109
1 Remembering Back to Abstract 2 Examples 3 Theorems 4 Isomorphic Vector Spaces 5 Inverses Examples 6 Constructing Isomorphisms Chapter 9 Linear Transformations 9.2 Isomorphisms
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Chapter 9 Linear Transformations9.2 Isomorphisms
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Definition
Who remembers how to define an isomorphism? Abstractpeople?
DefinitionA linear transformation T : V →W is an isomorphism if itis both one-to-one and onto.
Implication
If T is an isomorphism, then there exists an inverse functionto T , S : W → V that is necessarily a linear transformationand so it is also an isomorphism.
In this case, we say V and W are isomorphic vector spaces.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Definition
Who remembers how to define an isomorphism? Abstractpeople?
DefinitionA linear transformation T : V →W is an isomorphism if itis both one-to-one and onto.
Implication
If T is an isomorphism, then there exists an inverse functionto T , S : W → V that is necessarily a linear transformationand so it is also an isomorphism.
In this case, we say V and W are isomorphic vector spaces.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Definition
Who remembers how to define an isomorphism? Abstractpeople?
DefinitionA linear transformation T : V →W is an isomorphism if itis both one-to-one and onto.
Implication
If T is an isomorphism, then there exists an inverse functionto T , S : W → V that is necessarily a linear transformationand so it is also an isomorphism.
In this case, we say V and W are isomorphic vector spaces.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Definition
Who remembers how to define an isomorphism? Abstractpeople?
DefinitionA linear transformation T : V →W is an isomorphism if itis both one-to-one and onto.
Implication
If T is an isomorphism, then there exists an inverse functionto T , S : W → V that is necessarily a linear transformationand so it is also an isomorphism.
In this case, we say V and W are isomorphic vector spaces.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
Example
Show that the linear transformation T : P2 → R3 with
T (a2x2 + a1x + a0) =
a2 − 2a1a1 − 2a0a0 − a2
is an isomorphism.
We are given that this is a linear transformation. How wouldwe prove this?
We will need to be able to take a function and determine ifit is an isomorphism by first justifying if it is a lineartransformation in the future ...
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
Example
Show that the linear transformation T : P2 → R3 with
T (a2x2 + a1x + a0) =
a2 − 2a1a1 − 2a0a0 − a2
is an isomorphism.
We are given that this is a linear transformation. How wouldwe prove this?
We will need to be able to take a function and determine ifit is an isomorphism by first justifying if it is a lineartransformation in the future ...
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
Example
Show that the linear transformation T : P2 → R3 with
T (a2x2 + a1x + a0) =
a2 − 2a1a1 − 2a0a0 − a2
is an isomorphism.
We are given that this is a linear transformation. How wouldwe prove this?
We will need to be able to take a function and determine ifit is an isomorphism by first justifying if it is a lineartransformation in the future ...
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
dim(R3) =
3
dim(P2) = 3
What do we need to do in order to show to prove we havean isomorphism?
We need only show T is one-to-one here because ... if thedimension is the same, then if T is one-to-one, it is abijection.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
dim(R3) = 3
dim(P2) =
3
What do we need to do in order to show to prove we havean isomorphism?
We need only show T is one-to-one here because ... if thedimension is the same, then if T is one-to-one, it is abijection.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
dim(R3) = 3
dim(P2) = 3
What do we need to do in order to show to prove we havean isomorphism?
We need only show T is one-to-one here because ... if thedimension is the same, then if T is one-to-one, it is abijection.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
dim(R3) = 3
dim(P2) = 3
What do we need to do in order to show to prove we havean isomorphism?
We need only show T is one-to-one here because ... if thedimension is the same, then if T is one-to-one, it is abijection.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
dim(R3) = 3
dim(P2) = 3
What do we need to do in order to show to prove we havean isomorphism?
We need only show T is one-to-one here because ...
if thedimension is the same, then if T is one-to-one, it is abijection.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
dim(R3) = 3
dim(P2) = 3
What do we need to do in order to show to prove we havean isomorphism?
