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Lecture 8: The matrix of a linear transformation.Applications
Danny W. Crytser
April 7, 2014
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Example
Let T : R2 → R3 be the linear transformation defined by
T
([x1x2
])=
x1 + x22x13x2
.
Let’s find the matrix A such that T (x) = Ax for all x ∈ R2. Notethat A must have 2 columns (domain R2) and 3 rows (domain R3).The two basis vectors in the domain are e1, e2. Their images are
T (e1) =
120
, T (e2) =
103
.So A has those vectors as columns
A =
1 12 00 3
.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 3: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/3.jpg)
Example
Let T : R2 → R3 be the linear transformation defined by
T
([x1x2
])=
x1 + x22x13x2
.Let’s find the matrix A such that T (x) = Ax for all x ∈ R2.
Notethat A must have 2 columns (domain R2) and 3 rows (domain R3).The two basis vectors in the domain are e1, e2. Their images are
T (e1) =
120
, T (e2) =
103
.So A has those vectors as columns
A =
1 12 00 3
.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 4: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/4.jpg)
Example
Let T : R2 → R3 be the linear transformation defined by
T
([x1x2
])=
x1 + x22x13x2
.Let’s find the matrix A such that T (x) = Ax for all x ∈ R2. Notethat A must have 2 columns (domain R2) and 3 rows (domain R3).
The two basis vectors in the domain are e1, e2. Their images are
T (e1) =
120
, T (e2) =
103
.So A has those vectors as columns
A =
1 12 00 3
.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 5: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/5.jpg)
Example
Let T : R2 → R3 be the linear transformation defined by
T
([x1x2
])=
x1 + x22x13x2
.Let’s find the matrix A such that T (x) = Ax for all x ∈ R2. Notethat A must have 2 columns (domain R2) and 3 rows (domain R3).The two basis vectors in the domain are e1, e2.
Their images are
T (e1) =
120
, T (e2) =
103
.So A has those vectors as columns
A =
1 12 00 3
.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 6: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/6.jpg)
Example
Let T : R2 → R3 be the linear transformation defined by
T
([x1x2
])=
x1 + x22x13x2
.Let’s find the matrix A such that T (x) = Ax for all x ∈ R2. Notethat A must have 2 columns (domain R2) and 3 rows (domain R3).The two basis vectors in the domain are e1, e2. Their images are
T (e1) =
120
, T (e2) =
103
.
So A has those vectors as columns
A =
1 12 00 3
.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 7: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/7.jpg)
Example
Let T : R2 → R3 be the linear transformation defined by
T
([x1x2
])=
x1 + x22x13x2
.Let’s find the matrix A such that T (x) = Ax for all x ∈ R2. Notethat A must have 2 columns (domain R2) and 3 rows (domain R3).The two basis vectors in the domain are e1, e2. Their images are
T (e1) =
120
, T (e2) =
103
.So A has those vectors as columns
A =
1 12 00 3
.Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 8: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/8.jpg)
Example
Let T : R2 → R3 be the linear transformation defined by
T
([x1x2
])=
x1 + x22x13x2
.Let’s find the matrix A such that T (x) = Ax for all x ∈ R2. Notethat A must have 2 columns (domain R2) and 3 rows (domain R3).The two basis vectors in the domain are e1, e2. Their images are
T (e1) =
120
, T (e2) =
103
.So A has those vectors as columns
A =
1 12 00 3
.Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 9: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/9.jpg)
We can check to see we got the right answer:
A
[x1x2
]= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2. There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn, then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 10: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/10.jpg)
We can check to see we got the right answer:
A
[x1x2
]
= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2. There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn, then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 11: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/11.jpg)
We can check to see we got the right answer:
A
[x1x2
]= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2. There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn, then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 12: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/12.jpg)
We can check to see we got the right answer:
A
[x1x2
]= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2. There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn, then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 13: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/13.jpg)
We can check to see we got the right answer:
A
[x1x2
]= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2. There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn, then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 14: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/14.jpg)
We can check to see we got the right answer:
A
[x1x2
]= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2.
There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn, then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 15: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/15.jpg)
We can check to see we got the right answer:
A
[x1x2
]= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2. There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn,
then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 16: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/16.jpg)
We can check to see we got the right answer:
A
[x1x2
]= x1
120
+ x2
103
=
x1 + x22x13x2
= T
([x1x2
]).
Thus T (x) = Ax for all x ∈ R2. There’s a fancy term for thematrix we’ve cooked up.
Definition
If T : Rn → Rm is a linear transformation and e1, e2, . . . , en arethe standard basis vectors in Rn, then the matrix
A =[T (e1) T (e2) . . . T (en)
]which satisfies T (x) = Ax for all x ∈ Rn is called the standardmatrix for T .
