Lecture 27: 8.2 Isomorphism

Lecture 27: 8.2

Isomorphism

Wei-Ta Chu

Theorem 8.2.2

Theorem 8.2.2: If V is a finite-dimensional vector space, and if T: V → V is a linear operator, then the following statements re equivalent

T is one-to-one

ker(T) = {0} ker(T) = {0}

Example

Show that if V is a finite-dimensional vector space and c is any nonzero scalar, then the linear operator T:V →V defined by T(v)=cv is one-to-one and onto

Solution: The operator T is onto (and hence one-to-one) for if v is any vector in V then that vector is the image of for if v is any vector in V then that vector is the image of the vector (1/c)v

Example

Example

Let be the sequence space, and consider the linear “shifting operators” on V defined by

T₁(u₁, u₂, …, u_n, …) = (0, u₁, u₂, …, u_n, …)

T₂(u₁, u₂, …, u_n, …) = (u₂, u₃, …, u_n, …)

(a) Show that T is one-to-one but not onto. (a) Show that T₁ is one-to-one but not onto. (b) Show that T₂ is onto but not one-to-one.

Solution (a): The operator T₁ is one-to-one because

distinct sequences in obviously have distinct images. This operator is not onto because no vector in maps into the sequence (1, 0, 0, …, 0, …), for example.

Example

Solution (b): The operator T₂ is not one-to-one because, for example, the vectors (1, 0, 0, …, 0, …) and (2, 0, 0, …, 0, …) both map to (0, 0, 0, …, 0, …). This operator is onto because every possible sequence of real numbers can be obtained with an appropriate choice of the numbers u , be obtained with an appropriate choice of the numbers u₂,

Example

The linear transformations T₁: P_{3 →R}4 and T₂: M_{22 →R}4 defined by

are both one-to-one and onto.

Verify by showing that their kernels contain only the zero vector.

Example

Let T:P_{n →Pn+1} be the linear transformation

T(p)=T(p(x))=xp(x)

If p=p(x)=c₀+c₁x+…+c_nxn and q=q(x)=d₀+d₁x+…+d_nxn

are distinct polynomials, then they are different in at least one coefficient.

one coefficient.

Thus, T(p)=c₀x+c₁x2_+…+c

nxn+1 and

T(q)=d₀x+d₁x2+…+d_nxn+1 also differ in at least one coefficient.

If follows that T is one-to-one since it maps distinct

polynomials p and q into distinct polynomials T(p) and

Dimension and Linear

Transformations

A linear transformation T:V →W in the case where V and

W are finite-dimensional:

If dim(W) < dim(V), then T cannot be one-to-one If dim(V) < dim(W), then T cannot be onto.

Isomorphism

Definition: If a linear transformation T: V→ W is both

one-to-one and onto, then T is said to be an isomorphism, and the vector spaces V and W are said to be isomorphic. The word isomorphic is derived from the Greek words

iso, meaning “identical” and morphe, meaning “form”. iso, meaning “identical” and morphe, meaning “form”.

Isomorphic vector spaces have the same “algebraic form,” even though they may consist of different kinds of

Isomorphism

The isomorphism matches

up vector operations in P₂ and R3_.

Operation in P₂ Operation in R3

Theorem 8.2.3

Theorem 8.2.3: Every real n-dimensional vector space is isomorphic to Rn_.

Proof: Let V be a real n-dimensional vector space. To prove that V is isomorphic to Rn _{we must find a linear}

transformation T:V→Rn that is one-to-one and onto. transformation T:V→Rn that is one-to-one and onto. Let v₁, v₂, …, v_n be any basis for V, let u=k₁v₁ + k₂v₂ +

…+k_nv_n be the representation of a vector u in V as a linear combination of the basis vectors

Define the transformation T:V→Rn by T(u)=(k₁, k₂, …,

Theorem 8.2.3

To prove the linearity, let u and v be vectors in V, let c be a scalar, and let u=k₁v₁+k₂v₂+…+k_nv_n and

v=d₁v₁+d₂v₂+…+d_nv_n be the representations of u and v as linear combinations of the basis vectors.

Then, Then,

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Theorem 8.2.3

To show T is one-to-one, we must show that if u and v are distinct vectors in V, then so are their images in Rn_{. But if}

u is 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 u and v have distinct images under T. The transformation T is onto, for if w=(k₁, k₂, …k_n) is any vector in Rn_{, then it follows that w is the image under T of}

Example

The mapping

from P_n-1 to Rn _{is one-to-one, onto, and linear. This is}

called the natural isomorphism from P_n-1 to Rn because, 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 P_n-1 into the standard basis for

Rn:

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Example

The matrices form a basis for the vector space M₂₂ of 2 by 2 matrices. An isomorphism T: M_{22 → R}4 _{can be}

constructed by first writing a matrix A in M₂₂ in terms of the basis vectors as

and then defining T as T(A) = (a₁, a₂, a₃, a₄)

Inner Product Space Isomorphisms

In the case where V is a real n-dimensional inner product space, both V and Rn _{have a geometric structure arising}

from their respective inner products.