We need only show T is one-to-one here because ... if thedimension is the same, then if T is one-to-one, it is abijection.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
So how can we show a function is one-to-one?
Easiest way, if we can do it, is to show thatT (a2x
2 + a1x + a0) = 0 has only the trivial solution(remember the Big Theorem)?
We need the following system of equations to have only onesolution:
a2 − 2a1 = 0
a1 − 2a0 = 0
−a2 + a0 = 0
What do we do with this?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
So how can we show a function is one-to-one?
Easiest way, if we can do it, is to show thatT (a2x
2 + a1x + a0) = 0 has only the trivial solution(remember the Big Theorem)?
We need the following system of equations to have only onesolution:
a2 − 2a1 = 0
a1 − 2a0 = 0
−a2 + a0 = 0
What do we do with this?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
So how can we show a function is one-to-one?
Easiest way, if we can do it, is to show thatT (a2x
2 + a1x + a0) = 0 has only the trivial solution(remember the Big Theorem)?
We need the following system of equations to have only onesolution:
a2 − 2a1 = 0
a1 − 2a0 = 0
−a2 + a0 = 0
What do we do with this?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
So how can we show a function is one-to-one?
Easiest way, if we can do it, is to show thatT (a2x
2 + a1x + a0) = 0 has only the trivial solution(remember the Big Theorem)?
We need the following system of equations to have only onesolution:
a2 − 2a1 = 0
a1 − 2a0 = 0
−a2 + a0 = 0
What do we do with this?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
1 −2 0 00 1 −2 0−1 0 1 0
∼
1 0 0 00 1 0 00 0 1 0
What does this tell us?
Since the only solution is trivial, T is one-to-one andtherefore an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
1 −2 0 00 1 −2 0−1 0 1 0
∼1 0 0 0
0 1 0 00 0 1 0
What does this tell us?
Since the only solution is trivial, T is one-to-one andtherefore an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 1
1 −2 0 00 1 −2 0−1 0 1 0
∼1 0 0 0
0 1 0 00 0 1 0
What does this tell us?
Since the only solution is trivial, T is one-to-one andtherefore an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 2
Example
Show that the linear transformation T : P2 → R3 with
T (p(x)) =
p(−1)p(0)p(1)
is an isomorphism.
Since we are talking about the same vector spaces, we willagain only worry about showing the transformation isone-to-one. So, we need to show T (p(x)) = 0. Question is,how?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 2
Example
Show that the linear transformation T : P2 → R3 with
T (p(x)) =
p(−1)p(0)p(1)
is an isomorphism.
Since we are talking about the same vector spaces, we willagain only worry about showing the transformation isone-to-one. So, we need to show T (p(x)) = 0. Question is,how?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 2
a2 − a1 + a0 = 0
a0 = 0
a2 + a1 + a0 = 0
1 −1 1 00 0 1 01 1 1 0
∼1 0 0 0
0 1 0 00 0 1 0
Since the only solution is trivial, T is one-to-one andtherefore an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 2
a2 − a1 + a0 = 0
a0 = 0
a2 + a1 + a0 = 0
1 −1 1 00 0 1 01 1 1 0
∼
1 0 0 00 1 0 00 0 1 0
Since the only solution is trivial, T is one-to-one andtherefore an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 2
a2 − a1 + a0 = 0
a0 = 0
a2 + a1 + a0 = 0
1 −1 1 00 0 1 01 1 1 0
∼1 0 0 0
0 1 0 00 0 1 0
Since the only solution is trivial, T is one-to-one andtherefore an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 2
a2 − a1 + a0 = 0
a0 = 0
a2 + a1 + a0 = 0
1 −1 1 00 0 1 01 1 1 0
∼1 0 0 0
0 1 0 00 0 1 0
Since the only solution is trivial, T is one-to-one andtherefore an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 3
Example
Determine if T : Rn×n → Rn×n with T (A) = BAB−1 with Ba fixed invertible n × n matrix is an isomorphism.