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Matrices and visualization of linear transformations
There are a few different types of linear transformations R2 → R2
that we can describe with words (“rotate the planecounterclockwise by π/2”) and then we get the matrix just by
tracking the image of the basis vectors e1 =
[10
]and
e2 =
[01
].
The book has a big catalog of such transformations.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Matrices and visualization of linear transformations
There are a few different types of linear transformations R2 → R2
that we can describe with words (“rotate the planecounterclockwise by π/2”) and then we get the matrix just by
tracking the image of the basis vectors e1 =
[10
]and
e2 =
[01
]. The book has a big catalog of such transformations.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 19: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/19.jpg)
Matrices and visualization of linear transformations
There are a few different types of linear transformations R2 → R2
that we can describe with words (“rotate the planecounterclockwise by π/2”) and then we get the matrix just by
tracking the image of the basis vectors e1 =
[10
]and
e2 =
[01
]. The book has a big catalog of such transformations.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 20: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/20.jpg)
Reflection through x1 = x2
Suppose that T reflects through the line x1 = x2.
What does Tlook like? What is the matrix of T? The image of e1 is e2 andvice versa.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Reflection through x1 = x2
Suppose that T reflects through the line x1 = x2. What does Tlook like? What is the matrix of T?
The image of e1 is e2 andvice versa.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Reflection through x1 = x2
Suppose that T reflects through the line x1 = x2. What does Tlook like? What is the matrix of T? The image of e1 is e2 andvice versa.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Reflection through x1 = x2
Suppose that T reflects through the line x1 = x2. What does Tlook like? What is the matrix of T? The image of e1 is e2 andvice versa.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Contractions. Expansions
You can stretch/squeeze the plane in one direction while keepingthe other direction fixed.
For example, let’s say we stretched theplane along the x-axis by a factor of 2, but didn’t distort it in they -axis. What would the matrix look like? The image of e1 is 2e1and e2 doesn’t change. The graphic on the right represents thissituation, where we have k = 2.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Contractions. Expansions
You can stretch/squeeze the plane in one direction while keepingthe other direction fixed. For example, let’s say we stretched theplane along the x-axis by a factor of 2, but didn’t distort it in they -axis. What would the matrix look like?
The image of e1 is 2e1and e2 doesn’t change. The graphic on the right represents thissituation, where we have k = 2.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 26: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/26.jpg)
Contractions. Expansions
You can stretch/squeeze the plane in one direction while keepingthe other direction fixed. For example, let’s say we stretched theplane along the x-axis by a factor of 2, but didn’t distort it in they -axis. What would the matrix look like? The image of e1 is 2e1and e2 doesn’t change. The graphic on the right represents thissituation, where we have k = 2.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 27: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/27.jpg)
Contractions. Expansions
You can stretch/squeeze the plane in one direction while keepingthe other direction fixed. For example, let’s say we stretched theplane along the x-axis by a factor of 2, but didn’t distort it in they -axis. What would the matrix look like? The image of e1 is 2e1and e2 doesn’t change. The graphic on the right represents thissituation, where we have k = 2.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Shear transformations
Shear transformations add some multiple of one basis vector toanother basis vector (not to be confused with row operations).
Forexample e1 7→ e1 + 2e2 and e2 7→ e2.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Shear transformations
Shear transformations add some multiple of one basis vector toanother basis vector (not to be confused with row operations). Forexample e1 7→ e1 + 2e2 and e2 7→ e2.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 30: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/30.jpg)
Shear transformations
Shear transformations add some multiple of one basis vector toanother basis vector (not to be confused with row operations). Forexample e1 7→ e1 + 2e2 and e2 7→ e2.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 31: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/31.jpg)
Onto
Definition
A function T : Rn → Rm is said to be onto if the range is thecodomain, that is, for each vector y ∈ Rm there is at least onex ∈ Rn with T (x) = y.
The preceding definition is supposed to address the potentialdifference between range and codomain.
Example
Let T : R2 → R2 be given by T (x1, x2) = (x1, 0). Then T is notonto: the range is the x-axis, an object in mathematics noteworthyfor not being the entire plane. The reflection across the linex1 = x2 given by T (x1, x2) = (x2, x1) is onto: every vector in R2 isthe reflection of some other vector (that other vector is itsreflection)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto
Definition
A function T : Rn → Rm is said to be onto if the range is thecodomain, that is, for each vector y ∈ Rm there is at least onex ∈ Rn with T (x) = y.
The preceding definition is supposed to address the potentialdifference between range and codomain.
