Linear algebra

43
Linear algebra 1 Linear operators and representations Motivating example: Web start-up Vector space and basis Eigenvector-eigenvalue analysis + ^ [ 1 2 ] = [ 1 ,1 1 , 2 2 ,1 2 , 2 ][ 1 2 ] ^ =

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Linear algebra. Motivating example: Web start-up. Eigenvector-eigenvalue analysis. Vector space and basis. Linear operators and representations. +. Example: Modeling a freemium cloud data storage business. Free. Premium. +. +. +. - PowerPoint PPT Presentation

Transcript of Linear algebra

Page 1: Linear algebra

Linear algebra

1

Linear operators and representations

Motivating example: Web start-up

Vector space and basis

Eigenvector-eigenvalue analysis

+

�̂� �⃑�

𝑣

[𝑣1′𝑣2′ ]=[𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2] [𝑣1𝑣2]

�̂� �⃑�=𝜆𝑣

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𝑥𝐹 𝑥𝑃

𝑥𝐹 (𝑡+∆ 𝑡 )=𝑥𝐹 (𝑡 )+𝜌𝑥 𝑃 (𝑡 )−𝜐𝑥𝐹 (𝑡 )+𝛿𝑥𝑃 (𝑡 )−𝛼𝑥𝐹 (𝑡 )

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𝑥𝑃 (𝑡+∆ 𝑡 )=𝑥𝑃 (𝑡 )+𝜌 𝑥𝑃 (𝑡 )+𝜐 𝑥𝐹 (𝑡 )−𝛿𝑥 𝑃 (𝑡 )−𝛼 𝑥𝐹 (𝑡 )

Page 7: Linear algebra

Linear algebra

7

Linear operators and representations

Motivating example: Web start-up

Vector space and basis

Eigenvector-eigenvalue analysis

+

�̂� �⃑�

𝑣

[𝑣1′𝑣2′ ]=[𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2] [𝑣1𝑣2]

�̂� �⃑�=𝜆𝑣

Page 8: Linear algebra

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Vector

𝑣

A vector is an arrow. The position of the head in relation to the tail is expressed in terms of a magnitude and direction.

Page 9: Linear algebra

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A set of vectors

𝑣

�⃑��⃑�

𝑦�⃑�

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A set of vectors

𝑣

�⃑��⃑�

𝑦�⃑�

Page 11: Linear algebra

11

𝑣

�⃑��⃑�

𝑦

𝑣+𝑤

�⃑�

“Head-to-tail” addition of and produced a resultant vector not belonging to our original set of vectors

A set of vectors vs. a vector spaceThis scaling (doubling length in this example) of produced 2, which belongs to our original set of vectors

A vector space is a set of vectors that is “closed” under scaling and vector addition. Neither scaling nor vector addition produces a result not already included in the “space.”

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BasisA vector space is a set of vectors that are “closed” under scaling and vector addition.

A set of vectors , , . . .

Linear combination: addition of vectors with scalings

Used a set of vectors to prescribe a vector space!

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Basis set: Can’t remove any vector without changing space

A vector space is a set of vectors that are “closed” under scaling and vector addition.

A set of vectors , , . . .

Linear combination: addition of vectors with scalings

Basis for V vector space V2-dimensionalN

S

EW

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Basis: Coordinate system

Linear combination: addition of vectors with scalings

A vector space is a set of vectors that are “closed” under scaling and vector addition.

A set of vectors , , . . .

Page 15: Linear algebra

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Basis: Coordinate system

Linear combination: addition of vectors with scalings

A vector space is a set of vectors that are “closed” under scaling and vector addition.

A set of vectors , , . . .

𝑣=𝑣1𝑏1+𝑣2𝑏2

𝑏1 𝑏2

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Basis: Coordinate system

Linear combination: addition of vectors with scalings

A vector space is a set of vectors that are “closed” under scaling and vector addition.

A set of vectors , , . . .