Thoughts?
In section 9.1, we first proved this is a linear transformation,then we proved it is a bijection. So we therefore have that Tis an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 3
Example
Determine if T : Rn×n → Rn×n with T (A) = BAB−1 with Ba fixed invertible n × n matrix is an isomorphism.
Thoughts?
In section 9.1, we first proved this is a linear transformation,then we proved it is a bijection. So we therefore have that Tis an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Example 3
Example
Determine if T : Rn×n → Rn×n with T (A) = BAB−1 with Ba fixed invertible n × n matrix is an isomorphism.
Thoughts?
In section 9.1, we first proved this is a linear transformation,then we proved it is a bijection. So we therefore have that Tis an isomorphism.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ? dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ? dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?
T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ? dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ? dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ? dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ? dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ?
dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ? dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ? dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ? dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ? dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ?
dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ? dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ? dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
A Theorem
TheoremIf V is a finite dimensional vector space and W is isomorphicto V , then dim(V ) = dim(W ).
Proof.Since V and W are isomorphic, what must exist?T : V →W where T is a bijection.
I If T is one-to-one, what do we know about thedimension of V compared to W ? dim(V ) ≤ dim(W )
I If T is onto, what do we know about the dimension ofV compared to W ? dim(V ) ≥ dim(W )
Since T is both one-to-one and onto, both must hold,implying that dim(V ) = dim(W )
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Another Theorem
TheoremSuppose V and W are finite dimensional vector spaces withbases V = {v1, v2, . . . , vm} and W = {w1,w2, . . . ,wm}. Wedefine T : V →W by
T (c1v1 + c2v2 + . . . + cmvm) = c1w1 + c2w2 + . . . + cmwm
Then T is an isomorphism and hence V and W areisomorphic.
The proof is a bit tedious, but fortunately, it is in the text sowe will omit it here. What is important is the implication weget from taking these two theorems. Anyone want to guess?
Implication
If we have finite dimensional vector spaces, they areisomorphic if and only if they have the same dimension.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Another Theorem
TheoremSuppose V and W are finite dimensional vector spaces withbases V = {v1, v2, . . . , vm} and W = {w1,w2, . . . ,wm}. Wedefine T : V →W by
T (c1v1 + c2v2 + . . . + cmvm) = c1w1 + c2w2 + . . . + cmwm
Then T is an isomorphism and hence V and W areisomorphic.
The proof is a bit tedious, but fortunately, it is in the text sowe will omit it here. What is important is the implication weget from taking these two theorems. Anyone want to guess?
Implication
If we have finite dimensional vector spaces, they areisomorphic if and only if they have the same dimension.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Another Theorem
TheoremSuppose V and W are finite dimensional vector spaces withbases V = {v1, v2, . . . , vm} and W = {w1,w2, . . . ,wm}. Wedefine T : V →W by
T (c1v1 + c2v2 + . . . + cmvm) = c1w1 + c2w2 + . . . + cmwm
Then T is an isomorphism and hence V and W areisomorphic.
The proof is a bit tedious, but fortunately, it is in the text sowe will omit it here. What is important is the implication weget from taking these two theorems. Anyone want to guess?
Implication
If we have finite dimensional vector spaces, they areisomorphic if and only if
they have the same dimension.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Another Theorem
TheoremSuppose V and W are finite dimensional vector spaces withbases V = {v1, v2, . . . , vm} and W = {w1,w2, . . . ,wm}. Wedefine T : V →W by
T (c1v1 + c2v2 + . . . + cmvm) = c1w1 + c2w2 + . . . + cmwm
Then T is an isomorphism and hence V and W areisomorphic.
The proof is a bit tedious, but fortunately, it is in the text sowe will omit it here. What is important is the implication weget from taking these two theorems. Anyone want to guess?
Implication
If we have finite dimensional vector spaces, they areisomorphic if and only if they have the same dimension.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Isomorphic Vector Spaces
Are the following vector spaces isomorphic?
