be vectors in V . Then H = Span {v1,v2}is a subspace of V .

Checking: (1): 0 = 0v1 + 0v2, so 0 ∈ H.

(2): Let u,v ∈ H, i.e. both are l.c. of {v1,v2}:

u = s1v1 + s2v2, v = t1v1 + t2v2,

for some suitable numbers s1, s2, t1, t2. Then:

u+ v = (s1 + t1)v1 + (s2 + t2)v2

is again a l.c. of {v1,v2}, and thus collected by H.

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(3): Let u ∈ H, c a number. Then:

cu = (cs1)v1 + (cs2)v2

is a l.c. of {v1,v2}, and hence collected by H.

As conditions (1),(2),(3) are all valid, H is a subspace of V .

The above proof allows obvious generalization to:

Thm 1 (P.210): For any v1, . . . ,vp ∈ V , the collection ofvectors H = Span {v1, . . . ,vp} is a subspace of V .

Note: We will call H to be the subspace spanned (or gener-ated) by {v1, . . . ,vp}, and {v1, . . . ,vp} is called a spanningset (or generating set) of H.

19

Subspaces of a Matrix

Let A be an m×n matrix. Associated to this matrix A,we define:

1. The null space of A, denoted by NulA;

2. The column space of A, denoted by ColA;

3. The row space of A, denoted by RowA.

Note that:

• NulA and RowA will be subspaces of Rn;

• ColA will be a subspace of Rm.

20

The Null Space of a Matrix

Def: Let A be an m× n matrix. The null space of A is:

NulA := {x | x in Rn and Ax = 0}.

i.e. the solution set of the homogeneous system Ax = 0.

• size of each solution: no. of columns in A.so if A is of size m× n, NulA is inside Rn.

Example: Does v belong to NulA?

A =

[1 1 −10 −3 2

], v =

123

21

Thm 2 (P.215): NulA is a subspace of Rn.

Checking: (1) A0 = 0, so 0 is in NulA.

(2) Let Au = 0 = Av. Then consider u+ v:

A(u+ v) = Au+Av = 0+ 0 = 0,

so u+ v ∈ NulA.

(3) Let Au = 0, c any number. Then consider cu:

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

so cu ∈ NulA.

22

Exercise: Describe the null space of A:

A =

1 2 2 12 5 10 31 3 8 2

.

***

From the above exercise, obviously we will have:

Thm: NulA = {0} when Ax = 0 has unique solution. WhenAx = 0 has {x1, . . . ,xk} as a set of basic solutions, we haveNulA = Span{x1, . . . ,xk}.

23

The Column Space of a Matrix

Def: Let A = [a1 . . . an ] be an m× n matrix. Then thecolumn space of A is:

ColA := Span {a1, . . . ,an}.

Fact: ColA is a subspace of Rm.

Exercises: Check if v ∈ ColA:

A =

1 2 1 22 4 0 61 2 2 1

(i) v =

121

(ii) v =

147

.

***

24

Thm: v ∈ ColA ⇔ [ A | v ] is consistent.

Example: Find a condition on v such that v ∈ ColA.:

A =

1 2 5−1 −1 −33 2 74 0 4

v =

y1y2y3y4

.

Sol: Consider the augmented matrix [ A | v ]:1 2 5 | y1−1 −1 −3 | y23 2 7 | y34 0 4 | y4

→

1 2 5 | y10 1 2 | y2 + y10 −4 −8 | y3 − 3y10 −8 −16 | y4 − 4y1

25

1 2 5 | y1−1 −1 −3 | y23 2 7 | y34 0 4 | y4

→

1 2 5 | y10 1 2 | y1 + y20 0 0 | y1 + 4y2 + y30 0 0 | 4y1 + 8y2 + y4

For the system to be consistent, we need:{

y1 + 4y2 + y3 = 0

4y1 + 8y2 + y4 = 0

26

Remark: When A is representing a matrix transformationT : Rn → Rm, we have:

v ∈ range of T ↔ can find x ∈ Rn such that T (x) = v

↔ [ A | v ] is consistent.

