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Linear combination of eigenvs

  • Thread starter krisrai
  • Start date
  • #1
15
0

Homework Statement



1. Initially, we assume the breeder’s plants are all growing in a field where they will be
cross-polenated randomly, with genes that can come from anywhere (even neighbouring
fields of flowers). A given plant would thus be crossed randomly, so that its offspring
would get an R or W gene with equal probability. As a result the offspring of the plants
will be as follows:

the offspring of the red plants will be 50% red, 50% pink and 0% white
the offspring of the pink plants will be 25% red, 50% pink and 25% white
the offspring of the white plants will be 0% red, 50% pink and 50% white


Create a transition matrix A so that, given the state vector ~St at time t, the fraction of
the breeder’s crop that is of each color the next year (time t + 1) can be found as
St+1 = A*St
where
St= [rt+1 pt+1 wt+1]^T
Use the probabilities above to populate the columns of the matrix.

2. It turns out that red flowers are the most popular at the florists, so the breeder begins
with an initial state vector, S0, with r0 = 1/2, p0 = 1/4, and w0 = 1/4. Using your matrixfrom question 1, determine the proportions of each type of flower in years t = 1, and t = 2.


3. Write the initial state vector 2 as linear combinations of the eigenvectors.


Homework Equations


LinearMultiplication....


The Attempt at a Solution



1. A=
[1/2 1/4 0
1/2 1/2 1/2
0 1/4 1/2]

2. S(t+1)=AS(t)

so to find S(3) I first need my S(2) and S(1)
which I found:

t=0, S(1)= A*S(0)

[1/2 1/4 0 [1/2 [5/16
1/2 1/2 1/2 X 1/4 = 1/2
0 1/4 1/2] 1/4] 3/16]
*these I know are correct because the proportions add up to 1 or 100%


and t=1 S(2)=AS(1)

[1/2 1/4 0 [5/16 [9/32
1/2 1/2 1/2 X 1/2 = 1/2
0 1/4 1/2] 3/16] 11/32]

*Could someone please tell me where Im going wrong? My proportions of each colour do not add up to one


so for my t=2 S(3)=AS(2) Im scared to do because my flowers are disproportional



3. So here I found my eigenvalues and associated eigenvectors for my matrix A
Lambda1=1
X1=
[1
2
1]

Lambda2=1/2
X2=
[1
0
-1]

Lambda3= 0
X3=
[1
-2
1]

would I be able to write my linear combination of eigen vectors as:
r(t)= 1/2 + j+k+m
p(t)= 1/4 +2j -2m
w(t)= 1/4 +j-k+m

Help would be really appreciated. I need to be able to find my S(10) from this linear equation and I dont know how to do it.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,777
911

Homework Statement



1. Initially, we assume the breeder’s plants are all growing in a field where they will be
cross-polenated randomly, with genes that can come from anywhere (even neighbouring
fields of flowers). A given plant would thus be crossed randomly, so that its offspring
would get an R or W gene with equal probability. As a result the offspring of the plants
will be as follows:

the offspring of the red plants will be 50% red, 50% pink and 0% white
the offspring of the pink plants will be 25% red, 50% pink and 25% white
the offspring of the white plants will be 0% red, 50% pink and 50% white


Create a transition matrix A so that, given the state vector ~St at time t, the fraction of
the breeder’s crop that is of each color the next year (time t + 1) can be found as
St+1 = A*St
where
St= [rt+1 pt+1 wt+1]^T
Use the probabilities above to populate the columns of the matrix.

2. It turns out that red flowers are the most popular at the florists, so the breeder begins
with an initial state vector, S0, with r0 = 1/2, p0 = 1/4, and w0 = 1/4. Using your matrixfrom question 1, determine the proportions of each type of flower in years t = 1, and t = 2.


3. Write the initial state vector 2 as linear combinations of the eigenvectors.


Homework Equations


LinearMultiplication....


The Attempt at a Solution



1. A=
[1/2 1/4 0
1/2 1/2 1/2
0 1/4 1/2]

2. S(t+1)=AS(t)

so to find S(3) I first need my S(2) and S(1)
which I found:

t=0, S(1)= A*S(0)

[1/2 1/4 0 [1/2 [5/16
1/2 1/2 1/2 X 1/4 = 1/2
0 1/4 1/2] 1/4] 3/16]
*these I know are correct because the proportions add up to 1 or 100%


and t=1 S(2)=AS(1)

[1/2 1/4 0 [5/16 [9/32
1/2 1/2 1/2 X 1/2 = 1/2
0 1/4 1/2] 3/16] 11/32]

*Could someone please tell me where Im going wrong? My proportions of each colour do not add up to one
(1/4)(1/2)+ (1/2)(3/16)= 1/8+ 3/32= ? (NOT 11/32!)

so for my t=2 S(3)=AS(2) Im scared to do because my flowers are disproportional



3. So here I found my eigenvalues and associated eigenvectors for my matrix A
Lambda1=1
X1=
[1
2
1]

Lambda2=1/2
X2=
[1
0
-1]

Lambda3= 0
X3=
[1
-2
1]

would I be able to write my linear combination of eigen vectors as:
r(t)= 1/2 + j+k+m
p(t)= 1/4 +2j -2m
w(t)= 1/4 +j-k+m

Help would be really appreciated. I need to be able to find my S(10) from this linear equation and I dont know how to do it.
 
  • #3
15
0
thank you i must have multiplied the other lines wrong much:)
i figured out number 3 this thread is SOLVED
 

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