I Pn and Rn?
I Pn and Rn+1?
I R7×4 and R28?
I Rn×m and Rnm?
I P5 and R2×3?
TheoremIf V is a vector space and dim(V ) = n, then V is isomorphicto Rn.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Isomorphic Vector Spaces
Are the following vector spaces isomorphic?
I Pn and Rn?
I Pn and Rn+1?
I R7×4 and R28?
I Rn×m and Rnm?
I P5 and R2×3?
TheoremIf V is a vector space and dim(V ) = n, then V is isomorphicto Rn.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Isomorphic Vector Spaces
Are the following vector spaces isomorphic?
I Pn and Rn?
I Pn and Rn+1?
I R7×4 and R28?
I Rn×m and Rnm?
I P5 and R2×3?
TheoremIf V is a vector space and dim(V ) = n, then V is isomorphicto Rn.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Isomorphic Vector Spaces
Are the following vector spaces isomorphic?
I Pn and Rn?
I Pn and Rn+1?
I R7×4 and R28?
I Rn×m and Rnm?
I P5 and R2×3?
TheoremIf V is a vector space and dim(V ) = n, then V is isomorphicto Rn.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Isomorphic Vector Spaces
Are the following vector spaces isomorphic?
I Pn and Rn?
I Pn and Rn+1?
I R7×4 and R28?
I Rn×m and Rnm?
I P5 and R2×3?
TheoremIf V is a vector space and dim(V ) = n, then V is isomorphicto Rn.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Isomorphic Vector Spaces
Are the following vector spaces isomorphic?
I Pn and Rn?
I Pn and Rn+1?
I R7×4 and R28?
I Rn×m and Rnm?
I P5 and R2×3?
TheoremIf V is a vector space and dim(V ) = n, then V is isomorphicto Rn.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1
Example
Find the inverse of the isomorphism T : P2 → R3 with
T (a2x2 + a1x + a0) =
a2 − 2a1a1 − 2a0a0 − a2
What do we need to find?
Given v =
abc
∈ R3, we need to find the polynomial
p(x) = a2x2 + a1x + a0 such that T (p(x)) = v.
a2 − 2a1 = a
a1 − 2a0 = b
−a2 + a0 = c
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1
Example
Find the inverse of the isomorphism T : P2 → R3 with
T (a2x2 + a1x + a0) =
a2 − 2a1a1 − 2a0a0 − a2
What do we need to find?
Given v =
abc
∈ R3, we need to find the polynomial
p(x) = a2x2 + a1x + a0 such that T (p(x)) = v.
a2 − 2a1 = a
a1 − 2a0 = b
−a2 + a0 = c
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1
Example
Find the inverse of the isomorphism T : P2 → R3 with
T (a2x2 + a1x + a0) =
a2 − 2a1a1 − 2a0a0 − a2
What do we need to find?
Given v =
abc
∈ R3, we need to find the polynomial
p(x) = a2x2 + a1x + a0 such that T (p(x)) = v.
a2 − 2a1 = a
a1 − 2a0 = b
−a2 + a0 = c
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1
Example
Find the inverse of the isomorphism T : P2 → R3 with
T (a2x2 + a1x + a0) =
a2 − 2a1a1 − 2a0a0 − a2
What do we need to find?
Given v =
abc
∈ R3, we need to find the polynomial
p(x) = a2x2 + a1x + a0 such that T (p(x)) = v.
a2 − 2a1 = a
a1 − 2a0 = b
−a2 + a0 = c
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1Now what?
1 −2 0 a0 1 −2 b−1 0 1 c
∼1 −2 0 a
0 1 −2 b0 −2 1 a + c
∼
1 0 −4 a + 2b0 1 −2 b0 0 −3 a + 2b + c
∼
1 0 0 −13(a + 2b + 4c)
0 1 0 −13(2a + b + 2c)
0 0 1 −13(a + 2b + c)
So, we have S : R3 → P2 defined by
S
abc
= −1
3((a+2b+4c)x2+(2a+b+2c)x+(a+2b+c))
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1Now what?