𝑣=𝑣1𝑏1+𝑣2𝑏2

𝑏1 𝑏2

�⃑�1�⃑�2 𝑣=𝑣1 �⃑�1+𝑣2 �⃑�2

Page 17: Linear algebra

Linear algebra

17

Linear operators and representations

Motivating example: Web start-up

Vector space and basis

Eigenvector-eigenvalue analysis

+

�̂� �⃑�

𝑣

[𝑣1′𝑣2′ ]=[𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2] [𝑣1𝑣2]

�̂� �⃑�=𝜆𝑣

Page 18: Linear algebra

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Operator

𝑣

�̂� �⃑�Given a vector, an operator outputs a vector, possibly scaled and/or rotated

A function associates objects from a domain with objects in a codomain, sometimes in terms of elementary arithmetic operations.

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Linear operators

𝛼𝑣

�̂�𝛼 �⃑�=𝛼 �̂� �⃑�

Scaling Addition

𝑣 �⃑�𝑣+𝑤

�̂� �⃑�

𝑣

�̂�𝑤�⃑�

�̂� (𝛼�⃑�+𝛽𝑤 )=𝛼 �̂� �⃑�+𝛽 �̂� �⃑�

Page 20: Linear algebra

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Representing linear operators

�̂� (𝛼�⃑�+𝛽𝑤 )=𝛼 �̂� �⃑�+𝛽 �̂� �⃑�

𝑣=𝑣1𝑏1+𝑣2𝑏2

𝑏1 𝑏2

¿𝑣1 �̂�𝑏1+𝑣2 �̂�𝑏2𝑣 ′= �̂� �⃑�

𝑣 ′=𝑣1′ 𝑏1+𝑣2

′ 𝑏2

¿𝑣1 [( �̂�𝑏1)1𝑏1+ ( �̂� 𝑏1 )2𝑏2 ]¿ �̂� (𝑣1𝑏1+𝑣2𝑏2 )

+𝑣2 [( �̂� 𝑏2 )1𝑏1+( �̂�𝑏2 )2𝑏2 ]¿𝑣1 [𝔸1 , 1𝑏1+𝔸 2 ,1𝑏2 ]

+𝑣2 [𝔸 1, 2𝑏1+𝔸2 , 2𝑏2 ]

¿ (𝑣1𝔸1 , 1+𝑣2𝔸 1, 2 )𝑏1+(𝑣1𝔸 2, 1+𝑣2𝔸 2 ,2 )𝑏2

𝑣1′ 𝑏1+𝑣2′ 𝑏2

𝑣1′=𝔸 1 ,1𝑣1+𝔸 1 ,2𝑣2𝑣2′=𝔸2 ,1𝑣1+𝔸2 , 2𝑣2

[𝑣1′𝑣2′ ]=[𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2] [𝑣1𝑣2]

𝔸1 , 2

Page 21: Linear algebra

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Representing linear operators

𝑣=𝑣1𝑏1+𝑣2𝑏2

𝑏1 𝑏2

𝑣 ′=𝑣1′ 𝑏1+𝑣2

′ 𝑏2

𝑣1′=𝔸 1 ,1𝑣1+𝔸 1 ,2𝑣2𝑣2′=𝔸2 ,1𝑣1+𝔸2 , 2𝑣2

[𝑣1′𝑣2′ ]=[𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2] [𝑣1𝑣2]

�̂� (𝛼�⃑�+𝛽𝑤 )=𝛼 �̂� �⃑�+𝛽 �̂� �⃑� 𝑣 ′= �̂� �⃑� Abstract action on vector

Relationship between coefficients

Representation in the context of a particular basis

𝑣 ′→ [𝑣1′𝑣2′ ] 𝑣→[𝑣1𝑣2]

�̂�→ [𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2]

“The vector v-prime is represented by the column vector v-prime-sub-1, v-prime-sub-2”

“The operator A is represented by the matrix A”

Page 22: Linear algebra

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Vector transformation algorithm implies matrix multiplication

