Hello,




More of my philosophy about eigenvectors and eigenvalues and Markov chains in mathematics and more of my thoughts..

I am a white arab from Morocco, and i think i am smart since i have also invented many scalable algorithms and algorithms..


I think i am highly smart since I have passed two certified IQ tests and i have scored "above" 115 IQ, and i mean that it is "above" 115 IQ, and as you know, i have just invented my philosophy and the new ideas of my philosophy and posted them here and
i have just invented many proverbs and posted them here and i have invented many poems of Love and poems and posted them here... , so now i will quickly create an interesting mathematical tutorial about eigenvectors and Markov chains in mathematics, and
notice carefully my way of creating it, and here it is:


Say v is an eigenvector of a matrix A with eigenvalue λ. Then Av=λv.


So there is an important thing to know in mathematics and it is that mathematical eigenvectors show in phenomenons that exhibit a "stable" behavior with time, and in Markov chains we search also for
an eigenvector where the system will stabilize, so we have to solve:

A*vector(v) = I*vector(v) [1]

The identity matrix I has 1 as its only eigenvalue. All vectors are associated eigenvectors since

I*v = v = (1)v

for all v.


A is the transition matrix
and v a vector.

So since the Eigenvalue is 1, that means we have to solve the
following system of equation that will give you the eigenvector
where the system will "stabilize" its behavior:

(A - I)*vector(x)= vector(0)

I is the identity matrix.


So now here is my tutorial about Markov chains in mathematics:

In mathematics, many Markov chains automatically find their own way to an equilibrium distribution as the chain wanders through time. This happens for many Markov chains, but not all. I will talk about the conditions required for the chain to find its
way to an equilibrium distribution.

So if in mathematics we give a Markov chain on a finite state space and asks if it converges to an equilibrium distribution as t goes to infinity. An equilibrium distribution will always exist for a finite state space. But you need to check whether the
chain is irreducible and aperiodic. If so, it will converge to equilibrium. If the chain is irreducible but periodic, it cannot converge to an equilibrium distribution that is independent of start state. If the chain is reducible, it may or may not
converge.

So i will give an example:

Suppose that for a course you are currently taking there are two volumes on the market and represent them by A and B. Suppose further that the probability that a teacher using volume A keeps the same volume
next year is 0.4 and the probability that it will change for volume B
is 0.6. Furthermore the probability that a professor using B this
year changes to next year for A is 0.2 and the probability that it
again uses volume B is 0.8. We notice that the transition matrix is:

| 0.4 0.6 |
| |
| 0.2 0.8 |


So the interesting question for any businessman is whether his
market share will stabilize over time. In other words, does it exist
a probability vector (t1, t2) such that:

(t1, t2) * (transition matrix above) = (t1, t2) [2]

and notice that the mathematical system [2] looks like the mathematical system [1] above.

So notice that the transition matrix above is irreducible and aperiodic,
so it will converge to an equilibrium distribution that is (t1, t2) that
i will mathematically find, so the system of equations of [2] above is:

0.4 * t1 + 0.2 * t2 = t1
0.6 * t1 + 0.8 * t2 = t2

this gives:

-0.6 * t1 + 0.2 * t2 = 0
0.6 * t1 - 0.2 * t2 = 0

But we know that (t1, t2) is a probability vector, so we have:

t1 + t2 = 1

So we have to solve the following system of equations:

t1 + t2 = 1
0.6 * t1 - 0.2 * t2 = 0

So i have just solved it with R, and this gives the vector:

(0.25 , 0.75)

Which means that in the long term, volume A will grab 25% of the market
while volume B will grab 75% of the market unless the advertising campaign does change the transition probabilities.



Thank you,
Amine Moulay Ramdane.

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