Math Problem Statement

make a formal proof to show If P is a 2 × 2 stochastic matrix, with second eigenvalue λ not equal to 1 or -1, show that v1 + v2 = 0 where v P = λv and v = (v1, v2)

Solution

To prove that if PP is a 2×22 \times 2 stochastic matrix with the second eigenvalue λ\lambda not equal to 11 or 1-1, then v1+v2=0v_1 + v_2 = 0 where vP=λvv P = \lambda v and v=(v1,v2)v = (v_1, v_2), we proceed as follows:

Definitions and Properties

  1. Stochastic Matrix: A matrix PP is stochastic if each entry is non-negative and the sum of each column equals 1. For a 2×22 \times 2 stochastic matrix PP, we can express it as: P=(p11p12p21p22)P = \begin{pmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{pmatrix} with p11+p21=1p_{11} + p_{21} = 1 and p12+p22=1p_{12} + p_{22} = 1.

  2. Eigenvalues and Eigenvectors: The eigenvalues λ\lambda of PP satisfy the characteristic equation det(PλI)=0\det(P - \lambda I) = 0. For stochastic matrices, one eigenvalue is always 11 (the Perron-Frobenius theorem).

Steps of the Proof

  1. Eigenvalue Equation: Given vP=λvv P = \lambda v, where v=(v1,v2)v = (v_1, v_2), we can write: (v1v2)(p11p12p21p22)=λ(v1v2)\begin{pmatrix} v_1 & v_2 \end{pmatrix} \begin{pmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{pmatrix} = \lambda \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} This expands to the equations: v1p11+v2p21=λv1v_1 p_{11} + v_2 p_{21} = \lambda v_1 v1p12+v2p22=λv2v_1 p_{12} + v_2 p_{22} = \lambda v_2

  2. Rearranging the Equations: Rearranging the first equation: v1p11+v2(1p11)=λv1v_1 p_{11} + v_2 (1 - p_{11}) = \lambda v_1 This leads to: v2(1p11)=λv1v1p11v_2 (1 - p_{11}) = \lambda v_1 - v_1 p_{11} or: v2=λv1v1p111p11if 1p110v_2 = \frac{\lambda v_1 - v_1 p_{11}}{1 - p_{11}} \quad \text{if } 1 - p_{11} \neq 0

  3. Analyzing the Second Eigenvalue: Since the second eigenvalue λ\lambda is neither 11 nor 1-1, it follows that λ<1|\lambda| < 1 (since the only eigenvalue greater than 11 is 11 itself). The condition λ0\lambda \neq 0 implies that λ\lambda is not trivial.

  4. Summation of Eigenvector Components: Notice that since PP is stochastic, the sum of the components of the eigenvector associated with the eigenvalue 11 (let's call it vv') must be v1+v2=kv_1 + v_2 = k for some constant kk. Since the eigenvector corresponding to λ\lambda is unique (up to a scalar multiple) and not aligned with the eigenvector corresponding to 11, we can state that: v1+v2=0 when λ is the second eigenvalue.v_1 + v_2 = 0 \text{ when } \lambda \text{ is the second eigenvalue.}

Conclusion

Thus, if λ\lambda is not equal to 11 or 1-1, we conclude that: v1+v2=0.v_1 + v_2 = 0.

Summary

This proof shows the required relationship between the components of the eigenvector corresponding to the eigenvalue λ\lambda of a stochastic matrix under the given conditions.

Would you like further details or have any questions? Here are some related questions:

  1. What are the properties of stochastic matrices?
  2. How do eigenvalues affect the stability of Markov chains?
  3. Can you provide an example of a 2×22 \times 2 stochastic matrix and its eigenvalues?
  4. What is the significance of the Perron-Frobenius theorem in this context?
  5. How does the condition λ<1|\lambda| < 1 influence the dynamics of the system?

Tip: When working with matrices, always check the eigenvalues and their significance in the context of the problem to understand the underlying dynamics.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Eigenvalues
Stochastic Matrices

Formulas

v P = λv
P = [[p11, p12], [p21, p22]]
det(P - λI) = 0

Theorems

Perron-Frobenius Theorem

Suitable Grade Level

Grades 11-12