1 −2 0 a0 1 −2 b−1 0 1 c
∼
1 −2 0 a0 1 −2 b0 −2 1 a + c
∼
1 0 −4 a + 2b0 1 −2 b0 0 −3 a + 2b + c
∼
1 0 0 −13(a + 2b + 4c)
0 1 0 −13(2a + b + 2c)
0 0 1 −13(a + 2b + c)
So, we have S : R3 → P2 defined by
S
abc
= −1
3((a+2b+4c)x2+(2a+b+2c)x+(a+2b+c))
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1Now what?
1 −2 0 a0 1 −2 b−1 0 1 c
∼1 −2 0 a
0 1 −2 b0 −2 1 a + c
∼
1 0 −4 a + 2b0 1 −2 b0 0 −3 a + 2b + c
∼
1 0 0 −13(a + 2b + 4c)
0 1 0 −13(2a + b + 2c)
0 0 1 −13(a + 2b + c)
So, we have S : R3 → P2 defined by
S
abc
= −1
3((a+2b+4c)x2+(2a+b+2c)x+(a+2b+c))
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1Now what?
1 −2 0 a0 1 −2 b−1 0 1 c
∼1 −2 0 a
0 1 −2 b0 −2 1 a + c
∼
1 0 −4 a + 2b0 1 −2 b0 0 −3 a + 2b + c
∼
1 0 0 −13(a + 2b + 4c)
0 1 0 −13(2a + b + 2c)
0 0 1 −13(a + 2b + c)
So, we have S : R3 → P2 defined by
S
abc
= −1
3((a+2b+4c)x2+(2a+b+2c)x+(a+2b+c))
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1Now what?
1 −2 0 a0 1 −2 b−1 0 1 c
∼1 −2 0 a
0 1 −2 b0 −2 1 a + c
∼
1 0 −4 a + 2b0 1 −2 b0 0 −3 a + 2b + c
∼
1 0 0 −13(a + 2b + 4c)
0 1 0 −13(2a + b + 2c)
0 0 1 −13(a + 2b + c)
So, we have S : R3 → P2 defined by
S
abc
= −1
3((a+2b+4c)x2+(2a+b+2c)x+(a+2b+c))
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverses Example 1Now what?
1 −2 0 a0 1 −2 b−1 0 1 c
∼1 −2 0 a
0 1 −2 b0 −2 1 a + c
∼
1 0 −4 a + 2b0 1 −2 b0 0 −3 a + 2b + c
∼
1 0 0 −13(a + 2b + 4c)
0 1 0 −13(2a + b + 2c)
0 0 1 −13(a + 2b + c)
So, we have S : R3 → P2 defined by
S
abc
= −1
3((a+2b+4c)x2+(2a+b+2c)x+(a+2b+c))
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 2
Example
Find the inverse of T : P2 → R3 with
T (p(x)) =
p(−1)p(0)p(1)
What do we need to find?
Given v =
abc
∈ R3, we need to find the polynomial
p(x) = a2x2 + a1x + a0 such that T (p(x)) = v.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 2
Example
Find the inverse of T : P2 → R3 with
T (p(x)) =
p(−1)p(0)p(1)
What do we need to find?
Given v =
abc
∈ R3, we need to find the polynomial
p(x) = a2x2 + a1x + a0 such that T (p(x)) = v.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 2
Example
Find the inverse of T : P2 → R3 with
T (p(x)) =
p(−1)p(0)p(1)
What do we need to find?