Example
Let T : R2 → R2 be given by T (x1, x2) = (x1, 0). Then T is notonto: the range is the x-axis, an object in mathematics noteworthyfor not being the entire plane. The reflection across the linex1 = x2 given by T (x1, x2) = (x2, x1) is onto: every vector in R2 isthe reflection of some other vector (that other vector is itsreflection)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 33: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/33.jpg)
Onto
Definition
A function T : Rn → Rm is said to be onto if the range is thecodomain, that is, for each vector y ∈ Rm there is at least onex ∈ Rn with T (x) = y.
The preceding definition is supposed to address the potentialdifference between range and codomain.
Example
Let T : R2 → R2 be given by T (x1, x2) = (x1, 0). Then T is notonto: the range is the x-axis, an object in mathematics noteworthyfor not being the entire plane.
The reflection across the linex1 = x2 given by T (x1, x2) = (x2, x1) is onto: every vector in R2 isthe reflection of some other vector (that other vector is itsreflection)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 34: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/34.jpg)
Onto
Definition
A function T : Rn → Rm is said to be onto if the range is thecodomain, that is, for each vector y ∈ Rm there is at least onex ∈ Rn with T (x) = y.
The preceding definition is supposed to address the potentialdifference between range and codomain.
Example
Let T : R2 → R2 be given by T (x1, x2) = (x1, 0). Then T is notonto: the range is the x-axis, an object in mathematics noteworthyfor not being the entire plane. The reflection across the linex1 = x2 given by T (x1, x2) = (x2, x1) is onto: every vector in R2 isthe reflection of some other vector (that other vector is itsreflection)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 35: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/35.jpg)
Onto
Definition
A function T : Rn → Rm is said to be onto if the range is thecodomain, that is, for each vector y ∈ Rm there is at least onex ∈ Rn with T (x) = y.
The preceding definition is supposed to address the potentialdifference between range and codomain.
Example
Let T : R2 → R2 be given by T (x1, x2) = (x1, 0). Then T is notonto: the range is the x-axis, an object in mathematics noteworthyfor not being the entire plane. The reflection across the linex1 = x2 given by T (x1, x2) = (x2, x1) is onto: every vector in R2 isthe reflection of some other vector (that other vector is itsreflection)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 36: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/36.jpg)
Onto, ctd.
The quality of being onto has to do with existence of solutions: alinear transformation T given by T (x) = Ax is onto if Ax = b isconsistent for all b ∈ Rm.
Reviewing the following theorem allowsus to describe onto linear transformations with echelon forms. Wealready saw a version of this theorem
Theorem
Let A be an m × n matrix. Then Ax = b is consistent for allb ∈ Rm if and only if every row in the echelon form of A (notaugmented) has a nonzero entry.
We can reformulate it in terms of linear transformations.
Theorem
Let T (x) = Ax. Then T is onto if and only if every row in theechelon form of A (non-augmented) has a nonzero entry.Thishappens if and only if the columns of A span Rm.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto, ctd.
The quality of being onto has to do with existence of solutions: alinear transformation T given by T (x) = Ax is onto if Ax = b isconsistent for all b ∈ Rm. Reviewing the following theorem allowsus to describe onto linear transformations with echelon forms.
Wealready saw a version of this theorem
Theorem
Let A be an m × n matrix. Then Ax = b is consistent for allb ∈ Rm if and only if every row in the echelon form of A (notaugmented) has a nonzero entry.
We can reformulate it in terms of linear transformations.
Theorem
Let T (x) = Ax. Then T is onto if and only if every row in theechelon form of A (non-augmented) has a nonzero entry.Thishappens if and only if the columns of A span Rm.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 38: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/38.jpg)
Onto, ctd.
The quality of being onto has to do with existence of solutions: alinear transformation T given by T (x) = Ax is onto if Ax = b isconsistent for all b ∈ Rm. Reviewing the following theorem allowsus to describe onto linear transformations with echelon forms. Wealready saw a version of this theorem
Theorem
Let A be an m × n matrix. Then Ax = b is consistent for allb ∈ Rm if and only if every row in the echelon form of A (notaugmented) has a nonzero entry.
We can reformulate it in terms of linear transformations.
Theorem
Let T (x) = Ax. Then T is onto if and only if every row in theechelon form of A (non-augmented) has a nonzero entry.Thishappens if and only if the columns of A span Rm.
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Onto, ctd.
The quality of being onto has to do with existence of solutions: alinear transformation T given by T (x) = Ax is onto if Ax = b isconsistent for all b ∈ Rm. Reviewing the following theorem allowsus to describe onto linear transformations with echelon forms. Wealready saw a version of this theorem
Theorem
Let A be an m × n matrix. Then Ax = b is consistent for allb ∈ Rm if and only if every row in the echelon form of A (notaugmented) has a nonzero entry.
We can reformulate it in terms of linear transformations.
