A machine is either working (state 1) or not workind (state 2). If it is working one day the probability that it will be broken the next day is 0.1. If it is not working one day the probability that it will be working the next day is 0.8. Let Tn be the state of the machine n days from now. Assume the Markov assumption is satisfied so that Tn is a Markov Chain.

-Find the eigenvalues and eigenvectors of P. And find a formula for P^n.

If I recall Markov's chain, the matrix for the above would be

.9 .1
.8 .2

Is this called the transition matrix ?

The steady state vector, or eigenvector, is found by multiplying an initial probability vector by that transition vector and then using that resulting vector in your next multiplication.
This will eventually converge to a fixed vector, that is

[a b] times
.9 .1
.8 .2 equals [a b]

then .9a + .1b = a
and .8a + .2b = b
also a + b = 1 by the laws of probability.
solving this I got [.5 .5]

hope this helps, haven't done this stuff in years. I think the last time I taught this was over 20 years ago.

I made a fundamental error in my matrix multiplication

my equations should have been
.9a + .8b = a
and .1a + .2b = b

[a b] should have been [8/9 1/9]

To find the eigenvalues and eigenvectors of matrix P, we first need to define the matrix P.

Let's denote the transition probability matrix as P, where Pij represents the probability of transitioning from state i to state j. In this problem, we have two states: state 1 (working) and state 2 (not working).

The transition probability matrix P can be written as:

P = | P11 P12 |
| P21 P22 |

Given the information in the problem, we can fill in the values in the matrix P:

P = | 1 - 0.1 |
| 0.8 1 |

Now, we can proceed to find the eigenvalues and eigenvectors.

1. Eigenvalues (λ):
The eigenvalues (λ) are the solutions to the equation det(P - λI) = 0, where I is the identity matrix of the same order as P.

So, we have the equation:

| 1 - λ -0.1 | | x | = 0
| 0.8 1 - λ | | y |

Expanding the determinant using cofactor expansion along the first row, we get:

(1 - λ)(1 - λ) - (-0.1)(0.8) = (1 - λ)^2 - 0.08 = λ^2 - 2λ + 0.92 = 0

Solving this quadratic equation, we find two eigenvalues:

λ1 = 1.84 and λ2 = 0.16

2. Eigenvectors (v):
To find the eigenvectors, we substitute each eigenvalue back into the equation (P - λI)v = 0, and solve for v.

For λ1 = 1.84:

(1 - 1.84)x - 0.1y = 0. (equation 1)
0.8x + (1 - 1.84)y = 0. (equation 2)

Simplifying equations 1 and 2 gives us:

-0.84x - 0.1y = 0. (equation 3)
0.8x - 0.84y = 0. (equation 4)

From equation 3, we can express x in terms of y:

x = -0.1y / -0.84 = 0.119y

Substituting the value of x in equation 4, we have:

0.8(0.119y) - 0.84y = 0
0.095y - 0.84y = 0
-0.745y = 0
y = 0

Substituting the value of y = 0 into equation 3, we get:

-0.84x - 0.1(0) = 0
-0.84x = 0
x = 0

Therefore, one eigenvector corresponding to λ1 = 1.84 is:

v1 = [0, 0]

For λ2 = 0.16:

(1 - 0.16)x - 0.1y = 0. (equation 5)
0.8x + (1 - 0.16)y = 0. (equation 6)

Simplifying equations 5 and 6 gives us:

0.84x - 0.1y = 0. (equation 7)
0.8x + 0.84y = 0. (equation 8)

From equation 7, we can express x in terms of y:

x = 0.1y / 0.84 = 0.119y

Substituting the value of x in equation 8, we have:

0.8(0.119y) + 0.84y = 0
0.095y + 0.84y = 0
0.935y = 0
y = 0

Substituting the value of y = 0 into equation 7, we get:

0.84x - 0.1(0) = 0
0.84x = 0
x = 0

Therefore, one eigenvector corresponding to λ2 = 0.16 is:

v2 = [0, 0]

Thus, we have found the eigenvalues (λ1 = 1.84, λ2 = 0.16) and the eigenvectors (v1 = [0, 0], v2 = [0, 0]).

Regarding the formula for P^n, since the eigenvalues of P are both 1, P^n does not converge to a fixed matrix. Instead, it will oscillate between two matrices, depending on the initial state.