eigenvalues and eigenvectors

Question

That is right, you already stated teh value of E(x)^2 at the bottom being finite, so as you mentioned infinite - finite value is infinite

Suppose a phone company has two plans, regular and high use.Suppose that they find that each month on average 10% of those on the regular plan move to high use, while 90% stay on the regular plan.Meanwhile 60% of those on high use move to regular while 40% stay.In the following model we assume that the effects of new customers or customers leaving are negligible and are not included as a possibility.Then if Rk and Hk represent the proportion of customers in the regular and high use plans respectively during month k, then we have[Hk+1] [0.1 0.6] [Hk]

     [Rk+1]=[0.9 0.4] [Rk](

a) Find the eigenvalues and eigenvectors of A=[0.1 0.6

     0.9 0.4], showing working.

(b) Suppose the initial state was H0= 0.5 and R0= 0.5. Find a and b such that (R0, H0) =av1+bv2, where v1 is your eigenvector for the largest eigenvalue, and v2 is your eigenvector for the other eigenvalue.

Answer 1

So, the matrix A we have is

[0.1 0.6]

[0.9 0.4]

To solve for Eigenvalues, we solve the equation

Det(A-λI)=0, where I is the Identity Matrix, and λ represents the variable we use to solve our characteristic equation for eigenvalues.

[0.1 0.6] [λ 0] [0.1-λ 0.6]

[0.9 0.4] - [0 λ] = [0.9 0.4-λ]

Then, we take the determinant of this matrix

Recall that the determinant of a 2x2 matrix

[a b]

[c d]

is equal to ad-cb

So, the determinant of the above is

(0.1-λ)(0.4-λ) - (0.9)(0.6)

=0.5(2λ²-λ-1). We must solve this, and the roots of this will be our eigenvalues. You may use any method you like to solve this quadratic, such as the quadratic formula. I will be omitting the solving of the quadratic formula, and go straight to the roots I found of λ₁=-0.5 and λ₂=1λ

λ₁=-0.5

For each λ, you find its own vectors, which you do by substituting λ back into A-λI

A-λ₁I=

[0.1-(-0.5) 0.6]

[0.9 0.4-(-0.5]

=

[0.6 0.6]

[0.9 0.9]

So, we're looking for a vector v which, when right multiplied to A-Iλ₁, makes this equal to 0. We can see here that we have a homogeneous system of linear equations, which can be solved by Gaussian Elimination. However, I will leave that to you, as it can be seen here that the entries are equal to one another. This must mean that the variable in column 1 is equal to the negative variable in column 2, if we are to set this to 0.

x₁+x₂=0, and with x₁=-x₂, we can let x₂=1, such that x₁=-1, and our eigenvector for this, v₁=(-1)

(1)

Now, we must do the same thing for the second eigenvalue, λ₂=1

A-λ₂I=

[0.1-(1) 0.6]

[0.9 0.4-(1)]

=

[-0.9 0.6]

[0.9 -0.6]

Again, we solve this with row operations to find

x₁-(2/3)x₂=0

x₁=(2/3)x₂

So, v₂=(2/3)

(1)

I hope this is able to help and that you're able to follow the formatting! let me know if there's more I can assist with! Good luck!