2 Consider the matrix 053125 048125 04625041250025 013125 07

2. Consider the matrix 0.53125 0.48125 0.4625-0.4125-0.025 0.13125 0731250.1375 0.1375 0.025 P= 0.0625-0.1375 0.725 0.075 0.05 0.13125-0.61875-0.3375 1.0875 0.075 0.2 0.85 0.175 0.175 0.15 We will find an approximation of the inverse of P by using a trick similar to what we did in problem #1. Follow the steps:

Solution

ALERT: The calculations were not done on SAGE (not available)

This solution highlights the theoretical aspects (eigenvalues, nilpotence, convergence,etc).

0.46875 -0.48125 -0.4625 0.4125 0.025
-0.13125 0.16875 0.1375 -0.1375 -0.025
0.0625 0.1375 0.275        -0.075    -0.05
-0.13125 0.61875 0.3375 -0.0875 -0.075

-0.175 0.175 0.15 -0.2 0.15

The eigenvalues are

Clearly all the eigenvalues have modulus <1

(c) As the eigenvalues of B=I-P are all less than 1 in absolute value , the series

I +B +B² +...........+Bⁿ+......will converge to (I-B)^-1 (in analogy with the geometric series) = P^-1

Hence P(I +B +B² +...........+Bⁿ) will converge to P P^-1 , namely , I

(d), (e) Implement sage with n=20

(f) This approximation works because B is (nearly , approximately) nilpotent, even though it is not absolutely nilpotent.

It will in general not work for any P (ie. working with B=I-P wont lead to convergence).

The following may be used. Starting with P , normalize P by dividing by the highest (abolute) eigenvalue and work with the resulting matrix.