Therefore, ColA is the same as the range of T in this case,as:

Range of T := {v ∈ Rm | T (x) = v for some x ∈ Rn}.

Thm: ColA = Rm iff every row of A has a pivot position.(i.e. T : x 7→ Ax is onto.)

27

The Row Space of a Matrix Identify:

Row ↔ Column by taking transpose

Def: Let A =

r1...rm

be an m× n matrix. The row space of

A is:RowA := Span{rT1 , . . . , rTm}.

i.e. RowA = ColAT .

Note: RowA is a subspace of Rn.

28

Let A → B by an ERO, e.g.

A =

r1r2...

→

−3r2 + r1r2...

=

r′1r′2...

= B.

Then r′1T, r′2

T, . . . ∈ RowA. So:

RowB ⊆ RowA.

But B → A by the reverse ERO, then we also have:

RowA ⊆ RowB.

29

Thm: Let A → B by EROs. Then RowA = RowB.

Recall: A → B = PA where P (size: m×m) is invertible.

Thm: When P is invertible, RowA = Row (PA).

Example: Describe RowA when A =

[1 1 32 2 4

].

Sol: By definition:

RowA = Span {

113

,

224

}.

30

Example: Describe RowA when A =

[1 1 32 2 4

].

Sol: By EROs:

A =

[1 1 32 2 4

]→

[1 1 00 0 1

]= B.

As RowA = RowB, we have:

RowA = Span {

110

,

001

}.

31

Linear Transformations in general

Def: Let V , W be two vector spaces over R. A transforma-tion T : V → W is called linear if

(a) T (u+ v) = T (u) + T (v)

(b) T (cu) = cT (u)

for any choices of u, v in V and any choice of scalar c.

In the special case that V = W , i.e. T : V → V is linear, wewill call T to be a linear operator on V .

32

Def: Let T : V → W be linear. We define:

1. The kernel of T , denoted by kerT , to be:

kerT := {v ∈ V : T (v) = 0} .

(solution set of T (v) = 0.)

2. The image/range of T , denoted by ImT , to be:

ImT := {w ∈ W : T (v) = w for some v ∈ V } .

(which is the same as T (V ).)

Thm: kerT is a subspace of V . ImT is a subspace of W .

33

Remark: When V = Rn, W = Rm, i.e. T is given by amatrix transformation: x 7→ Ax. Then:

kerT = NulA and ImT = ColA.

Thm: T is one-to-one iff kerT = {0}.T is onto iff ImT = W .

Pf: (1-1) As T is linear, we have:

T (x1) = T (x2) ⇔ T (x1 − x2) = 0.

So T is one-to-one ⇔ always have x1 = x2 ⇔ kerT = {0}.

For onto property, directly from definition.

34

Example: Consider V = W = P(R). The differential oper-ator D will be a linear transformation:

D(p(t)) =d

dtp(t).

We have kerD = Span {1} and ImD = P.

• Thus D is onto but NOT one-to-one.

Example: The integral operator Ia : P → P with lowerlimit a:

Ia(p(t)) =

∫ t

a

p(x)dx.

will be one-to-one but NOT onto.

35

Linearly Independent/Dependent Sets

Def: An indexed set of vectors S = {v1, . . . ,vp} is calledlinearly independent (l.i.) if the vector equation:

c1v1 + . . .+ cpvp = 0

has unique (zero) solution:

c1 = . . . = cp = 0.

Def: S is called linearly dependent (l.d.) if it is not l.i., i.e.exist d1, . . . , dp, not all zeros, such that:

d1v1 + . . .+ dpvp = 0.

36

Note: When S contains infinitely many vectors:(i) S is said to be l.i. if every finite subset of S is always l.i.(ii) S is said to be l.d. if there exists a finite subset of S

being l.d.

Example: In P2(R), S = {1, 2t, t2} is l.i.

Sol: Consider the “vector” equation:

c1 · 1 + c2 · 2t+ c3 · t2 = 0(t) (as polynomials).

