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Lecture 28: 8.2 Isomorphism

Wei-Ta Chu

2012/1/2

Theorem 8.2.3

Theorem 8.2.3: Every real n-dimensional vector space is isomorphic to Rn. Proof: Let Vbe a real n-dimensional vector space. To prove that Vis isomorphic to Rnwe must find a linear transformation

T:V→

Rnthat is one

-to-one and onto. transformation

T:V→

Rnthat is one

-to-one and onto. Let v1, v2, ..., vnbe any basis for V, let u=k1v1+ k2v2+ ...+knvnbe the representation of a vector uin Vas a linear combination of the basis vectors Define the transformation T:V→Rnby T(u)=(k1, k2, ..., k n). We will show that Tis an isomorphism. 2

Theorem 8.2.3

To prove the linearity, let uand vbe vectors in V, let cbe a scalar, and let u=k1v1+k2v2+...+knvnand v=d1v1+d2v2+...+dnvnbe the representations of uand vas linear combinations of the basis vectors. Then, Then, 3

Therefore, Tis linear.

Theorem 8.2.3

To show Tis one-to-one, we must show that if uand vare distinct vectors in V, then so are their images in Rn. But if uis not equal to v, and if the representations of these

vector in terms of the basis vectors are described above, then we must have for at least one

i. Thus, then we must have for at least one i. Thus, which shows that uand vhave distinct images under T. The transformation Tis onto, for if w=(k1, k2, ...kn) is any vector in Rn, then it follows that wis the image under Tof the vector u=k1v1+k2v2+...+knvn 4

Example

The mapping

from Pn-1to Rnis one-to-one, onto, and linear. This is called the natural isomorphism from Pn-1to Rnbecause, as the following computations show, it maps the natural basis {1, x, x2, ..., xn-1} for P into the standard basis for basis {1, x, x2, ..., xn-1} for Pn-1 into the standard basis for Rn: 5

Example

The matrices form a basis for the vector space M22of 2 by

2 matrices. An isomorphism T: M22→ R4can be

constructed by first writing a matrix Ain M22in terms of the basis vectors as and then defining Tas T(A) = (a1, a2, a3, a4) More generally, this idea can be used to show that the vector space Mmnof mby nmatrices with real entries is isomorphic to Rmn. 6

Inner Product Space Isomorphisms

In the case where Vis a real n-dimensional inner product space, both Vand Rnhave a geometric structure arising from their respective inner products. It is reasonable to inquire if there exists an isomorphism form Vto

Rnthat preserves the geometric structure as well

form Vto

Rnthat preserves the geometric structure as well

as the algebraic structure. For example, we would want orthogonal vectors in Vto have orthogonal counterparts in Rn. 7

Inner Product Space Isomorphisms

In order for an isomorphism to preserve geometric

structure, it obviously has to preserve inner products, since notions of length, angle, and orthogonality are all based on the inner product. Thus, if Vand W are inner product spaces, then we call an

Thus, if

Vand W are inner product spaces, then we call an isomorphism T:V→Wan inner product space isomorphism if 8

Inner Product Space Isomorphisms

It can be proved that if Vis any real n-dimensional inner product space and Rnhas the Euclidean inner product, then there exists an inner product space isomorphism form Vto Rn. Under such an isomorphism, the inner product space V Under such an isomorphism, the inner product space V has the same algebraic and geometric structure as Rn. Every n-dimensional inner product space is a "carbon copy" of Rnwith the Euclidean inner product that differs only in the notation used to represent vectors. 9

Example

Let Rnbe the vector space of real n-tuples in comma- delimited form, let Mnbe the vector space of real nby 1 matrices, let Rnhave the Euclidean inner product , and let Mnhave the inner product in which uand vare expressed in column form. in which uand vare expressed in column form.

