mh1200 lecture 12b

7/23/2019 MH1200 Lecture 12b

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Basis and Dimension

7/23/2019 MH1200 Lecture 12b


Finite-dimensional spacesRecall we defined a finite-dimensional space as one that

is spanned by a finite number of vectors.

Today we will see how questions about a finite-dimensionalvector space can be translated into questions about

Euclidean space.

Then we can use all the techniques we have developed

for Euclidean space to solve them.

7/23/2019 MH1200 Lecture 12b


BasisLet be a vector space. Vectors are a finite

basis for if they are linearly independent and

v1, . . . , vrV

V

Note: Every vector can be written as a linear

combination of in a unique way.

span({v1, . . . , vr}) = V .

v ∈ V

1, . . . , r

7/23/2019 MH1200 Lecture 12b


Example

Consider the vector space of degree polynomials.2 ≤ 2

A basis for this space is given by .{1, x, x2}

We saw yesterday that these functions are linearly

independent and span .2

More generally, a basis for the vector space ofpolynomials of degree is given by .

P n

≤ n {1, x, x2

, . . . , xn}

This is the canonical basis for .n

7/23/2019 MH1200 Lecture 12b


Example


Another basis for this space is given by .{1 + x, x, x2− x}

These functions span because their span contains

the functions .

P 2

1 = 1 + x + (−1)x

x = x

1, x, x2

x2= x

2− x+ x

7/23/2019 MH1200 Lecture 12b


Example


Another basis for this space is given by .{1 + x, x, x2− x}

These functions are linearly independent. Suppose

a ·

1 + (a + b−

c)x + c · x2 = 0

a(1 + x) + bx + c(x2 − x) = 0.

.

By linear independence of this means1, x, x2

a = 0, a+ b− c = 0, c = 0 a = 0, b = 0, c = 0

7/23/2019 MH1200 Lecture 12b


ExampleConsider the vector space of 2-by-3 matrices with real

entries.

A basis for this space is given by the matrices

1 0 0

0 0 0 , 0 1 0

0 0 0 , 0 0 1

0 0 0 ,

0 0 0

1 0 0

,

0 0 0

0 1 0

,

0 0 0

0 0 1

.

This is the canonical basis for M (2, 3).

7/23/2019 MH1200 Lecture 12b


Construction of a basis

Theorem: Every finite-dimensional vector space has afinite basis.

V

Proof: Let be such that .v1, . . . , vr span({v1, . . . , vr}) = V

We know that such vectors exist as is finite-dimensional.V

If are linearly independent, then we have found a

basis.

v1, . . . , vr

Otherwise, some can be expressed as a linear

combination of the others.i

vi = a1v1 + . . .+ ai−1vi−1 + ai+1vi+1 + . . .+ arvr

7/23/2019 MH1200 Lecture 12b




V

Otherwise, some can be expressed as a linearcombination of the others.i

vi = a1v1 + . . .+ ai−1vi−1 + ai+1vi+1 + . . .+ arvr

Proof:

This means

span({v1, . . . , vr}) = span({v1, . . . , vi−1, vi+1, . . . , vr})

We can “throw away” without changing the span.i

7/23/2019 MH1200 Lecture 12b




V

Proof:

We can “throw away” without changing the span.i

If the vectors are linearly

independent, then we are done.1, . . . , i−1, i+1, . . . , r

Otherwise, we repeat this process.

This process must terminate, and when it does, we have

found a basis.

the “throw away” theorem

7/23/2019 MH1200 Lecture 12b


CoordinatesWe discussed the coordinates of a vector with respect to

a given basis in Euclidean space.

Let be a basis for .B = {(1, 0, 0), (1, 1, 0), (1, 1, 1)} R

3

(0,−1, 1) = 1 · (1, 0, 0) + (−2) · (1, 1, 0) + 1 · (1, 1, 1)

Example:

The coordinates of with respect to thebasis are

v = (0,−

1, 1)

(v)B = (1,−2, 1).

7/23/2019 MH1200 Lecture 12b


CoordinatesWe can do the same thing in a general finite-dimensional

vector space .V

Fix some basis for . We have already seen

such a basis exists.

v1, . . . , vr V

Every vector can be written as a unique linear

combination of the basis vectors .u ∈

v1, . . . , vr

u = a1v1 + · · ·+ arvr

(a1, . . . , ar) ∈ Rr

The coordinate vector of with respect to the

basis is .1, . . . , r

7/23/2019 MH1200 Lecture 12b


ExampleConsider the vector space of degree polynomials.P 2 ≤ 2

A basis for this space is given by .{1 + x, x, x2− x}

What is the coordinate vector of with respect to thisbasis?

x

2

It is because(0, 1, 1)

x2 = 0 · (1 + x) + 1 · x + 1 · (x2 − x).