By equating coefficients of 1, t, t2, we get c1 = c2 = c3 = 0.

Rmk: The above equation is a polynomial “identity”.

37

Example: In P2, S = {1 + t, 1− t2, t+ t2} is l.d. since:

c1(1 + t) + c2(1− t2) + c3(t+ t2) = 0(t)

has a non-zero solution (c1, c2, c3) = (1,−1,−1).

Example: In P, S = {1, t, t2, t3, . . .} is l.i.

Sol: Take any finite subset of S:

S′ = {ti1 , ti2 , . . . , tik}, 0 ≤ i1 < i2 < . . . < ik.

Obviously S′ is always l.i., so S will be l.i.

38

Example: In function space, S = {sin2 x, cos2 x} l.i.

Sol: Consider the “vector” equation:

c1 sin2 x+ c2 cos

2 x = 0(x) (as functions).

Put x = 0, π2 , we obtain “number” equations:{

c1 · 0 + c2 · 1 = 0

c1 · 1 + c2 · 0 = 0

from which we get c1 = c2 = 0 already. So S must be l.i.

39

Example: In function space, S = {1, sin2 x, cos2 x} is l.d.

Sol: We have non-trivial solution to the “vector” equation:

(−1) · 1 + (1) · sin2 x+ (1) · cos2 x = 0(x),

which is true for all x ∈ D.

Exercise: Let c1, c2, c3 be distinct numbers. In functionspace, is {ec1x, ec2x, ec3x} l.i.?

***

40

Bases, Coordinates, and Dimensions

• When {v1, . . . ,vp} is l.d., we can remove one vj fromthe set without changing its span.

e.g. Consider {v1,v2,v3} with v2 = 3v1 − 4v3.

the l.c. x = a1v1 + a2v2 + a3v3 can be rewritten as:

x = a1v1 + a2(3v1 − 4v3) + a3v3

= (a1 + 3a2)v1 + (a3 − 4a2)v3.

which is a vector in Span {v1,v3}.

• If removing any vector from {v1, . . . ,vp} will change itsspan, the set must be l.i.

41

Basis for a Subspace

Def: Let H be a subspace of V . An indexed set B of vectorsin H is called a basis for H if:

(i) B is linearly independent; and

(ii) SpanB = H.

When H = {0}, we choose B = ϕ as the basis.

Remarks: (i) Since SpanB = H, every vector in H can bewritten as a l.c. of vectors in B.(ii) Since B is l.i., removing any vector from it will makethe span smaller. So we can regard a basis as a “minimalcollection of vectors” from H that can span H.

42

Examples: (i) {e1, . . . , en} is a basis for Rn.

(ii) Both {[10

],

[01

]} and {

[1−1

],

[11

]} are bases for R2.

(iii) {1, t, t2, . . . , tn} forms a basis for Pn.

P has a basis {1, t, t2, . . .}.

(iv) The following set of matrices is a basis for the vectorspace of 2× 2 matrices.

B = {[1 00 0

],

[0 10 0

],

[0 01 0

],

[0 00 1

]}.

43

(v) Check that B is a basis for R3.

B = {

30−6

,

−417

,

−215

}.Sol: Consider the equation:

c1

30−6

+ c2

−417

+ c3

−215

= b.

44

↔

3 −4 −20 1 1−6 7 5

c1c2c3

=

b1b2b3

.

We find that the coefficient matrix is invertible. So:

• B is linearly independent.(all basic variables, unique solution for c1, c2, c3.)

• SpanB = ColA = R3.(every row of A has a pivot position.)

Thus, B is a basis of R3.

Thm: Let [a1 . . . an ] be an n×n invertible matrix. ThenB = {a1, . . . ,an} will form a basis for Rn.

45

Bases for NulA

Express the general solution of Ax = 0 in parametric form:

x = s1x1 + . . .+ skxk, where s1, . . . , sk ∈ R.

So B = {x1, . . . ,xk} can span NulA.

We note that B must be l.i., for example:

x =

−s− tst

= s

−110

+ t

−101

.

x = 0 iff s = t = 0.