𝑣 ′= �̂� �⃑�

𝑣 ′ ′=�̂� 𝑣 ′

𝑣

�̂� 𝑣 ′→[𝔹1 ,1 𝔹1 , 2

𝔹2 ,1 𝔹2 , 2][ 𝔸 1 ,1𝑣1+𝔸 1 ,2𝑣2𝔸 2, 1𝑣1+𝔸 2 ,2𝑣2  ]

𝑣→[𝑣1𝑣2]�̂�→ [𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2]𝑣1′=𝔸 1 ,1𝑣1+𝔸 1 ,2𝑣2𝑣2′=𝔸2 ,1𝑣1+𝔸2 , 2𝑣2

¿ [𝔹1, 1 (𝔸 1 ,1𝑣1+𝔸1 ,2𝑣2 )+𝔹1 ,2 (𝔸 2 ,1𝑣1+𝔸 2 ,2𝑣2 )𝔹2, 1 (𝔸 1 ,1𝑣1+𝔸1 ,2𝑣2 )+𝔹2 , 2 (𝔸 2 ,1𝑣1+𝔸 2 ,2𝑣2 )]

¿ [ (𝔹1 ,1𝔸1 ,1+𝔹1 ,2𝔸2 , 1 )𝑣1+ (𝔹1 , 1𝔸 1, 2+𝔹1, 2𝔸 2 ,2 )𝑣2(𝔹2 ,1𝔸 1 ,1+𝔹2 ,2𝔸2 , 1 )𝑣1+ (𝔹2 ,1𝔸 1 ,2+𝔹2 ,2𝔸2 , 2 )𝑣2]

¿ [𝔹1, 1𝔸 1, 1+𝔹1, 2𝔸 2 ,1 𝔹1 , 1𝔸1 , 2+𝔹1 , 2𝔸 2 ,2

𝔹2, 1𝔸 1, 1+𝔹2 ,2𝔸 2 ,1 𝔹2 , 1𝔸1 , 2+𝔹2, 2𝔸 2 ,2][𝑣1𝑣2]�̂� �̂�𝑣→[𝔹1 ,1 𝔹1 ,2

𝔹2 ,1 𝔹2 ,2] [𝔸1 ,1 𝔸1 , 2

𝔸2 , 1 𝔸2 , 2] [𝑣1𝑣2]

Page 23: Linear algebra

Linear algebra

23

Linear operators and representations

Motivating example: Web start-up

Vector space and basis

Eigenvector-eigenvalue analysis

+

�̂� �⃑�

𝑣

[𝑣1′𝑣2′ ]=[𝔸 1 ,1 𝔸 1 ,2

𝔸 2 ,1 𝔸 2 ,2] [𝑣1𝑣2]

�̂� �⃑�=𝜆𝑣

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𝑥𝐹 𝑥𝑃

𝑥𝐹 (𝑡+∆ 𝑡 )=𝑥𝐹 (𝑡 )+𝜌𝑥 𝑃 (𝑡 )−𝜐𝑥𝐹 (𝑡 )+𝛿𝑥𝑃 (𝑡 )−𝛼𝑥𝐹 (𝑡 )

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𝑥𝑃 (𝑡+∆ 𝑡 )=𝑥𝑃 (𝑡 )+𝜌 𝑥𝑃 (𝑡 )+𝜐 𝑥𝐹 (𝑡 )−𝛿𝑥 𝑃 (𝑡 )−𝛼 𝑥𝐹 (𝑡 )

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Free Premium𝑥𝐹 𝑥𝑃𝑥𝐹 (𝑡+∆ 𝑡 )=𝑥𝐹 (𝑡 )+𝜌𝑥 𝑃 (𝑡 )−𝜐𝑥𝐹 (𝑡 )+𝛿𝑥𝑃 (𝑡 )−𝛼𝑥𝐹 (𝑡 )𝑥𝑃 (𝑡+∆ 𝑡 )=𝑥𝑃 (𝑡 )+𝜌 𝑥𝑃 (𝑡 )+𝜐 𝑥𝐹 (𝑡 )−𝛿𝑥 𝑃 (𝑡 )−𝛼 𝑥𝐹 (𝑡 )