7/23/2019 MH1200 Lecture 12b


Coordinate MappingFix some basis for .1, . . . , r V

u = a1v1 + · · ·+ arvr

(a1, . . . , ar) ∈ Rr

then the coordinate vector of with respect to the

basis is .v1, . . . , vr

u

The basis defines a mapping= {v1, . . . , vr}

coordB : V → Rr

where if .coordB(u) = (a1, . . . , ar) u = a1v1 + · · ·+ arvr

This is a function, because the representation is unique.

If

7/23/2019 MH1200 Lecture 12b


Coordinate MappingThe basis defines a mappingB = {v1, . . . , vr}

coordB : V → Rr


This is a function, because the representation is unique.

Claim: The coordinate mapping is one-to-one.

coordB(u1) = coordB(u2)If then .1 = 2

Claim: The coordinate mapping is onto.

for any (a1, . . . , ar) ∈ Rr

a1v1 + · · ·+ arvr ∈

7/23/2019 MH1200 Lecture 12b



coordB : V → Rr


This mapping is one-to-one and onto. It is a bijection between and .V R

r

This mapping has even more nice properties.

coordB(u + v) = coordB(u) + coordB(v)

u = a1v1 + · · · + arvr

v = b1v1 + · · · + brvru + v = (a1 + b1)v1 + · · · + (ar + br)vr

7/23/2019 MH1200 Lecture 12b



coordB : V → Rr


This mapping is a bijection between and .V Rr

§

coordB(u + v) = coordB(u) + coordB(v)§

§ coordB(c · u) = c · coordB(u) c ∈ Rfor any

u = a1v1 + · · ·+ arvrIf then c · u = c(a1v1 + · · · + arvr)

= ca1v1 + · · · + carvr

7/23/2019 MH1200 Lecture 12b


ExampleConsider the vector space of degree polynomials.P 2 ≤ 2

Fix the basis .B = {1,x ,x2}

coordB(3− 2x + 5x2) = (3,−2, 5)

coordB(−

1 + x + x2) = (−

1, 1, 1)

(3− 2x + 5x2) + (−1 + x + x2) = 2− x + 6x2

(3,−2, 5) + (−1 + 1 + 1) = (2,−1, 6)

Adding polynomials in is just like adding vectors in .2 R3

7/23/2019 MH1200 Lecture 12b


IsomorphismLet and be two vector spaces. A function

is an isomorphism if

V W f : V → W

is a bijection between and .V §

§

§

f W

f (u + v) = f (u) + f (v) for all u, v ∈ V

f (c · v) = c · f (v) for all c ∈ R, v ∈ V

Two spaces are isomorphic if there is an isomorphism

between them.

Isomorphic spaces are essentially the same.

7/23/2019 MH1200 Lecture 12b


Linear combinationsLet and be two vector spaces. If is

an isomorphism then

V W f : V →W

f (c1v1 + c2v2 + · · · + crvr) = c1f (v1) + c2f (v2) + · · · + crf (vr)

This follows by progressively applying the properties of an

isomorphism.

f ((c1v1) + (c2v2 + · · · + crvr)) = f (c1v1) + f (c2v2 + · · · + c

rvr)

= c1f (v1) + f (c2v2 + · · · + crvr)

= c1f (v1) + c2f (v2) + · · · + crf (vr)

7/23/2019 MH1200 Lecture 12b



coordB : V → Rr


This mapping is a bijection between and .V Rr

§

coordB(u + v) = coordB(u) + coordB(v)§

§ coordB(c · u) = c · coordB(u) c ∈ Rfor any

If has a basis with many elements then is

isomorphic to .

V V r

Rr

Reason: The coordinate mapping is an isomorphism.

7/23/2019 MH1200 Lecture 12b


ConsequencesExample: The vector space of polynomials of degree

is isomorphic to .2 ≤ 2

R3

An isomorphism is given by the coordinate mapping with

respect to the basis .{1, x, x2}

The coordinate mapping lets us transfer questions about

a general finite-dimensional vector space to questions about

Euclidean space.

We are already familiar with Euclidean space, and know how

to answer the questions there!

7/23/2019 MH1200 Lecture 12b


Consequences

Theorem: Let be a vector space with zero element andbasis . Then

V B = {v1, . . . , vr}

coordB(0) = 0̄ ∈ Rr

Proof: coordB(0) = coordB(0 · 0)

= 0 · coordB(0)

= 0̄

Question: Does the same hold for any isomorphism?