Theorem
Let T (x) = Ax. Then T is onto if and only if every row in theechelon form of A (non-augmented) has a nonzero entry.Thishappens if and only if the columns of A span Rm.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto, ctd.
The quality of being onto has to do with existence of solutions: alinear transformation T given by T (x) = Ax is onto if Ax = b isconsistent for all b ∈ Rm. Reviewing the following theorem allowsus to describe onto linear transformations with echelon forms. Wealready saw a version of this theorem
Theorem
Let A be an m × n matrix. Then Ax = b is consistent for allb ∈ Rm if and only if every row in the echelon form of A (notaugmented) has a nonzero entry.
We can reformulate it in terms of linear transformations.
Theorem
Let T (x) = Ax. Then T is onto if and only if every row in theechelon form of A (non-augmented) has a nonzero entry.
Thishappens if and only if the columns of A span Rm.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto, ctd.
The quality of being onto has to do with existence of solutions: alinear transformation T given by T (x) = Ax is onto if Ax = b isconsistent for all b ∈ Rm. Reviewing the following theorem allowsus to describe onto linear transformations with echelon forms. Wealready saw a version of this theorem
Theorem
Let A be an m × n matrix. Then Ax = b is consistent for allb ∈ Rm if and only if every row in the echelon form of A (notaugmented) has a nonzero entry.
We can reformulate it in terms of linear transformations.
Theorem
Let T (x) = Ax. Then T is onto if and only if every row in theechelon form of A (non-augmented) has a nonzero entry.Thishappens if and only if the columns of A span Rm.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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One-to-one
Definition
A function T : Rn → Rm is said to be one-to-one if T (x) = T (x′)implies x = x′ for vectors x, x′ ∈ Rn.
That is, T is one-to-one iftwo vectors in the domain have the same image under T onlywhen they are equal.
Example
The map T : R3 → R3 given by T (x1, x2, x3) = (x1, x2, 0) is notone-to-one: T (0, 0, 1) = T (0, 0, 0) = 0 and yet the vectors are notequal. The map T : R2 → R3 given by
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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One-to-one
Definition
A function T : Rn → Rm is said to be one-to-one if T (x) = T (x′)implies x = x′ for vectors x, x′ ∈ Rn. That is, T is one-to-one iftwo vectors in the domain have the same image under T onlywhen they are equal.
Example
The map T : R3 → R3 given by T (x1, x2, x3) = (x1, x2, 0) is notone-to-one: T (0, 0, 1) = T (0, 0, 0) = 0 and yet the vectors are notequal. The map T : R2 → R3 given by
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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One-to-one
Definition
A function T : Rn → Rm is said to be one-to-one if T (x) = T (x′)implies x = x′ for vectors x, x′ ∈ Rn. That is, T is one-to-one iftwo vectors in the domain have the same image under T onlywhen they are equal.
Example
The map T : R3 → R3 given by T (x1, x2, x3) = (x1, x2, 0) is notone-to-one:
T (0, 0, 1) = T (0, 0, 0) = 0 and yet the vectors are notequal. The map T : R2 → R3 given by
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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One-to-one
Definition
A function T : Rn → Rm is said to be one-to-one if T (x) = T (x′)implies x = x′ for vectors x, x′ ∈ Rn. That is, T is one-to-one iftwo vectors in the domain have the same image under T onlywhen they are equal.
Example
The map T : R3 → R3 given by T (x1, x2, x3) = (x1, x2, 0) is notone-to-one: T (0, 0, 1) = T (0, 0, 0) = 0 and yet the vectors are notequal.
The map T : R2 → R3 given by
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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One-to-one
Definition
A function T : Rn → Rm is said to be one-to-one if T (x) = T (x′)implies x = x′ for vectors x, x′ ∈ Rn. That is, T is one-to-one iftwo vectors in the domain have the same image under T onlywhen they are equal.
Example
The map T : R3 → R3 given by T (x1, x2, x3) = (x1, x2, 0) is notone-to-one: T (0, 0, 1) = T (0, 0, 0) = 0 and yet the vectors are notequal. The map T : R2 → R3 given by
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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One-to-one, ctd.
The quality of being one-to-one has to do with uniqueness ofsolutions: the linear transformation T given by T (x) = Ax ifwhenever Ax = b is consistent it has unique solutions.
Theorem
Let A be a matrix. Then T is one-to-one if and only if everycolumn in the echelon form of A (non-augmented) has a pivot.This happens if and only if the columns of A are linearlyindependent.
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One-to-one, ctd.
The quality of being one-to-one has to do with uniqueness ofsolutions: the linear transformation T given by T (x) = Ax ifwhenever Ax = b is consistent it has unique solutions.
Theorem
Let A be a matrix. Then T is one-to-one if and only if everycolumn in the echelon form of A (non-augmented) has a pivot.