[𝑥𝐹 (𝑡+∆ 𝑡 )𝑥𝑃 (𝑡+∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ][𝑥𝐹 (𝑡 )𝑥𝑃 (𝑡 ) ]

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�⃑� (𝑡 )=𝑥𝐹 (𝑡 ) �⃑� +𝑥𝑃 (𝑡 ) �⃑�

Page 30: Linear algebra

0 0.25 0.5 0.75 1 1.25 1.5 1.750

0.25

0.5

0.75

1

1.25

1.5

1.75 �̂� �⃑�=𝜆𝑣

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

𝑥 𝑃

𝑁

𝑥 𝐹

𝑁

Easy-

lookin

g-one-d

imen

siona

l pro

blem

+

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� �⃑�=𝜆𝑣�̂� �⃑�=𝜆 𝐼 �⃑�

�̂� �⃑�− 𝜆 �̂� �⃑�= 0⃑( �̂�− 𝜆𝐼 ) �⃑�= 0⃑

([𝔸1 ,1 𝔸1 , 2

𝔸2 , 1 𝔸2 , 2]− 𝜆 [1 00 1])[𝑣𝐹

𝑣𝑃 ]=[00 ]([𝑎 𝑏𝑐 𝑑]− 𝜆[1 0

0 1 ])[𝑣 𝐹

𝑣 𝑃 ]=[00]

STOP

Check that

is consistent in a matrix representation

[𝑎− 𝜆 𝑏𝑐 𝑑− 𝜆][𝑣𝐹

𝑣 𝑃 ]=[00](𝑎− 𝜆 )𝑣 𝐹+𝑏𝑣𝑃=0𝑐 𝑣𝐹+(𝑑−𝜆 )𝑣 𝑃=0

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� �⃑�=𝜆𝑣(𝑑− 𝜆 ) (𝑎− 𝜆 )𝑣𝐹+ (𝑑− 𝜆 )𝑏𝑣𝑃=0

𝑏𝑐 𝑣𝐹+𝑏 (𝑑− 𝜆 )𝑣 𝑃=0- [ ][ (𝑎−𝜆 ) (𝑑− 𝜆 )−𝑏𝑐 ]𝑣𝐹=0

(𝑎− 𝜆 ) (𝑑− 𝜆 )−𝑏𝑐=0𝑎𝑑− 𝜆𝑎− 𝜆𝑑+𝜆2−𝑏𝑐=0

𝜆2− (𝑎+𝑑 ) 𝜆+(𝑎𝑑−𝑏𝑐 )=0

𝜆±=(𝑎+𝑑 )±√ (𝑎+𝑑 )2−4 (1 ) (𝑎𝑑−𝑏𝑐 )

2 (1 )

𝜆±=(𝑎+𝑑 )±√𝑎2+2𝑎𝑑+𝑑2−4 𝑎𝑑+4𝑏𝑐

2

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� �⃑�=𝜆𝑣𝜆±=

(𝑎+𝑑 )±√𝑎2+2𝑎𝑑+𝑑2−4 𝑎𝑑+4𝑏𝑐2

𝜆±=(𝑎+𝑑 )±√ (𝑎−𝑑 )2+4𝑏𝑐

2

𝜆±=(2−𝜐−𝛼−𝛿 )±√ (𝛿−𝜐−𝛼 )2+4 (𝜌+𝛿 )𝜐

2There are 2 possibly special scaling factors. Does each l actually correspond to a special ?

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� �⃑�=𝜆𝑣𝜆±=

(𝑎+𝑑 )±√ (𝑎−𝑑 )2+4𝑏𝑐2

There are 2 possibly special scaling factors. Does each l actually correspond to a special ?