Li I d d

7/23/2019 MH1200 Lecture 12b


Linear IndependenceTheorem: Let be a finite-dimensional vector space and fix

a basis for . Then vectorsare linearly dependent if and only if

V

B =

{v1, . . . , vr} V u1, . . . , uk ∈

V

coordB(u1), . . . , coordB(uk) ∈ Rr

are linearly dependent.

Proof:

Suppose are linearly dependent, thusu1, . . . , uk ∈ V

c1u1 + · · · ckuk = 0

where some .ci 6= 0

Apply the function to both sides.coordB

coordB

(0

) = ¯0

7/23/2019 MH1200 Lecture 12b


Theorem: Let be a finite-dimensional vector space and fix

a basis for . Then vectors

are linearly dependent if and only if

V

B = {v1, . . . , vr} V u1, . . . , uk ∈ V



Proof:

c1u1 +· · ·

ckuk = 0 where some .ci 6= 0

Apply the function to both sides.coordB coordB(0) = 0̄

Simplifying the left hand side:

coordB(c1u1 + · · ·

+ ckuk) = coordB(c1u1) + coordB(c2u2 + · · ·

+ ckuk)

= c1coordB(u1) + coordB(c2u2 + · · · + ckuk)

= c1coordB(u1) + · · · + ckcoordB(uk)

This shows are lin. dependent.coordB(u1), . . . , coordB(uk)

7/23/2019 MH1200 Lecture 12b


Theorem: Let be a finite-dimensional vector space and fix

a basis for . Then vectors

are linearly dependent if and only if

V

B = {v1, . . . , vr} V u1, . . . , uk ∈ V



Proof:

Suppose are lin. dependent.coordB(u1), . . . , coordB(uk)

0̄ = c1 · coordB(u1) + · · · + ck · coordB(uk)

= coordB(c1u1) + · · · + coordB(ckuk)

= coordB(c1u1 + · · · + ckuk)

We know that and as is

one-to-one this means

coordB(0) = 0̄ coordB : V → Rr

c1u1 + · · · + ckuk = 0

7/23/2019 MH1200 Lecture 12b


Comment

Notice that in the last proof we only used the fact thatis an isomorphism.

coordB

We did not use any specific details about the action ofcoord

B

The last theorem actually holds with respect to any

isomorphism.

7/23/2019 MH1200 Lecture 12b


SpanTheorem: Let be a finite-dimensional vector space and fix

a basis for . Thenif and only if

V

B = {v1, . . . , vr} V v ∈ span({u1, . . . , uk})

for any .v, u1, . . . , uk ∈ V

Proof: If thenv ∈ span({u1, . . . , uk}) v = c1u1 + . . .+ ckuk

coordB(v) = coordB(c1u1 + · · ·

+ ckuk)= c1coordB(u1) + · · · + ckcoordB(uk)

and so .coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)})

coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)}),

7/23/2019 MH1200 Lecture 12b


SpanTheorem: Let be a finite-dimensional vector space and fix

a basis for . Thenif and only if

V

B = {v1, . . . , vr} V v ∈ span({u1, . . . , uk})

coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)})

for any .v, u1, . . . , uk ∈ V

Proof: If

coordB

(v

) = c1

coordB

(u1

) + · · ·

+ ck

coordB

(uk

)= coordB(c1u1 + · · · + ckuk)

Again as is one-to-one, this meanscoordB

v = c1u1 + · · · + ckuk

coordB(v) ∈ span({coordB(u1), . . . , coordB(uk)}),

7/23/2019 MH1200 Lecture 12b


Dimension TheoremTheorem: Let be a finite dimensional vector space and

let and be two bases

for .

V

B1 = {u1, . . . , ur} B2 = {v1, . . . , vs}V Then .r = s

Proof: Suppose that . From the dimension theorem

in Euclidean space we know that the largest linearly

independent set in is of size .

r < s

Rr

r

Consider . As are linearly

independent, so arecoordB1

: V → Rr

coordB1(v1), . . . , coordB1

(vs)

v1, . . . , vs

∈ Rr

.

This is a contradiction as r < s.

7/23/2019 MH1200 Lecture 12b


Dimension TheoremTheorem: Let be a finite dimensional vector space and

let and be two bases

for .

V

B1 = {u1, . . . , ur} B2 = {v1, . . . , vs}V Then .r = s

Proof: We can make an analogous argument if This

completes the proof.s < r.

7/23/2019 MH1200 Lecture 12b


Dimension

Definition: Let be a finite-dimensional vector space. Thedimension of is the number of elements in a basis for

V

V V.

This definition makes sense by the dimension theorem.

Remark: If is a vector space of dimension then it is

isomorphic to .V d

Rd

7/23/2019 MH1200 Lecture 12b


Example: PolynomialsConsider the vector space of polynomials of degree

at most

P n

.