This happens if and only if the columns of A are linearlyindependent.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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One-to-one, ctd.
The quality of being one-to-one has to do with uniqueness ofsolutions: the linear transformation T given by T (x) = Ax ifwhenever Ax = b is consistent it has unique solutions.
Theorem
Let A be a matrix. Then T is one-to-one if and only if everycolumn in the echelon form of A (non-augmented) has a pivot.This happens if and only if the columns of A are linearlyindependent.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto/one-to-one: echelon form
We can summarize all of this in one biggish theorem:
Theorem
Let A be an m × n matrix. The linear transformationT : Rn → Rm given by T (x) = Ax is
1 onto if and only if every row of the echelon form of A has apivotif and only if the columns of A span Rm
2 one-to-one if and only if every column of the echelon form ofA has a pivotif and only if the columns of A are linearlyindependent
You can see that the only way that T can be both onto andone-to-one is if m = n.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto/one-to-one: echelon form
We can summarize all of this in one biggish theorem:
Theorem
Let A be an m × n matrix. The linear transformationT : Rn → Rm given by T (x) = Ax is
1 onto if and only if every row of the echelon form of A has apivotif and only if the columns of A span Rm
2 one-to-one if and only if every column of the echelon form ofA has a pivotif and only if the columns of A are linearlyindependent
You can see that the only way that T can be both onto andone-to-one is if m = n.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto/one-to-one: echelon form
We can summarize all of this in one biggish theorem:
Theorem
Let A be an m × n matrix. The linear transformationT : Rn → Rm given by T (x) = Ax is
1 onto if and only if every row of the echelon form of A has apivot
if and only if the columns of A span Rm
2 one-to-one if and only if every column of the echelon form ofA has a pivotif and only if the columns of A are linearlyindependent
You can see that the only way that T can be both onto andone-to-one is if m = n.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Onto/one-to-one: echelon form
We can summarize all of this in one biggish theorem:
Theorem
Let A be an m × n matrix. The linear transformationT : Rn → Rm given by T (x) = Ax is
1 onto if and only if every row of the echelon form of A has apivotif and only if the columns of A span Rm
2 one-to-one if and only if every column of the echelon form ofA has a pivot
if and only if the columns of A are linearlyindependent
You can see that the only way that T can be both onto andone-to-one is if m = n.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 54: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/54.jpg)
Onto/one-to-one: echelon form
We can summarize all of this in one biggish theorem:
Theorem
Let A be an m × n matrix. The linear transformationT : Rn → Rm given by T (x) = Ax is
1 onto if and only if every row of the echelon form of A has apivotif and only if the columns of A span Rm
2 one-to-one if and only if every column of the echelon form ofA has a pivotif and only if the columns of A are linearlyindependent
You can see that the only way that T can be both onto andone-to-one is if m = n.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 55: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/55.jpg)
Onto/one-to-one: echelon form
We can summarize all of this in one biggish theorem:
Theorem
Let A be an m × n matrix. The linear transformationT : Rn → Rm given by T (x) = Ax is
1 onto if and only if every row of the echelon form of A has apivotif and only if the columns of A span Rm
2 one-to-one if and only if every column of the echelon form ofA has a pivotif and only if the columns of A are linearlyindependent
You can see that the only way that T can be both onto andone-to-one is if m = n.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 56: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/56.jpg)
Onto/one-to-one: echelon form
We can summarize all of this in one biggish theorem:
Theorem
Let A be an m × n matrix. The linear transformationT : Rn → Rm given by T (x) = Ax is
1 onto if and only if every row of the echelon form of A has apivotif and only if the columns of A span Rm
2 one-to-one if and only if every column of the echelon form ofA has a pivotif and only if the columns of A are linearlyindependent
You can see that the only way that T can be both onto andone-to-one is if m = n.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Example
Example
Let A =
1 72 34 2
and define T : R2 → R3 by T (x) = Ax.
Is T
onto? Is T one-to-one?
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Example
Example
Let A =
1 72 34 2
and define T : R2 → R3 by T (x) = Ax. Is T
onto? Is T one-to-one?
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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APPLICATIONS
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Voltage loops
1 A closed loop in a network has three things affiliated to it:
some voltage sources, some resistors, and an oriented(clockwise or counter-clockwise) current.
2 Each of these has a weight measuring how much voltage,resistance, or current there is (one current for the whole loop).
3 A voltage source is positive for a loop if the the current flowsfrom the positive (long) terminal to the negative (short)terminal.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Voltage loops
1 A closed loop in a network has three things affiliated to it:some voltage sources,
some resistors, and an oriented(clockwise or counter-clockwise) current.
2 Each of these has a weight measuring how much voltage,resistance, or current there is (one current for the whole loop).
