Find a linear transformation $ T $ that sends $ A $to the parallelogram $ B $ [closed]

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Suppose we are given the unit square $ A $ in the plane with corners $ (0, 0), (1, 0), (1, 1) and (0, 1) $

Find a linear transformation $ T $ that sends $ A $to the parallelogram $ B $ with corners $ (0, 0), (1, 2), (2, 2) and (1, 0) $.

Answer:

$ T(0,0)=(0,0) \ T(1,0)=(1,2), \ T(1,1)=(2,2), \ T(0,1)=(1,0) $

From this how to find the linear transformation $ T(x,y) $ ?

Help me finding $ T(x,y) $.

linear-algebra

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asked Jul 17 at 16:09

yourmath

1,8021617

closed as off-topic by John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister Jul 18 at 13:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-1
down vote

favorite

Suppose we are given the unit square $ A $ in the plane with corners $ (0, 0), (1, 0), (1, 1) and (0, 1) $

Find a linear transformation $ T $ that sends $ A $to the parallelogram $ B $ with corners $ (0, 0), (1, 2), (2, 2) and (1, 0) $.

Answer:

$ T(0,0)=(0,0) \ T(1,0)=(1,2), \ T(1,1)=(2,2), \ T(0,1)=(1,0) $

From this how to find the linear transformation $ T(x,y) $ ?

Help me finding $ T(x,y) $.

linear-algebra

share|cite|improve this question

asked Jul 17 at 16:09

yourmath

1,8021617

closed as off-topic by John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister Jul 18 at 13:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

Suppose we are given the unit square $ A $ in the plane with corners $ (0, 0), (1, 0), (1, 1) and (0, 1) $

Find a linear transformation $ T $ that sends $ A $to the parallelogram $ B $ with corners $ (0, 0), (1, 2), (2, 2) and (1, 0) $.

Answer:

$ T(0,0)=(0,0) \ T(1,0)=(1,2), \ T(1,1)=(2,2), \ T(0,1)=(1,0) $

From this how to find the linear transformation $ T(x,y) $ ?

Help me finding $ T(x,y) $.

linear-algebra

share|cite|improve this question

asked Jul 17 at 16:09

yourmath

1,8021617

Suppose we are given the unit square $ A $ in the plane with corners $ (0, 0), (1, 0), (1, 1) and (0, 1) $

Find a linear transformation $ T $ that sends $ A $to the parallelogram $ B $ with corners $ (0, 0), (1, 2), (2, 2) and (1, 0) $.

Answer:

$ T(0,0)=(0,0) \ T(1,0)=(1,2), \ T(1,1)=(2,2), \ T(0,1)=(1,0) $

From this how to find the linear transformation $ T(x,y) $ ?

Help me finding $ T(x,y) $.

linear-algebra

share|cite|improve this question

asked Jul 17 at 16:09

yourmath

1,8021617

share|cite|improve this question

share|cite|improve this question

asked Jul 17 at 16:09

yourmath

1,8021617

asked Jul 17 at 16:09

yourmath

1,8021617

asked Jul 17 at 16:09

yourmath

1,8021617

1,8021617

closed as off-topic by John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister Jul 18 at 13:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister Jul 18 at 13:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Gibbs, Jyrki Lahtonen, Shailesh, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

Since $T(0,0)=(0,0)$, as required for a linear transformation, and $T(1,1)=T(1,0)+T(0,1)$, the definition is consistent and it suffices to consider the linear tranformation such that

• $T(1,0)=(1,2)$


• $T(0,1)=(1,0)$

that is in matrix form with reference to the standard basis

$$T(x,y)=beginbmatrix1&1\2&0endbmatrixbeginbmatrixx\yendbmatrix$$

share|cite|improve this answer

answered Jul 17 at 16:14

gimusi

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus $ T(x,y)=(x+y,2x) $ . My question is- Is $ T(x,y) $ is unique ?
  – yourmath
  Jul 17 at 18:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes exactly. Of course it is unique for the theorem of existence and uniqueness of linear transformations. I will give you some references.
  – gimusi
  Jul 17 at 18:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  last question: what is the transformation $ T $ that send $ A $ to $ A $ ? Is it $ T(x,y)=(x,y) $ ?
  – yourmath
  Jul 17 at 18:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity transformation, represented by the I matrix.
  – gimusi
  Jul 17 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @yourmath For the uniqueness refer to math.stackexchange.com/q/289397/505767
  – gimusi
  Jul 17 at 19:52
  

