Linear Algebra; Stochastic matrix and Steady State vectors

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SUMMARY

Every 2 x 2 stochastic matrix, represented as P = |1-a b| |a 1-b|, where a and b are constants between 0 and 1, has at least one steady-state vector. The existence of a steady-state vector is confirmed by solving the equation Px = x, which leads to the system (P - I)x = 0. If a = b = 0, there are two linearly independent steady-state vectors; otherwise, there is only one. This conclusion is derived from the properties of eigenvectors and determinants in linear algebra.

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MustangGt94
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Homework Statement



Question: 18. Show that every 2 x 2 stochastic matrix has at least one steady-state vector. Any such matrix can be written as

P = |1-a b |
| a 1-b |

where a and b are constants between 0 and 1. (There are two linearly independent steady-state vectors if a = b = 0. Otherwise, there is only one.)

The Attempt at a Solution



I am guessing that they want me to show that the above matrix has a solution to the equation Px = x so that the steady state vector exists since there is a solution. My solution was that since a and b are other constants between 0 and 1 the matrix would have a solution?

Thank you for the help!
 
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Hi Mustang,

P = \begin{pmatrix} 1-a & b\\ a & 1-b\end{pmatrix}

Consider that if Px = x, then Px - x = 0, i.e., (P - I)x = 0, where I is the 2x2 identity matrix. As a steady state vector (just a certain type of "eigenvector", if you are familiar with the term) is necessarily nonzero, recall how one can use the determinant to determine where the system Ax=0 has a nonzero solution.
 
Last edited:
Ah! I see what you mean, thanks a lot Unco, much appreciated!
 

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