46

Thm: When Ax = 0 has non-zero solutions, a set of basicsolutions B = {x1, . . . ,xk} will form a basis for NulA.

When Ax = 0 has unique zero solution, i.e. NulA = {0}, wesay that ϕ is the basis for NulA.

Exercises: Find a basis for NulA where:

(i) A =

[1 22 1

], (ii) A =

1 1 2 2 32 2 1 1 33 3 2 2 2

.

***

47

Bases for RowA

Recall when A → B by EROs, we have RowA = RowB. Solet B be a REF of A. Clearly the non-zero rows of B willspan RowB = RowA, and also they form a l.i. set, e.g.:

B =

1 1 1 10 2 3 40 0 0 50 0 0 0

↔ B = {

1111

,

0234

,

0005

}

c1

1111

+ c2

0234

+ c3

0005

=

0000

,

48

unique solution: c1 = c2 = c3 = 0.

Thm 13 (P.247): Let B be a REF of A. Then the non-zerorows of B will form a basis for RowA.

Exercise: Find a basis for RowA:

A =

1 4 6 81 1 3 22 5 9 101 0 2 0

.

***

49

Bases for ColA

We use the trick: ColA = RowAT .

Example: Find a basis for ColA:

A =

1 2 1 12 3 1 21 3 1 31 2 1 1

.

Sol: Perform EROs on AT :

AT =

1 2 1 12 3 3 21 1 1 11 2 3 1

→

1 2 1 10 1 0 00 0 1 00 0 0 0

50

A basis for ColA will be:

B = {

1211

,

0100

,

0010

}.Warning: Row operations on A will change ColA!.

• i.e. EROs needed to be performed on AT .(then RowAT = ColA will not change.)

51

If we want to use only the columns from A to form such abasis, we have the following result:

Thm 6 (P.228): The pivot columns of A will form a basisfor ColA.

Exercise: Find a basis for ColA, using columns from A:

A =

1 2 1 12 3 1 21 3 1 31 2 1 1

.

***

52

Idea of Proof of Thm 6:

1. A l.c. of columns of A:

c1a1 + . . .+ cnan = b

corresponds to a solution of Ax = b: xi = ci

2. Row operations [ A | b ] → [ A′ | b′ ] will not change thissolution, i.e. same dependence relation for new columns:

c1a′1 + . . .+ cna

′n = b′

3. In RREF(A) the pivot columns are automatically l.i.,spanning ColRREF(A). These two properties can bebrought back to pivot columns in A by reverse EROs.

53

Recall: Studying V using a basis B

Basis: An indexed set B of vectors that is:

(a) linearly independent;

(b) spanning the space V , i.e. SpanB = V .

Thm 7 (P.232): Let B = {b1, . . . ,bp} be a basis for a sub-space H (or V ). Then for each x ∈ H, there exists a uniqueset of scalars (c1, . . . , cp) such that:

x = c1b1 + . . .+ cpbp.

54

Proof: The existence of {c1, . . . , cp} is guaranteed by condi-tion (b). Now suppose that d1, . . . , dp are numbers with thesame properties, i.e.

x = d1b1 + . . .+ dpbp.

Take the difference, then:

0 = (c1 − d1)b1 + . . .+ (cp − dp)bp.

As B is l.i. (condition (a)), ci − di = 0 for each i.

Hence the numbers c1, . . . , cp are uniquely determined.

55

Coordinate Vector Relative to a Basis

Def: Let B = {b1, . . . ,bp} be a basis for H. The coordinatevector of x relative to B (or B-coordinate vector of x) is thevector in Rp formed by the numbers c1, . . . , cp:

x = c1b1 + . . .+ cpbp ↔ [x]B :=

c1...cp

.

Note that x is in H ⊂ V , but [x]B is in Rp.

56

Example: Let B = {[

2−1

],

[−2−1

]} be a basis for R2. De-

scribe geometrically the B-coordinate vector of x =

[x1

x2

].