  
  
  
  

add a comment | 

up vote
2
down vote

You know what $T(1,0)$ and $T(0,1)$ are whence
$$
T(x,y)=xT(1,0)+yT(0,1)
$$

share|cite|improve this answer

answered Jul 17 at 16:12

Foobaz John

18.1k41245

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Since $T(0,0)=(0,0)$, as required for a linear transformation, and $T(1,1)=T(1,0)+T(0,1)$, the definition is consistent and it suffices to consider the linear tranformation such that

• $T(1,0)=(1,2)$


• $T(0,1)=(1,0)$

that is in matrix form with reference to the standard basis

$$T(x,y)=beginbmatrix1&1\2&0endbmatrixbeginbmatrixx\yendbmatrix$$

share|cite|improve this answer

answered Jul 17 at 16:14

gimusi

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus $ T(x,y)=(x+y,2x) $ . My question is- Is $ T(x,y) $ is unique ?
  – yourmath
  Jul 17 at 18:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes exactly. Of course it is unique for the theorem of existence and uniqueness of linear transformations. I will give you some references.
  – gimusi
  Jul 17 at 18:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  last question: what is the transformation $ T $ that send $ A $ to $ A $ ? Is it $ T(x,y)=(x,y) $ ?
  – yourmath
  Jul 17 at 18:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity transformation, represented by the I matrix.
  – gimusi
  Jul 17 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @yourmath For the uniqueness refer to math.stackexchange.com/q/289397/505767
  – gimusi
  Jul 17 at 19:52
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Since $T(0,0)=(0,0)$, as required for a linear transformation, and $T(1,1)=T(1,0)+T(0,1)$, the definition is consistent and it suffices to consider the linear tranformation such that

• $T(1,0)=(1,2)$


• $T(0,1)=(1,0)$

that is in matrix form with reference to the standard basis

$$T(x,y)=beginbmatrix1&1\2&0endbmatrixbeginbmatrixx\yendbmatrix$$

share|cite|improve this answer

answered Jul 17 at 16:14

gimusi

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus $ T(x,y)=(x+y,2x) $ . My question is- Is $ T(x,y) $ is unique ?
  – yourmath
  Jul 17 at 18:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes exactly. Of course it is unique for the theorem of existence and uniqueness of linear transformations. I will give you some references.
  – gimusi
  Jul 17 at 18:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  last question: what is the transformation $ T $ that send $ A $ to $ A $ ? Is it $ T(x,y)=(x,y) $ ?
  – yourmath
  Jul 17 at 18:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity transformation, represented by the I matrix.
  – gimusi
  Jul 17 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @yourmath For the uniqueness refer to math.stackexchange.com/q/289397/505767
  – gimusi
  Jul 17 at 19:52
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Since $T(0,0)=(0,0)$, as required for a linear transformation, and $T(1,1)=T(1,0)+T(0,1)$, the definition is consistent and it suffices to consider the linear tranformation such that

• $T(1,0)=(1,2)$


• $T(0,1)=(1,0)$

that is in matrix form with reference to the standard basis

$$T(x,y)=beginbmatrix1&1\2&0endbmatrixbeginbmatrixx\yendbmatrix$$

share|cite|improve this answer

answered Jul 17 at 16:14

gimusi

65.4k73584

Since $T(0,0)=(0,0)$, as required for a linear transformation, and $T(1,1)=T(1,0)+T(0,1)$, the definition is consistent and it suffices to consider the linear tranformation such that