A basis is given by {1, x, x2

, . . . , xn}.

The dimension of is .P n n + 1

7/23/2019 MH1200 Lecture 12b


Example: MatricesLet be the vector space of m-by-n matrices

with real entries.

M (m,n)

A basis for this space is given by the matrices

E ij(k, `) =

(1 if k = i, ` = j

0 otherwise

where 1 ≤ i ≤ m, 1 ≤ j ≤ n.

In total, there are elements in this basis.mn

The dimension of isM (m,n) .

S b

7/23/2019 MH1200 Lecture 12b


SubspacesTheorem: Let be a vector space of dimension with a

basis and let If

rV

U ⊆ V.B = {v1, . . . , vr},

{coordB(u) : u ∈ U } ⊆ Rr

is a subspace of then is a subspace ofRr

U V.

S =

Proof: As is a subspace of it contains .S R 0̄

Thus as this is the only vector satisfying0 ∈ U coordB(v) = 0̄.

S b

7/23/2019 MH1200 Lecture 12b



basis and let If

rV

U ⊆ V.B = {v1, . . . , vr},



U V.

S =

Proof: Now we see that is closed under scalar mult.U

Let and As is a subspace,

alsou ∈ U, c ∈ R s = coordB(u) ∈ S. S

c · s ∈ S.

Note that , and is theonly vector with this property as is one-to-one.

coordB(c ·

u) = c ·

coordB(u) = c ·

s c ·

u

coordB

Thus c · u ∈ U.

S b

7/23/2019 MH1200 Lecture 12b



basis and let If

rV

U ⊆ V.B = {v1, . . . , vr},



U V.

S =

Proof: Now we see that is closed under vector addition.U

Take and letu1, u2 ∈ U s1 = coordB(u1), s2 = coordB(u2) ∈ S.

S As is a subspace also s1 + s2 ∈ S.

Note that coordB(u1 + u2) = coordB(u1) + coordB(u2) = s1 + s2

and this is the only vector with this property (as one-to-one).

Thus u1 + u2 ∈ U.

E l

7/23/2019 MH1200 Lecture 12b


ExampleConsider the vector space of 2-by-2 matrices.M (2, 2)

S =

a11 a12

a21 a22

: a11 + a22 = 0

Let

be the subset of of matrices whose diagonalelements sum to zero.

M (2, 2)

We could show this directly and verify that contains

the all-zero matrix and satisfies the closure conditions.S

Is a subspace?S

7/23/2019 MH1200 Lecture 12b


Consider the vector space of 2-by-2 matrices.M (2, 2)

S =a11 a12

a21 a22

: a11 + a22 = 0

Let

be the subset of of matrices whose diagonal

elements sum to zero.

M (2, 2)

A basis for is given by the set of matricesM (2, 2)

E = 1 0

0 0 , 0 1

0 0 , 0 0

1 0 , 0 0

0 1 .

Note that

coordE

a11 a12

a12 a22

= (a11, a12, a21, a22)

7/23/2019 MH1200 Lecture 12b



S =

a11 a12

a21 a22

: a11 + a22 = 0

Let



M (2, 2)

Consider the set .T = {coordE (A) : A ∈ S } ⊆ R4

By the previous theorem, if is a subspace then so is .T S

But is simply the null space of the matrixand therefore is a subspace.

1 0 0 1

E l ti d

7/23/2019 MH1200 Lecture 12b



S =

a11 a12

a21 a22

: a11 + a22 = 0

Let



M (2, 2)

Example continued

What is a basis for S ?

To do this, we can first find a basis for

T = {coordE (A) : A ∈ S }

= {(a11, a12, a21, a22) : a11 + a22 = 0}

7/23/2019 MH1200 Lecture 12b




= {(a11, a12, a21, a22) : a11 + a22 = 0}

We can do this by finding the special solutions:

{(−1, 0, 0, 1), (0, 1, 0, 0), (0, 0, 1, 0)}

Now we look for vectors in which map to these special

solutions under .S

coordE

coordE

0 10 0

= (0, 1, 0, 0)

coordE

0 01 0

= (0, 0, 1, 0)

coordE

−

1 00 1

= (

−1, 0, 0, 1)

7/23/2019 MH1200 Lecture 12b




= {(a11

, a12

, a21

, a22

) : a11

+ a22

= 0}

The matrices form a basis for .

−1 0

0 1

,

0 1

0 0

,

0 0

1 0

S

They are linearly independent as their coordinate imagesare.

They span as their coordinate images span .S T

The dimension of is 3.S

mh1200 lecture 12b

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