The mapping T: Rn→ Mndefined by

is an inner product space isomorphism, so the distinction between Rnand Mnis essentially notational. 10 Lecture 28: 8.3 Compositions and Inverse TransformationsTransformations

Wei-Ta Chu

2012/1/2

Composition of Linear

Transformations

Definition: If T1: U→ Vand T2: V→ Ware linear transformations, then the composition of T2with T1, denoted by T2T1(which is read "T2circle T1"), is the function defined by the formula(T

T)(u) =

T(T(u))

(T2

T1)(u) =

T2(T1(u))

where uis a vector in U.

12U V W

uT1(u)T2(T1(u))T 2T1 T 1T2

Theorem 8.3.1

Theorem 8.3.1: If T1: U→ Vand T2: V→ Ware linear transformations, then (T2T1): U→ Wis also a linear transformation Proof: If uand vare vectors in Uand cis a scalar, then(T

T)(u+v) =

T(T(u+v)) =

T(T(u) +

T(v)) (T2

T1)(u+v) =

T2(T1(u+v)) =

T2(T1(u) +

T1(v))

= T2(T1(u)) + T2(T1(v)) = (T2T1)(u) + (T2T1)(v) and (T2T1)(cu) = T2(T1(cu)) = T2(cT1(u)) =cT2(T1(u)) = c(T2T1)(u) 13

Example

Let T1:P1→P2and T2: P2→P2be the linear

transformations given by the formulas T1(p(x))=xp(x) and T

2(p(x))=p(2x+4)

Then the composition (T2T1): P1→P2is given by the formula (

TT)(p(x))

formula (

T2T1)(p(x))

=T2(T1(p(x)))=T2(xp(x))=(2x+4)p(2x+4)

In particular, if p(x)=c0+c1x, then

(T2T1)(p(x)) = (T2T1)(c0+c1x) =(2x+4) (c0+c1(2x+4))=c0(2x+4) + c1(2x+4)2 14

Example

If T:V→Vis any linear operator, and if I:V→Vis the identity operator, then for all vectors vin V, we have (TI)(v)=T(I(v))=T(v) (IT)(v)=I(T(v))=T(v)It follows that TIand I

Tare the same as

It follows that

TIand I

Tare the same as

T; that is TI=Tand IT= T More compositions: T1:U→V, T2:V→W, T3:W→Y. The composition (T3T2T1)(u) = T3(T2(T1(u))) 15

Inverse Linear Transformations

A matrix operator TA:Rn→Rnis one-to-one if and only if the matrix Ais invertible. If wis the image of a vector x under the operator TA, then xis the image under TA-1of the vector w. If T:V→W is a linear transformation, then the range of T,

If T:V→W

is a linear transformation, then the range of T, denoted by R(T), is the subspace of Wconsisting of all images under Tof vectors in V. If Tis one-to-one, then each vector win R(T) is the image of a unique vector vin V. The inverse of T(denoted by T-1) maps wback into v. 16

Inverse Linear Transformations

T-1: R(T)→Vis a linear transformation. Moreover, T -1(T(v)) = T-1(w) = v

T(T-1(w))=T(v)=w

Example

The linear transformation T:Pn→Pn+1is given by

T(p)=T(p(x))=xp(x), which is one-to-one

In this case the range of Tis not all of Pn+1but rather the

subspace of Pn+1consisting of polynomials with a zero constant term. constant term. T(c0+c1x+...+cnxn) = c0x+c1x2+...+cnxn+1

It follows that T-1:R(T)→Pnis given by

T -1(c0x+c1x2+...+cnxn+1) = c0+c1x+...+cnxn In the case where n3, T-1(2x-x2+2x3+3x4)=2-x+5x2+3x318

Example

Let T:R3→R3be the linear operator defined by

T(x1,x2,x3) = (3x1+x2, -2x1-4x2+3x3, 5x1+4x2-2x3)

Determine whether Tis one-to-one; if so, find T-1(x1,x2,x3)

Solution: The standard matrix for Tis

This matrix is invertible.