• $T(1,0)=(1,2)$


• $T(0,1)=(1,0)$

that is in matrix form with reference to the standard basis

$$T(x,y)=beginbmatrix1&1\2&0endbmatrixbeginbmatrixx\yendbmatrix$$

share|cite|improve this answer

answered Jul 17 at 16:14

gimusi

65.4k73584

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 16:14

gimusi

65.4k73584

answered Jul 17 at 16:14

gimusi

65.4k73584

answered Jul 17 at 16:14

gimusi

65.4k73584

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus $ T(x,y)=(x+y,2x) $ . My question is- Is $ T(x,y) $ is unique ?
  – yourmath
  Jul 17 at 18:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes exactly. Of course it is unique for the theorem of existence and uniqueness of linear transformations. I will give you some references.
  – gimusi
  Jul 17 at 18:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  last question: what is the transformation $ T $ that send $ A $ to $ A $ ? Is it $ T(x,y)=(x,y) $ ?
  – yourmath
  Jul 17 at 18:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity transformation, represented by the I matrix.
  – gimusi
  Jul 17 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @yourmath For the uniqueness refer to math.stackexchange.com/q/289397/505767
  – gimusi
  Jul 17 at 19:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus $ T(x,y)=(x+y,2x) $ . My question is- Is $ T(x,y) $ is unique ?
  – yourmath
  Jul 17 at 18:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes exactly. Of course it is unique for the theorem of existence and uniqueness of linear transformations. I will give you some references.
  – gimusi
  Jul 17 at 18:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  last question: what is the transformation $ T $ that send $ A $ to $ A $ ? Is it $ T(x,y)=(x,y) $ ?
  – yourmath
  Jul 17 at 18:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity transformation, represented by the I matrix.
  – gimusi
  Jul 17 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @yourmath For the uniqueness refer to math.stackexchange.com/q/289397/505767
  – gimusi
  Jul 17 at 19:52
  

  
  
  
  

Thus $ T(x,y)=(x+y,2x) $ . My question is- Is $ T(x,y) $ is unique ?
– yourmath
Jul 17 at 18:44

Thus $ T(x,y)=(x+y,2x) $ . My question is- Is $ T(x,y) $ is unique ?
– yourmath
Jul 17 at 18:44

1

1

Yes exactly. Of course it is unique for the theorem of existence and uniqueness of linear transformations. I will give you some references.
– gimusi
Jul 17 at 18:50

Yes exactly. Of course it is unique for the theorem of existence and uniqueness of linear transformations. I will give you some references.
– gimusi
Jul 17 at 18:50

last question: what is the transformation $ T $ that send $ A $ to $ A $ ? Is it $ T(x,y)=(x,y) $ ?
– yourmath
Jul 17 at 18:57

last question: what is the transformation $ T $ that send $ A $ to $ A $ ? Is it $ T(x,y)=(x,y) $ ?
– yourmath
Jul 17 at 18:57

The identity transformation, represented by the I matrix.
– gimusi
Jul 17 at 19:27

The identity transformation, represented by the I matrix.
– gimusi
Jul 17 at 19:27

@yourmath For the uniqueness refer to math.stackexchange.com/q/289397/505767
– gimusi
Jul 17 at 19:52

@yourmath For the uniqueness refer to math.stackexchange.com/q/289397/505767
– gimusi
Jul 17 at 19:52

add a comment | 

up vote
2
down vote

You know what $T(1,0)$ and $T(0,1)$ are whence
$$
T(x,y)=xT(1,0)+yT(0,1)
$$

share|cite|improve this answer

answered Jul 17 at 16:12

Foobaz John

18.1k41245

add a comment | 

up vote
2
down vote

You know what $T(1,0)$ and $T(0,1)$ are whence
$$
T(x,y)=xT(1,0)+yT(0,1)
$$

share|cite|improve this answer

answered Jul 17 at 16:12

Foobaz John

18.1k41245

add a comment | 

up vote
2
down vote

up vote
2
down vote

You know what $T(1,0)$ and $T(0,1)$ are whence
$$
T(x,y)=xT(1,0)+yT(0,1)
$$

share|cite|improve this answer

answered Jul 17 at 16:12

Foobaz John

18.1k41245

You know what $T(1,0)$ and $T(0,1)$ are whence
$$
T(x,y)=xT(1,0)+yT(0,1)
$$

share|cite|improve this answer

answered Jul 17 at 16:12

Foobaz John

18.1k41245

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 16:12

Foobaz John

18.1k41245

answered Jul 17 at 16:12

Foobaz John

18.1k41245

answered Jul 17 at 16:12

Foobaz John

18.1k41245

18.1k41245

add a comment | 

add a comment |