[𝑎− 𝜆± 𝑏𝑐 𝑑− 𝜆±] [𝑣 𝐹

±

𝑣 𝑃± ]=[00]

(𝑎− 𝜆±) 𝑣𝐹± +𝑏𝑣𝑃

±=0𝑏𝑣 𝑃

±= (𝜆±−𝑎 )𝑣𝐹±

𝑣 𝑃±=

𝜆±−𝑎𝑏 𝑣 𝐹

±

𝑣 𝑃±=

𝛼+𝜐− 𝛿±√ (𝛿−𝜐−𝛼 )2+4 (𝜌+𝛿 )𝜐2 (𝜌+𝛿 )

𝑣𝐹±

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� �⃑�=𝜆𝑣𝜆±=

(𝑎+𝑑 )±√ (𝑎−𝑑 )2+4𝑏𝑐2

𝑣 𝑃±=

𝛼+𝜐− 𝛿±√ (𝛿−𝜐−𝛼 )2+4 (𝜌+𝛿 )𝜐2 (𝜌+𝛿 )

𝑣𝐹±

𝑄±

There are 2 special scaling factors; each l corresponds to a special vector . Unless something is hokey, they point in different directions and can serve as a basis.

�⃑�+¿ ¿ 𝑣−

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� 𝑣±=𝜆± 𝑣±𝑣 𝑃

±=𝑄±𝑣𝐹±

�⃑� (𝑡 )=𝑥𝐹 (𝑡 ) �⃑� +𝑥𝑃 (𝑡 ) �⃑�𝑁 �⃑� +0 �⃑�=𝑁𝑐

+ ¿⃑𝑣+¿+𝑁𝑐−𝑣−¿ ¿

�⃑�+¿ ¿ 𝑣−

�⃑�=𝑐+¿ ¿¿

Inaugural trials

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�⃑�=𝑐+¿ ¿¿

�⃑�=¿𝑐+¿+𝑐−=1¿ 𝑐

+¿𝑄 +¿+𝑐 −𝑄−=0¿ ¿

�̂� 𝑣±=𝜆± 𝑣±𝑣 𝑃

±=𝑄±𝑣𝐹±

�⃑� (𝑡 )=𝑥𝐹 (𝑡 ) �⃑� +𝑥𝑃 (𝑡 ) �⃑�𝑁 �⃑� +0 �⃑�=𝑁𝑐

+ ¿⃑𝑣+¿+𝑁𝑐−𝑣−¿ ¿

Inaugural trials

�⃑�=𝑐+¿ ¿¿

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

𝑐+¿+𝑐−=1¿ 𝑐+¿𝑄 +¿+𝑐 −𝑄−=0¿ ¿

𝑐+¿𝑄 +¿=− 𝑐−𝑄−¿ ¿

𝑐+¿=−𝑐−

𝑄−

𝑄+¿¿¿−𝑐−

𝑄−

𝑄+¿+𝑐−=1¿

𝑐−¿𝑐−¿ 𝑐−=

𝑄+¿

𝑄+¿−𝑄−

¿¿ 𝑐

+¿=− 𝑄−

𝑄+¿−𝑄−

¿¿

�⃑� (𝑡 )=𝑁𝑐+¿⃑ 𝑣+¿ +𝑁 𝑐−𝑣−¿ ¿

�̂� 𝑣±=𝜆± 𝑣±𝑣 𝑃

±=𝑄±𝑣𝐹±

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

𝑐−=𝑄+¿

𝑄+¿−𝑄−

¿¿ 𝑐

+¿=− 𝑄−

𝑄+¿−𝑄−

¿¿

�⃑� (𝑡 )=𝑁𝑐+¿⃑ 𝑣+¿ +𝑁 𝑐−𝑣−¿ ¿

�̂� 𝑣±=𝜆± 𝑣±𝑣 𝑃

±=𝑄±𝑣𝐹±

�̂� �̂� �̂� �⃑� (𝑡 )=− 𝑁𝑄−

𝑄+¿−𝑄−

𝜆+¿𝜆+¿ 𝜆+ ¿ �̂�⃑

𝑣 +¿+ 𝑁𝑄+¿

𝑄+ ¿−𝑄− 𝜆− 𝜆− 𝜆− �̂� �⃑�−¿¿ ¿ ¿

¿ ¿¿