Given v =
abc
∈ R3, we need to find the polynomial
p(x) = a2x2 + a1x + a0 such that T (p(x)) = v.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 2
a2 − a1 + a0 = a
a0 = b
a2 + a1 + a0 = c
1 −1 1 a0 0 1 b1 1 1 c
∼1 0 0 1
2(a− 2b + c)0 1 0 1
2(c − a)0 0 1 b
So, the inverse S : R3 → P2 is defined by
S
abc
=1
2((a− 2b + c)x2 + (c − a)x + 2b)
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 2
a2 − a1 + a0 = a
a0 = b
a2 + a1 + a0 = c
1 −1 1 a0 0 1 b1 1 1 c
∼
1 0 0 12(a− 2b + c)
0 1 0 12(c − a)
0 0 1 b
So, the inverse S : R3 → P2 is defined by
S
abc
=1
2((a− 2b + c)x2 + (c − a)x + 2b)
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 2
a2 − a1 + a0 = a
a0 = b
a2 + a1 + a0 = c
1 −1 1 a0 0 1 b1 1 1 c
∼1 0 0 1
2(a− 2b + c)0 1 0 1
2(c − a)0 0 1 b
So, the inverse S : R3 → P2 is defined by
S
abc
=1
2((a− 2b + c)x2 + (c − a)x + 2b)
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 2
a2 − a1 + a0 = a
a0 = b
a2 + a1 + a0 = c
1 −1 1 a0 0 1 b1 1 1 c
∼1 0 0 1
2(a− 2b + c)0 1 0 1
2(c − a)0 0 1 b
So, the inverse S : R3 → P2 is defined by
S
abc
=1
2((a− 2b + c)x2 + (c − a)x + 2b)
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is
S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB
Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) =
B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B =
(B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) =
A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) =
B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 =
(BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) =
A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Inverse Example 3
Example
Find the inverse of T : Rn×n → Rn×n with T (A) = BAB−1
with B a fixed invertible n × n matrix
The inverse S : Rn×n → Rn×n is S(A) = B−1AB Why?
S(T (A)) = B−1(BAB−1)B = (B−1B)A(BB−1) = A
T (S(A)) = B(B−1AB)B−1 = (BB−1)A(B−1B) = A
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Example
Construct an isomorphism between the subspaces S of P3
consisting of polynomials p(x) such that p(−1) = 0 andp(1) = 0 and Rn for the appropriate n.
We don’t know what dimension n we need yet - we have tothink about what a typical polynomial will look like andwhat affect is had by the given conditions.
What does a typical polynomial look like?p(x) = a3x
3 + a2x2 + a1x + a0
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Example
Construct an isomorphism between the subspaces S of P3
consisting of polynomials p(x) such that p(−1) = 0 andp(1) = 0 and Rn for the appropriate n.
We don’t know what dimension n we need yet - we have tothink about what a typical polynomial will look like andwhat affect is had by the given conditions.
What does a typical polynomial look like?p(x) = a3x
3 + a2x2 + a1x + a0
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Example
Construct an isomorphism between the subspaces S of P3
consisting of polynomials p(x) such that p(−1) = 0 andp(1) = 0 and Rn for the appropriate n.
We don’t know what dimension n we need yet - we have tothink about what a typical polynomial will look like andwhat affect is had by the given conditions.
What does a typical polynomial look like?
p(x) = a3x3 + a2x
2 + a1x + a0
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Example
Construct an isomorphism between the subspaces S of P3
consisting of polynomials p(x) such that p(−1) = 0 andp(1) = 0 and Rn for the appropriate n.
We don’t know what dimension n we need yet - we have tothink about what a typical polynomial will look like andwhat affect is had by the given conditions.
What does a typical polynomial look like?p(x) = a3x
3 + a2x2 + a1x + a0
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Now, the conditions - what do they do for us?
p(−1) = −a3 + a2 − a1 + a0 = 0
p(1) = a3 + a2 + a1 + a0 = 0
What can we do with this?
[−1 1 −1 1 01 1 1 1 0
]∼[
1 0 1 0 00 1 0 1 0
]What does this tell us?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Now, the conditions - what do they do for us?
p(−1) = −a3 + a2 − a1 + a0 = 0
p(1) = a3 + a2 + a1 + a0 = 0
What can we do with this?