Sol: First we find [x]B:[x1

x2

]= c1

[2−1

]+ c2

[−2−1

]↔

c1 =

x1 − 2x2

4

c2 = −x1 + 2x2

4So the B-coordinate vector of x is:

[x]B =

[c1c2

]=

[ x1−2x2

4

−x1+2x2

4

].

57

x

y

x1

x2

x

original coordinate system

x =

[x1

x2

]

x

y

c2(−2,−1)

c1(2,−1)

(2,−1)(−2,−1)

x

B-coordinate system

[x]B =

[c1c2

]

58

Example: Let B = {1, 1 + t, 1 + t+ t2} be a basis for P2.

Consider p(t) = t − t2. To find the B-coordinate vector ofp(t), we express p(t) as a l.c. of vectors in B:

t− t2 = (−1) + 2(1 + t)− (1 + t+ t2).

Then the B-coordinate vector of p is:

[p]B =

−12−1

.

59

Example: Let H = SpanB where B = {1, cosx, cos 2x}.It is easy to check that B is l.i., so it is a basis for H.

Let f(x) = cos2 x. Since f(x) = 12 + 1

2 cos 2x, we have:

[f ]B =

12012

.

But if we take B′ = {cosx, 1, cos 2x}, we have:

[f ]B′ =

01212

.

So the ordering of vectors in a basis is important.

60

Exercise: Let B = {

11−1

,

1−11

,

−111

} be a basis for

R3. Find [e1]B, [e2]B, and [e3]B.

***

Def: The mapping x 7→ [x]B is called the coordinate map-ping determined by B, or simply B-coordinate mapping.

Thm 8 (P.235): Let B = {b1, . . . ,bp} be a basis for H.Then the B-coordinate mapping is a one-to-one and ontolinear transformation from H to Rp. Its inverse is also alinear transformation from Rp back to H.

Proof: Direct verification.

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Under this coordinate mapping, any linear problem in V canbe translated to a corresponding problem in Rp, and thenwe are equipped with the powerful matrix theory.

Example: Check that the set of vectors:

{t+ t3, 1− t+ t2, 5− 3t3, 4 + 2t2, 3− t+ t3},

is l.d. in P3.

Can check it directly by solving the polynomial identity.

Sol: Let B = {1, t, t2, t3}. By B-coordinate mapping, theyare sent to:

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0101

,

1−110

,

500−3

,

4020

,

3−101

5 vectors in R4 must be l.d., i.e. exists c1, . . . , c5, not allzeros, such that:

c1

0101

+c2

1−110

+c3

500−3

+c4

4020

+c5

3−101

=

0000

.

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Under inverse B-coordinate mapping, we obtain a depen-dence relation among the 5 polynomials in P3:

c1(t+ t3) + c2(1− t+ t2) + c3(5− 3t3)

+ c4(4 + 2t2) + c5(3− t+ t3) = 0(t).

So the 5 polynomials are l.d. in P3.

The technique used above generalizes to:

Thm 9 (P.241): Let B = {b1, . . . ,bp} be a basis for H.Then any subset in H containing q > p vectors must be l.d.

Proof: Send these q vectors by B-coordinate mapping to qvectors in Rp.

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Thm 10 (P.242): Let B and C be two bases for V consistingof n vectors and m vectors respectively. Then n = m.

Pf: First consider B as a basis for V .

• If m > n, C must be l.d. by previous theorem.

• Since C is l.i., we must have m ≤ n.

Now consider C as a basis for V .

• If n > m, B must be l.d. by previous theorem.

• Since B is l.i., we must have n ≤ m.

Therefore n = m.

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Dimension of a Vector Space

Def: When V has a basis B of n vectors, V will be calledfinite dimensional of dimension n or n-dimensional. We willwrite dimV = n.

Def: The dimension of the zero space {0} is defined to be 0.

Def: If V cannot be spanned by any finite set of vectors, Vwill be called infinite dimensional. We will write dimV = ∞.

Examples: dimRn = n, dimPn = n+ 1, dimP = ∞.