Example

Therefore, T-1(x1,x2,x3)

= (4x1-2x2-3x3, -11x1+6x2+9x3, -12x1+7x2+10x3) 20

Composition of One-to-One Linear

Transformations

Theorem 8.3.2: If T1:U→Vand T2:V→Ware one-to-one linear transformations, then

T2T1is one-to-one

(T2T1)-1=T1-1T2-1Proof(a): We want to show that T T maps distinct

Proof(a): We want to show that

T1maps distinct

vectors in Uinto distinct vectors in W. If uand vare distinct vectors in U, then T1(u) and T1(v) are distinct vectors in Vsince T1is one-to-one. The fact that T2is one-to-one imply that T2(T1(u)) and T2(T1(v)) are also distinct vectors. These can also expressed as (T2T1)(u) and (T2T1)(v) so T2T1maps uand vinto distinct vectors in W. 21

Composition of One-to-One Linear

Transformations

Proof(b): We want to show (T2T1)-1(w) = (T1-1T2-1) (w).

Let u= (T2T1)-1(w). Then, (T2T1)(u) = w, or

equivalently, T2(T1(u))=w. Taking T2-1of each side of this equation, then taking T -1of each side of the result, then equation, then taking

T1-1of each side of the result, then

u=T1-1(T2-1(w)), or equivalently, u= (T1-1T2-1)(w). The inverse of a composition is the composition of the inverse in the reverse order. 22

Exercises

Sec. 8.1: 3, 13, 16, 35, 45 (True-False)

Sec. 8.2: 6, 8, 15, 21 (True-False)

Sec. 8.3: 3, 14, 20, 24 (True-False)

Sec. 8.4: 9, 13, 22 (True

-False)

Sec. 8.4: 9, 13, 22 (True

-False)

Sec. 8.5: 4, 10, 13, 23 (True-False)

23
Lecture 28: 8.4 Matrices for General Linear TransformationsTransformations

Wei-Ta Chu

2012/1/2

Matrices of Linear Transformations

Suppose that Vis an n-dimensional vector space, Wis an m-dimensional vector space, and that T:V→Wis a linear transformation. Suppose further that Bis a basis for V, that B" is a basis for

W, and that for each vector

xin

V, the coordinate

for

W, and that for each vector

xin

V, the coordinate

matrices for xand T(x) are [x]Band [T(x)]B", respectively. 25
x [x]BT(x) [T(x)]B"

A vector in V

(n-dimensional)

A vector in RnA vector in RmA vector in W

(m-dimensional)

Matrices of Linear Transformations

Find an mby nmatrix such that multiplication by Amaps the vector [x]Binto the vector [T(x)]B"for each xin V. If we can do so, we can execute the linear transformation Tby using matrix multiplication. Find T(x) indirectly

Find T(x) indirectly

Step 1. Compute the coordinate vector [x]B

Step 2. Multiple [x]Bon the left by Ato produce [T(x)]B" Step 3. Reconstruct T(x) from its coordinate vector [T(x)]B" 26
x [x]BT(x) [T(x)]B" (1) (2)(3)

Multiply by A

Matrices of Linear Transformations

The key is to find an mby nmatrix Awith the property

A[x]B= [T(x)]B"

For this purpose, let B={u1, u2, ..., un} be a basis for the n-dimensional space Vand B"={v1, v2, ..., vm} a basis for the m-dimensional space W. the m-dimensional space W. A[u1]B=[T(u1)]B", A[u2]B=[T(u2)]B", ..., A[un]B=[T(un)]B", But, 27

Matrices of Linear Transformations

So, 28

Matrices of Linear Transformations

Substituting these results

The successive columns of Aare the coordinate vectors of T(u1), T(u2), ..., T(un) with respect to the basis B". 29

Matrices of Linear Transformations

We will call this the matrix for Trelative to the bases B and B" and will denote it by the symbol [T]B",B. The matrix has the property [T]B",B[x]B= [T(x)]B" In the special case where TA: Rn→Rmis multiplication by A, and where Band

B" are the standard bases for

Rnand

A, and where

Band

B" are the standard bases for

Rnand

Rm, respectively, then [T]B",B= A

Example

Let T:P1→P2be the linear transformation defined by T(p(x)) = xp(x). Find the matrix for Twith respect to the standard bases B=[u1, u2] and B"=[v1, v2, v3] where u1=1, u