[−1 1 −1 1 01 1 1 1 0
]∼[
1 0 1 0 00 1 0 1 0
]What does this tell us?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Now, the conditions - what do they do for us?
p(−1) = −a3 + a2 − a1 + a0 = 0
p(1) = a3 + a2 + a1 + a0 = 0
What can we do with this?
[−1 1 −1 1 01 1 1 1 0
]∼[
1 0 1 0 00 1 0 1 0
]What does this tell us?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Now, the conditions - what do they do for us?
p(−1) = −a3 + a2 − a1 + a0 = 0
p(1) = a3 + a2 + a1 + a0 = 0
What can we do with this?
[−1 1 −1 1 01 1 1 1 0
]∼
[1 0 1 0 00 1 0 1 0
]What does this tell us?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Now, the conditions - what do they do for us?
p(−1) = −a3 + a2 − a1 + a0 = 0
p(1) = a3 + a2 + a1 + a0 = 0
What can we do with this?
[−1 1 −1 1 01 1 1 1 0
]∼[
1 0 1 0 00 1 0 1 0
]
What does this tell us?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
Now, the conditions - what do they do for us?
p(−1) = −a3 + a2 − a1 + a0 = 0
p(1) = a3 + a2 + a1 + a0 = 0
What can we do with this?
[−1 1 −1 1 01 1 1 1 0
]∼[
1 0 1 0 00 1 0 1 0
]What does this tell us?
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
The subspace S is spanned by two linearly independentpolynomials p1(x) = x3 − x and p2(x) = x2 − 1.
dim(S) =2
S is isomorphic to R2
Now, using the theorem we didn’t prove, we haveT : S → R2 given by
T (c1p1(x) + c2p2(x)) =
[c1c2
]
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
The subspace S is spanned by two linearly independentpolynomials p1(x) = x3 − x and p2(x) = x2 − 1.
dim(S) =
2
S is isomorphic to R2
Now, using the theorem we didn’t prove, we haveT : S → R2 given by
T (c1p1(x) + c2p2(x)) =
[c1c2
]
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
The subspace S is spanned by two linearly independentpolynomials p1(x) = x3 − x and p2(x) = x2 − 1.
dim(S) =2
S is isomorphic to R2
Now, using the theorem we didn’t prove, we haveT : S → R2 given by
T (c1p1(x) + c2p2(x)) =
[c1c2
]
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
The subspace S is spanned by two linearly independentpolynomials p1(x) = x3 − x and p2(x) = x2 − 1.
dim(S) =2
S is isomorphic to
R2
Now, using the theorem we didn’t prove, we haveT : S → R2 given by
T (c1p1(x) + c2p2(x)) =
[c1c2
]
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
The subspace S is spanned by two linearly independentpolynomials p1(x) = x3 − x and p2(x) = x2 − 1.
dim(S) =2
S is isomorphic to R2
Now, using the theorem we didn’t prove, we haveT : S → R2 given by
T (c1p1(x) + c2p2(x)) =
[c1c2
]
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
The subspace S is spanned by two linearly independentpolynomials p1(x) = x3 − x and p2(x) = x2 − 1.
dim(S) =2
S is isomorphic to R2
Now, using the theorem we didn’t prove, we haveT : S → R2 given by
T (c1p1(x) + c2p2(x)) =
[c1c2
]
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Constructing Isomorphisms
The subspace S is spanned by two linearly independentpolynomials p1(x) = x3 − x and p2(x) = x2 − 1.
dim(S) =2
S is isomorphic to R2
Now, using the theorem we didn’t prove, we haveT : S → R2 given by
T (c1p1(x) + c2p2(x)) =
[c1c2
]
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Example
Determine if T : P→ R∞ given by
T (anxn + an−1x
n−1 + . . . + a1x + a0)
= (a0, a1, . . . , an−1, an, 0, 0, . . .)
is an isomorphism.
First, is T a linear transformation? It is, which we couldverify using the standard method, but we won’t here.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Example
Determine if T : P→ R∞ given by
T (anxn + an−1x
n−1 + . . . + a1x + a0)
= (a0, a1, . . . , an−1, an, 0, 0, . . .)
is an isomorphism.