�̂�𝑀 �⃑� (𝑡 )=− 𝑁𝑄−

𝑄+¿−𝑄−

𝜆+¿𝑀 𝑣+¿ + 𝑁𝑄+¿

𝑄+¿−𝑄−

𝜆−𝑀𝑣−¿

¿¿ ¿¿

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� 𝑣±=𝜆± 𝑣±𝑣 𝑃

±=𝑄±𝑣𝐹±

�̂�𝑀 �⃑� (𝑡 )=− 𝑁𝑄−

𝑄+¿−𝑄−

𝜆+¿𝑀 𝑣+¿ + 𝑁𝑄+¿

𝑄+¿−𝑄−

𝜆−𝑀𝑣−¿

¿¿ ¿¿

�⃑� (𝑡+𝑀∆ 𝑡 )=− 𝑁𝑄−

𝑄+¿−𝑄−

𝜆+¿𝑀¿ ¿¿

𝑥𝐹 (𝑡+𝑀∆ 𝑡 )=𝑁𝑄+¿𝜆−

𝑀−𝑄− 𝜆+¿ 𝑀

𝑄 +¿ −𝑄−

¿¿

¿

𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )=𝑁 𝑄+¿𝑄−

𝑄+¿−𝑄−

¿¿¿

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� 𝑣±=𝜆± 𝑣±𝑣 𝑃

±=𝑄±𝑣𝐹±

𝑥𝐹 (𝑡+𝑀∆ 𝑡 )=𝑁𝑄+¿𝜆−

𝑀−𝑄− 𝜆+¿ 𝑀

𝑄 +¿ −𝑄−

¿¿

¿

𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )=𝑁 𝑄+¿𝑄−

𝑄+¿−𝑄−

¿¿¿

𝜆±=(2−𝜐−𝛼−𝛿 )±√ (𝛿−𝜐−𝛼 )2+4 (𝜌+𝛿 )𝜐

2

𝑄±=𝛼+𝜐−𝛿±√ (𝛿−𝜐−𝛼 )2+4 (𝜌+𝛿 )𝜐

2 (𝜌+𝛿 )

, , , = 0.2, 0.2, 0.1, 0.1

Page 42: Linear algebra

0 0.25 0.5 0.75 1 1.25 1.5 1.750

0.25

0.5

0.75

1

1.25

1.5

1.75

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

𝑥 𝑃

𝑁

𝑥 𝐹

𝑁

�̂� 𝑣±=𝜆± 𝑣±

, , , = 0.2, 0.2, 0.1, 0.1

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𝑥𝐹

𝑥𝑃

[𝑥𝐹 (𝑡+𝑀∆ 𝑡 )𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )]=[1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿][1−𝜐−𝛼 𝜌+𝛿𝜐 1−𝛿]⋯ [1−𝜐−𝛼 𝜌+𝛿

𝜐 1−𝛿 ] [𝑥 𝐹 (𝑡 )𝑥 𝑃 (𝑡 )]

M copies of matrix

�̂� 𝑣±=𝜆± 𝑣±𝑣 𝑃

±=𝑄±𝑣𝐹±

𝑥𝐹 (𝑡+𝑀∆ 𝑡 )=𝑁𝑄+¿𝜆−

𝑀−𝑄− 𝜆+¿ 𝑀

𝑄 +¿ −𝑄−

¿¿

¿

𝑥𝑃 (𝑡+𝑀 ∆ 𝑡 )=𝑁 𝑄+¿𝑄−

𝑄+¿−𝑄−

¿¿¿

𝜆±=(2−𝜐−𝛼−𝛿 )±√ (𝛿−𝜐−𝛼 )2+4 (𝜌+𝛿 )𝜐

2

𝑄±=𝛼+𝜐−𝛿±√ (𝛿−𝜐−𝛼 )2+4 (𝜌+𝛿 )𝜐

2 (𝜌+𝛿 )

Eigenvectors

Eigenvalues

, , , = 0.2, 0.2, 0.1, 0.1