3 A voltage source is positive for a loop if the the current flowsfrom the positive (long) terminal to the negative (short)terminal.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Voltage loops
1 A closed loop in a network has three things affiliated to it:some voltage sources, some resistors, and an oriented(clockwise or counter-clockwise) current.
2 Each of these has a weight measuring how much voltage,resistance, or current there is (one current for the whole loop).
3 A voltage source is positive for a loop if the the current flowsfrom the positive (long) terminal to the negative (short)terminal.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Voltage loops
1 A closed loop in a network has three things affiliated to it:some voltage sources, some resistors, and an oriented(clockwise or counter-clockwise) current.
2 Each of these has a weight measuring how much voltage,resistance, or current there is (one current for the whole loop).
3 A voltage source is positive for a loop if the the current flowsfrom the positive (long) terminal to the negative (short)terminal.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Ohm’s law
OHM’S LAW: If the current of I amps passes across a resistor ofR ohms, then the voltage drops by V = RI volts.
Passing from A to B the sum of the voltage drops is 3I1 − 3I2.(Notice that the voltage source in the first loop is positive.)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Ohm’s law
OHM’S LAW: If the current of I amps passes across a resistor ofR ohms, then the voltage drops by V = RI volts.
Passing from A to B the sum of the voltage drops is 3I1 − 3I2.(Notice that the voltage source in the first loop is positive.)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Ohm’s law
OHM’S LAW: If the current of I amps passes across a resistor ofR ohms, then the voltage drops by V = RI volts.
Passing from A to B the sum of the voltage drops is 3I1 − 3I2.
(Notice that the voltage source in the first loop is positive.)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Ohm’s law
OHM’S LAW: If the current of I amps passes across a resistor ofR ohms, then the voltage drops by V = RI volts.
Passing from A to B the sum of the voltage drops is 3I1 − 3I2.(Notice that the voltage source in the first loop is positive.)
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 68: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/68.jpg)
Kirchhoff’s law
Kirchhoff’s law governs how much current and resistance (so, howmuch voltage dropped) can be in an electrical network with givenvoltage sources.
It’s basically a balancing between the voltage lostthrough voltage drops and the voltage put into loops from voltagesources.
KIRCHHOFF’S LAW: If you add up the voltage drops in a loopthat equals the sum of the voltage sources in the loop.
Remember: when adding voltage sources you have to check to seeif they’re positive (current runs positive terminal to negativeterminal) or negative (vice versa).
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Kirchhoff’s law
Kirchhoff’s law governs how much current and resistance (so, howmuch voltage dropped) can be in an electrical network with givenvoltage sources. It’s basically a balancing between the voltage lostthrough voltage drops and the voltage put into loops from voltagesources.
KIRCHHOFF’S LAW: If you add up the voltage drops in a loopthat equals the sum of the voltage sources in the loop.
Remember: when adding voltage sources you have to check to seeif they’re positive (current runs positive terminal to negativeterminal) or negative (vice versa).
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff’s law
Kirchhoff’s law governs how much current and resistance (so, howmuch voltage dropped) can be in an electrical network with givenvoltage sources. It’s basically a balancing between the voltage lostthrough voltage drops and the voltage put into loops from voltagesources.
KIRCHHOFF’S LAW: If you add up the voltage drops in a loopthat equals the sum of the voltage sources in the loop.
Remember: when adding voltage sources you have to check to seeif they’re positive (current runs positive terminal to negativeterminal) or negative (vice versa).
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff: example
Let’s look at this.
First loop: Voltage source=30. Voltage drop infirst loop is 41 + (3I1 − 3I2) + 4I1 = 11I1 − 3I2. Must equal voltagesources in first loop = 30. So the equation for the first loop is11I1 − 3I2 = 30.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff: example
Let’s look at this. First loop: Voltage source=30.
Voltage drop infirst loop is 41 + (3I1 − 3I2) + 4I1 = 11I1 − 3I2. Must equal voltagesources in first loop = 30. So the equation for the first loop is11I1 − 3I2 = 30.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff: example
Let’s look at this. First loop: Voltage source=30. Voltage drop infirst loop is 41 + (3I1 − 3I2) + 4I1 = 11I1 − 3I2.
Must equal voltagesources in first loop = 30. So the equation for the first loop is11I1 − 3I2 = 30.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff: example
Let’s look at this. First loop: Voltage source=30. Voltage drop infirst loop is 41 + (3I1 − 3I2) + 4I1 = 11I1 − 3I2. Must equal voltagesources in first loop = 30.