First, is T a linear transformation?
It is, which we couldverify using the standard method, but we won’t here.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Example
Determine if T : P→ R∞ given by
T (anxn + an−1x
n−1 + . . . + a1x + a0)
= (a0, a1, . . . , an−1, an, 0, 0, . . .)
is an isomorphism.
First, is T a linear transformation? It is, which we couldverify using the standard method, but we won’t here.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T one-to-one?
T (p(x)) = (0, 0, 0, . . .) if and only if p(x) = 0, the zeropolynomial.
What is ker(T )? ker(T ) = {p(x) = 0}
What does this tell us? T is one-to-one.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T one-to-one?
T (p(x)) = (0, 0, 0, . . .) if and only if p(x) = 0, the zeropolynomial.
What is ker(T )? ker(T ) = {p(x) = 0}
What does this tell us? T is one-to-one.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T one-to-one?
T (p(x)) = (0, 0, 0, . . .) if and only if p(x) = 0, the zeropolynomial.
What is ker(T )?
ker(T ) = {p(x) = 0}
What does this tell us? T is one-to-one.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T one-to-one?
T (p(x)) = (0, 0, 0, . . .) if and only if p(x) = 0, the zeropolynomial.
What is ker(T )? ker(T ) = {p(x) = 0}
What does this tell us? T is one-to-one.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T one-to-one?
T (p(x)) = (0, 0, 0, . . .) if and only if p(x) = 0, the zeropolynomial.
What is ker(T )? ker(T ) = {p(x) = 0}
What does this tell us?
T is one-to-one.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T one-to-one?
T (p(x)) = (0, 0, 0, . . .) if and only if p(x) = 0, the zeropolynomial.
What is ker(T )? ker(T ) = {p(x) = 0}
What does this tell us? T is one-to-one.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T onto?
The image of a degree n polynomial has at most n + 1non-zero terms in the sequence.
Since the degree of the polynomial is finite, the image T (P)cannot contain any sequence that contains infinitely manyzeros.
Since we cannot get a sequence of, say, (1, 1, 1, . . .) as ourimage, T is not onto.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T onto?
The image of a degree n polynomial has at most n + 1non-zero terms in the sequence.
Since the degree of the polynomial is finite, the image T (P)cannot contain any sequence that contains infinitely manyzeros.
Since we cannot get a sequence of, say, (1, 1, 1, . . .) as ourimage, T is not onto.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T onto?
The image of a degree n polynomial has at most n + 1non-zero terms in the sequence.
Since the degree of the polynomial is finite, the image T (P)cannot contain any sequence that contains infinitely manyzeros.
Since we cannot get a sequence of, say, (1, 1, 1, . . .) as ourimage, T is not onto.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Last Example
Is T onto?
The image of a degree n polynomial has at most n + 1non-zero terms in the sequence.
Since the degree of the polynomial is finite, the image T (P)cannot contain any sequence that contains infinitely manyzeros.
Since we cannot get a sequence of, say, (1, 1, 1, . . .) as ourimage, T is not onto.
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Comments
The image of T from the last example can be shown tocontain exactly those sequences with finitely many non-zeroterms. So T is an isomorphism from P onto this subspace ofR∞.
When we have two infinite spaces, we cannot often showwhether they are isomorphic because of the degree ofinfinity. We have to consider the difference betweencountable infinite and uncountably infinite ...
1 RememberingBack to Abstract
2 Examples
3 Theorems
4 IsomorphicVector Spaces
5 InversesExamples
6 ConstructingIsomorphisms
Comments
The image of T from the last example can be shown tocontain exactly those sequences with finitely many non-zeroterms. So T is an isomorphism from P onto this subspace ofR∞.
When we have two infinite spaces, we cannot often showwhether they are isomorphic because of the degree ofinfinity. We have to consider the difference betweencountable infinite and uncountably infinite ...
