Question

A transition matrix T tells us how to get from one state to another. That is Sn=TSn-1, where Sn is the distribution vector at time n and Sn-1 is the distribution vector at time n-1. We can conclude? Sn=TSn/2 Sn=TS0 Sn=TnSn-1 Sn=TnS0

A transition matrix T tells us how to get from one state to another. That is Sn=TSn-1, where Sn is the distribution vector at time n and Sn-1 is the distribution vector at time n-1. We can conclude?
Sn=TSn/2
Sn=TS0
Sn=TnSn-1
Sn=TnS0

Answer 1

`S_n = T^n S_0`

Explanation:

We have from

`S_n = T S_(n-1)`
`S_n= T^kS_(n-k)`
`S_n= T^n S_0`

Answer 2

`S_n=T^nS_0`

Explanation:

In the calculation of matrix, associative law can be applied as if it were a number.

`ABC=A(BC)=(AB)C`

Therefore, the recurrence formula `S_n=TS_(n-1)` can be solved like that of a geometric progression.

In the recurrence formula for a geometric progression `a_n=ra_(n-1)`, the general term is `a_n=r^na_0`. Similarly, the general term of `S_n` is `S_n=T^nS_0`.

i.e.
`S_1=TS_0`
`S_2=TS_1=T(TS_0)=("TT")S_0=T^2S_0`
`S_3=TS_2=T(T^2S_0)=T^3S_0`
and so on.

See also:
http://stattrek.com/matrix-algebra/matrix-theorems.aspx

Note that `color(red)"commutative law AB=BA cannot be applied for matrix multiplication."` However, other rules(associative law, distributive law) can be used in multiplication.