So the equation for the first loop is11I1 − 3I2 = 30.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff: example
Let’s look at this. First loop: Voltage source=30. Voltage drop infirst loop is 41 + (3I1 − 3I2) + 4I1 = 11I1 − 3I2. Must equal voltagesources in first loop = 30. So the equation for the first loop is11I1 − 3I2 = 30.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff: example
11I1 − 3I2 = 30 (1)
−3I1 + 6I2 − I3 = 5 (2)
−I2 + 3I3 = −25 (3)
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Kirchhoff: example
11I1 − 3I2 = 30 (4)
−3I1 + 6I2 − I3 = 5 (5)
−I2 + 3I3 = −25 (6)
Has a unique solution: I1 = 3 amps, I2 = 1 amps, I3 = −8 amps.
The negative I3 answer says that the current flows clockwise inloop 3.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Kirchhoff: example
11I1 − 3I2 = 30 (4)
−3I1 + 6I2 − I3 = 5 (5)
−I2 + 3I3 = −25 (6)
Has a unique solution: I1 = 3 amps, I2 = 1 amps, I3 = −8 amps.The negative I3 answer says that the current flows clockwise inloop 3.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
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Difference equations
In many situations you will be measuring some system and all theinformation about the system at time kwill be contained in somevector xk .
(Could be age, salary, population, microbe count,whatever).
Definition
Suppose your data take the form of vectors xk ∈ Rn, wherek = 0, 1, 2, . . .s. If there is an n × n matrix A such thatx1 = Ax0, x2 = Ax1 and generally xk+1 = Axk (*) , we say thatequation (*) is a linear difference equation (some people call it arecursion relation, because it gives new measurement in terms ofthe old measurement).
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 80: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/80.jpg)
Difference equations
In many situations you will be measuring some system and all theinformation about the system at time kwill be contained in somevector xk . (Could be age, salary, population, microbe count,whatever).
Definition
Suppose your data take the form of vectors xk ∈ Rn, wherek = 0, 1, 2, . . .s. If there is an n × n matrix A such thatx1 = Ax0, x2 = Ax1 and generally xk+1 = Axk (*) , we say thatequation (*) is a linear difference equation (some people call it arecursion relation, because it gives new measurement in terms ofthe old measurement).
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 81: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/81.jpg)
Difference equations and population
You can study population dynamics using differenceequations.
Let’s say that in the nation of Zembla there is one cityand one suburb. The population distribution in Zembla in year 0can be recorded in a vector in R2:
x0 =
[r0s0
]where r0 is the population in the city in year 0 and s0 is thepopulation in the suburb in year 0. The vectors
x1 =
[r1s1
], x2 =
[r2s2
]record the popluation distribution in year
1, year 2, etc.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 82: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/82.jpg)
Difference equations and population
You can study population dynamics using differenceequations.Let’s say that in the nation of Zembla there is one cityand one suburb. The population distribution in Zembla in year 0can be recorded in a vector in R2:
x0 =
[r0s0
]where r0 is the population in the city in year 0 and s0 is thepopulation in the suburb in year 0. The vectors
x1 =
[r1s1
], x2 =
[r2s2
]record the popluation distribution in year
1, year 2, etc.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 83: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/83.jpg)
Difference equations and population
You can study population dynamics using differenceequations.Let’s say that in the nation of Zembla there is one cityand one suburb. The population distribution in Zembla in year 0can be recorded in a vector in R2:
x0 =
[r0s0
]
where r0 is the population in the city in year 0 and s0 is thepopulation in the suburb in year 0. The vectors
x1 =
[r1s1
], x2 =
[r2s2
]record the popluation distribution in year
1, year 2, etc.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 84: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/84.jpg)
Difference equations and population
You can study population dynamics using differenceequations.Let’s say that in the nation of Zembla there is one cityand one suburb. The population distribution in Zembla in year 0can be recorded in a vector in R2:
x0 =
[r0s0
]where r0 is the population in the city in year 0 and s0 is thepopulation in the suburb in year 0.
The vectors
x1 =
[r1s1
], x2 =
[r2s2
]record the popluation distribution in year
1, year 2, etc.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 85: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/85.jpg)
Difference equations and population
You can study population dynamics using differenceequations.Let’s say that in the nation of Zembla there is one cityand one suburb. The population distribution in Zembla in year 0can be recorded in a vector in R2:
x0 =
[r0s0
]where r0 is the population in the city in year 0 and s0 is thepopulation in the suburb in year 0. The vectors
x1 =
[r1s1
], x2 =
[r2s2
]record the popluation distribution in year
1, year 2, etc.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 86: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/86.jpg)
Difference equations and population
Let’s say that in any one year 95 percent of city people remain inthe city and 5 percent of city people go to the suburb.