It is reasonable to inquire if there exists an isomorphism form V to Rn that preserves the geometric structure as well form V to Rn that preserves the geometric structure as well as the algebraic structure.

For example, we would want orthogonal vectors in V to have orthogonal counterparts in Rn.

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 V and W are inner product spaces, then we call an Thus, if V and W are inner product spaces, then we call an isomorphism T:V→W an inner product space

Inner Product Space Isomorphisms

It can be proved that if V is any real n-dimensional inner product space and Rn _{has the Euclidean inner product,}

then there exists an inner product space isomorphism form V to 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 Rn with the Euclidean inner product that differs only in the notation used to represent vectors.

Example

Let Rn be the vector space of real n-tuples in

comma-delimited form, let M_n be the vector space of real n by 1 matrices, let Rn have the Euclidean inner product

, and let M_n have the inner product in which u and v are expressed in column form. in which u and v are expressed in column form. The mapping T: Rn _{→ Mn} defined by

Lecture 27: 8.3

Compositions and Inverse

Transformations

Transformations

Wei-Ta Chu

Composition of Linear

Transformations

Definition: If T_{1: U → V and T2: V → W are linear} transformations, then the composition of T₂ with T₁, denoted by T₂。T₁ (which is read “T₂ circle T₁”), is the function defined by the formula

Theorem 8.3.1

Theorem 8.3.1: If T_{1: U → V and T2: V → W are linear} transformations, then (T₂。T_{1): U → W is also a linear} transformation

Proof: If u and v are vectors in U and c is 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))

= (T₂。T₁)(u) + (T₂。T₁)(v) and

(T₂。T₁)(cu) = T₂(T₁(cu)) = T₂(cT₁(u)) =cT₂(T₁(u)) = c(T₂。T₁)(u)

Example

Let T₁:P_1→P2 and T₂: P_2→P2 be the linear

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

T₂(p(x))=p(2x+4)

Example

If T:V→V is any linear operator, and if I:V →V is the identity operator, then for all vectors v in V, we have (T。I)(v)=T(I(v))=T(v)

(I。T)(v)=I(T(v))=T(v)

It follows that T。I and I。T are the same as T; It follows that T。I and I。T are the same as T; that is T。I =T and I。T = T

More compositions: T_{1:U→V, T2:V→W, T3:W→Y. The} composition (T₃。T₂。T₁)(u) = T₃(T₂(T₁(u)))

Inverse Linear Transformations

A matrix operator T_A:Rn_→Rn is one-to-one if and only if the matrix A is invertible. If w is the image of a vector x under the operator T_A, then x is the image under T_A-1 of 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 W consisting of all images under T of vectors in V. If T is one-to-one, then each vector w in R(T) is the image of a unique vector v in

V.

Inverse Linear Transformations

T-1_{: R(T)→V is a linear transformation. Moreover,} T-1_{(T(v)) = T}-1_{(w) = v}

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

Example

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

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

In this case the range of T is not all of P_n+1 but rather the subspace of P_n+1 consisting of polynomials with a zero constant term.

constant term.

T(c₀+c₁x+…+c_nxn) = c₀x+c₁x2+…+c_nxn+1

It follows that T-1_:R(T)→Pn is given by

T-1(c₀x+c₁x2+…+c_nxn+1) = c₀+c₁x+…+c_nxn

Example

Let T:R3_→R3 be the linear operator defined by

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

Determine whether T is one-to-one; if so, find T-1(x₁,x₂,x₃) Solution: The standard matrix for T is

This matrix is invertible.

Example

Therefore, T-1_(x

1,x2,x3)

Composition of One-to-One Linear

Transformations

Theorem 8.3.2: If T_{1:U→V and T2:V→W are one-to-one} linear transformations, then

T₂。T₁ is one-to-one (T₂。T₁)-1 = T₁-1 。T₂-1

Proof(a): We want to show that T 。T maps distinct Proof(a): We want to show that T₂。T₁ maps distinct vectors in U into distinct vectors in W. If u and v are distinct vectors in U, then T₁(u) and T₁(v) are distinct vectors in V since T₁ is one-to-one. The fact that T₂ is one-to-one imply that T₂(T₁(u)) and T₂(T₁(v)) are also

distinct vectors. These can also expressed as (T₂。T₁ )(u) and (T₂。T₁ )(v) so T₂。T₁ maps u and v into distinct

vectors in W.

Composition of One-to-One Linear

Transformations

Proof(b): We want to show (T₂。T₁)-1 _{(w) = (T}

1-1。T2-1)

(w).

Let u= (T₂。T₁)-1 _{(w). Then, (T}

2。T1)(u) = w, or

equivalently, T₂(T₁(u))=w. Taking T₂-1 _{of each side of this}

equation, then taking T -1 of each side of the result, then equation, then taking T₁-1 of each side of the result, then

u=T₁-1(T₂-1(w)), or equivalently, u= (T₁-1。T₂-1)(w).