Basis for Rn: E = {e1, . . . , en}Basis for Pn: {1, t, t2, . . . , tn}Basis for P: {1, t, t2, . . .}

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Let H be a subspace of V and dimV = n ≥ 1.

Thm 11 (P.243): Any l.i. set in V can be extended to abasis for V . In particular, if H is a subspace of V , we have:

dimH ≤ dimV,

and if dimH = dimV , we have H = V .

Proof: Let S be a l.i. set in V .

1. If SpanS = V , S is already a basis.

2. If not, there will be a vector u /∈ SpanS.

3. The set S ∪ {u} will have one more element, still l.i.

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(For the equation c1v1 + . . . + cpvp + cp+1u = 0; firstshow cp+1 = 0 using (2), then show c1 = . . . = cp = 0.)

4. Go back to (1) with new S′ = S ∪ {u}.Such addition of vectors must stop somewhere since l.i.set in V cannot contain more than n vectors.

(Process stops → the l.i. set S′ can span V .)

If dimH = dimV , first choose any basis B for H.a. dimH = |B| = n(= dimV ) by assumption.b. Extend B to B′, a basis for V , by adding suitable vectors.c. Any l.i. set in V cannot contain more than n vectors.d. Must have B′ = B (adding no vector in (b)), and hence

H = SpanB = SpanB′ = V .

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The proof of previous theorem says that:

Thm 12 (P.243): Let dimV = n ≥ 1. Then

(a) Any l.i. set with n vectors is a basis for V .

(b) Any spanning set of V with n vectors is a basis for V .

Proof: (a) We cannot add vector to S as before since anyset with n+ 1 vectors must be l.d., so S is a basis for V .

(b) If S is l.d., we can remove certain vector in S and obtaina smaller set S′ which still spans V . Keep removing vectorsuntil S′ is l.i., and thus forming a basis for V . This S′ con-tains dimV = n vectors, i.e. we actually remove nothingfrom S.

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So, any two of the conditions will characterize a basis S:

(a) |S| = n = dimV .

(b) S is l.i.

(c) S spans V .

0. (b) & (c): (original defintion)

1. (a) & (b): Any l.i. set of n = dimV vectors is a basis.

2. (a) & (c): Any spanning set of n = dimV vectors is abasis.

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Example: Describe all possible subspaces of R3.

Sol: Let H be a subspace of R3. Then dimH = 0, 1, 2, 3.

0. dimH = 0. H = {0} is the only possibility.

3. dimH = 3. Then H = R3.

1. dimH = 1. Then H has a basis B = {v}.→ every x ∈ H is a l.c. of {v};i.e. x = cv where c ∈ R.→ H is a line passing through origin, containing v.

2. dimH = 2. Then H has a basis B = {v1,v2}.→ every x ∈ H is of the form x = c1v1 + c2v2.→ H is a plane through 0, containing both v1,v2.

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Dimensions of NulA, RowA, and ColA

Def: The dimension of NulA is called the nullity of A.

Def: The dimension of RowA is called the row rank of A.

Def: The dimension of ColA is called the column rank of A.

Let A be an m× n matrix with p pivot positions. ThenA has n − p free variables, and so the general solution ofAx = 0 can be written as:

x = s1x1 + . . .+ sn−pxn−p, where s1, . . . , sn−p ∈ R.

Thm: nullity of A = dimNulA = n− p.

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Let A be an m× n matrix with p pivot positions. Thenthere are p non-zero rows in a REF of A.

Thm: row rank of A = dimRowA = p.

Let A be an m× n matrix with p pivot positions. Thenthere are p pivot columns of A:

Thm: column rank of A = dimColA = p.

Recall the definition that rankA = no. of pivot positionsin A. So:

Thm: row rank of A = column rank of A = rankA.

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In summary, we have the following “rank theorem”:

Thm 14 (P.249): Let A be an m× n matrix. Then:

dimRowA = dimColA = rankA

rankA+ dimNulA = n.

In the language of matrix transformations, we have:

Thm: Let T : Rn → Rm be a matrix transformation. Then:

dim ImT + dimkerT = n = dimRn.

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