Let’s say inthe same time frame, 97 percent of suburb people remain in thesuburb and 3 percent go to the city.Thus, after year 0 we can see what the population looks like inyear 1:
r1 = .95r0 + .03s0
ands1 = .5r0 + .97s0
Thus [r1s1
]= r0
[.95.05
]+
[.03.97
]=
[.95 .03.05 .97
].
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 87: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/87.jpg)
Difference equations and population
Let’s say that in any one year 95 percent of city people remain inthe city and 5 percent of city people go to the suburb. Let’s say inthe same time frame, 97 percent of suburb people remain in thesuburb and 3 percent go to the city.
Thus, after year 0 we can see what the population looks like inyear 1:
r1 = .95r0 + .03s0
ands1 = .5r0 + .97s0
Thus [r1s1
]= r0
[.95.05
]+
[.03.97
]=
[.95 .03.05 .97
].
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 88: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/88.jpg)
Difference equations and population
Let’s say that in any one year 95 percent of city people remain inthe city and 5 percent of city people go to the suburb. Let’s say inthe same time frame, 97 percent of suburb people remain in thesuburb and 3 percent go to the city.Thus, after year 0 we can see what the population looks like inyear 1:
r1 = .95r0 + .03s0
ands1 = .5r0 + .97s0
Thus [r1s1
]= r0
[.95.05
]+
[.03.97
]=
[.95 .03.05 .97
].
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 89: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/89.jpg)
Difference equations and population
Let’s say that in any one year 95 percent of city people remain inthe city and 5 percent of city people go to the suburb. Let’s say inthe same time frame, 97 percent of suburb people remain in thesuburb and 3 percent go to the city.Thus, after year 0 we can see what the population looks like inyear 1:
r1 = .95r0 + .03s0
ands1 = .5r0 + .97s0
Thus [r1s1
]=
r0
[.95.05
]+
[.03.97
]=
[.95 .03.05 .97
].
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 90: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/90.jpg)
Difference equations and population
Let’s say that in any one year 95 percent of city people remain inthe city and 5 percent of city people go to the suburb. Let’s say inthe same time frame, 97 percent of suburb people remain in thesuburb and 3 percent go to the city.Thus, after year 0 we can see what the population looks like inyear 1:
r1 = .95r0 + .03s0
ands1 = .5r0 + .97s0
Thus [r1s1
]= r0
[.95.05
]+
[.03.97
]
=
[.95 .03.05 .97
].
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 91: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/91.jpg)
Difference equations and population
Let’s say that in any one year 95 percent of city people remain inthe city and 5 percent of city people go to the suburb. Let’s say inthe same time frame, 97 percent of suburb people remain in thesuburb and 3 percent go to the city.Thus, after year 0 we can see what the population looks like inyear 1:
r1 = .95r0 + .03s0
ands1 = .5r0 + .97s0
Thus [r1s1
]= r0
[.95.05
]+
[.03.97
]=
[.95 .03.05 .97
].
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 92: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/92.jpg)
The Transition Matrix
Now we can write down
x1 =
[.95 .03.05 .97
]x0
and
x2 =
[.95 .03.05 .97
]x1
and, in general,
xk =
[.95 .03.05 .97
]xk−1.
If A is the matrix above, then we can write xk = Axk−1. You canuse this to predict the future.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 93: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/93.jpg)
The Transition Matrix
Now we can write down
x1 =
[.95 .03.05 .97
]x0
and
x2 =
[.95 .03.05 .97
]x1
and, in general,
xk =
[.95 .03.05 .97
]xk−1.
If A is the matrix above, then we can write xk = Axk−1. You canuse this to predict the future.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 94: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/94.jpg)
The Transition Matrix
Now we can write down
x1 =
[.95 .03.05 .97
]x0
and
x2 =
[.95 .03.05 .97
]x1
and, in general,
xk =
[.95 .03.05 .97
]xk−1.
If A is the matrix above, then we can write xk = Axk−1. You canuse this to predict the future.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 95: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/95.jpg)
The Transition Matrix
Now we can write down
x1 =
[.95 .03.05 .97
]x0
and
x2 =
[.95 .03.05 .97
]x1
and, in general,
xk =
[.95 .03.05 .97
]xk−1.
If A is the matrix above, then we can write xk = Axk−1.
You canuse this to predict the future.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications
![Page 96: The matrix of a linear transformation. Applications](https://reader035.vdocuments.mx/reader035/viewer/2022071520/613c76dac957d930775e2a34/html5/thumbnails/96.jpg)
The Transition Matrix
Now we can write down
x1 =
[.95 .03.05 .97
]x0
and
x2 =
[.95 .03.05 .97
]x1
and, in general,
xk =
[.95 .03.05 .97
]xk−1.
If A is the matrix above, then we can write xk = Axk−1. You canuse this to predict the future.
Dan Crytser Lecture 8: The matrix of a linear transformation. Applications