2=x; v1=1, v2=x, v3=x2Solution: From the given formula for

Twe obtain

Solution: From the given formula for

Twe obtain

T(u1)=T(1)=(x)(1)=x, and T(u2)=T(x)=(x)(x)=x2

The coordinate vectors for T(u1) and T(u2) relative to B" are 31

Example

Thus, the matrix for Twith respect to Band B" is

Example

Let T:P1→P2be the linear transformation in the previous example, and use the three-step procedure to perform the computation T(a+bx)=x(a+bx)=ax+bx2 Step 1. The coordinate matrix for x=a+bxrelative to the basis B={1,x} is Step 2: Multiplying [x]Bby the matrix [T]B",B, we obtain Step 3. Reconstructing T(x)=T(a+bx) from [T(x)]B"we obtain T(a+bx)=0+ax+bx2=ax+bx2 33

Example

Let T:R2→R3be the linear transformation defined by

Find the matrix for the transformation

Twith respect to

Find the matrix for the transformation

Twith respect to

the bases B={u1,u2} for R2and B"={v1,v2,v3} for R3, where 34

Example

From the formula for T,

Expressing these vectors as linear combination of v1, v2, and v

3, we obtain T(u

1)=v 1-2v

3, T(u

2)=3v 1+v 2-v 3. 3 1 1 3 2 1 2 3 Thus, 35
quotesdbs_dbs20.pdfusesText_26

[PDF] [PDF] (Lecture 28 Compositions and Inverse Transformations [???e????])

Lecture 28: 8.2 Isomorphism

Wei-Ta Chu

2012/1/2

Theorem 8.2.3

T:V→

Rnthat is one

T:V→

Rnthat is one

Theorem 8.2.3

Therefore, Tis linear.

Theorem 8.2.3

Example

The mapping

Example

2 matrices. An isomorphism T: M22→ R4can be

Inner Product Space Isomorphisms

Rnthat preserves the geometric structure as well

Rnthat preserves the geometric structure as well

Inner Product Space Isomorphisms

In order for an isomorphism to preserve geometric

Thus, if

Inner Product Space Isomorphisms

Example

The mapping T: Rn→ Mndefined by

Wei-Ta Chu

2012/1/2

Composition of Linear

Transformations

T)(u) =

T(T(u))

T1)(u) =

T2(T1(u))

12U V W

Theorem 8.3.1

T)(u+v) =

T(T(u+v)) =

T(T(u) +

T1)(u+v) =

T2(T1(u+v)) =

T2(T1(u) +

T1(v))

Example

Let T1:P1→P2and T2: P2→P2be the linear

2(p(x))=p(2x+4)

TT)(p(x))

T2T1)(p(x))

In particular, if p(x)=c0+c1x, then

Example

Tare the same as

It follows that

Tare the same as

Inverse Linear Transformations

If T:V→W

Inverse Linear Transformations

T(T-1(w))=T(v)=w

Example

T(p)=T(p(x))=xp(x), which is one-to-one

It follows that T-1:R(T)→Pnis given by

Example

Let T:R3→R3be the linear operator defined by

T(x1,x2,x3) = (3x1+x2, -2x1-4x2+3x3, 5x1+4x2-2x3)

Solution: The standard matrix for Tis

This matrix is invertible.

Example

Therefore, T-1(x1,x2,x3)

Composition of One-to-One Linear

Transformations

T2T1is one-to-one

Proof(a): We want to show that

T1maps distinct

Composition of One-to-One Linear

Transformations

Let u= (T2T1)-1(w). Then, (T2T1)(u) = w, or

T1-1of each side of the result, then

Exercises

Sec. 8.1: 3, 13, 16, 35, 45 (True-False)

Sec. 8.2: 6, 8, 15, 21 (True-False)

Sec. 8.3: 3, 14, 20, 24 (True-False)

Sec. 8.4: 9, 13, 22 (True