Homework Problems

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Type your solutions below each problem. Explain your answers in complete sentences. Before printing, close all subsections exept this one.

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Problem 1. Let M be the matrix

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Consider the system of equations

Mx = 0 ,

where 0 is the zero vector. Calculate det(M). What does the value of the determinant tell you about the existence and uniqueness of solutions to the system for various values of a and b? Discuss.

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Problem 2. Consider the matrix N defined below. The matrix is nilpotent ; that is, there exists a positive integer m such that

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where 0 denotes the zero matrix.

(a) Assuming that a is not zero, find the smallest value of m for which that happens.

(b) Check that the determinant of N is 0. In fact, the determinant of any nilpotent matrix must be 0. Explain why.

> N:=matrix(5,5,[0,a,1,2,0, 0,0,a,1,2, 0,0,0,a,1, 0,0,0,0,a, 0,0,0,0,0]);

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Problem 3. Consider the system of equations

Px=w ,

where P is the matrix defined below and q=[0,1,2,2,3].

(a) Find all values of b for which the system has a unique solution. Find this solution.

(b) For the remaining values of b, do solutions exist? Discuss.

> P:=matrix([[0,b,1,0,b],[1,0,0,b,0],[0,1,b,0,1],[b,0,0,1,0],[0,b,1,0,2*b]]);

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> q:=vector([0,1,2,2,3]);

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Problem 4. Consider the collection of vectors v1=[0,2,1,0,2], v2=[1,0,0,2,0], v3=[0,1,2,0,1], v4=[0,6,3,0,6], v5=[0,2,1,0,4]. Are they linearly independent? What is the maximum number of vectors in the collection that are linearly independent? Find a linearly independent subcollection of the maximum size.

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Problem 5. Markov Transition Matrices

Suppose that we have a device with two states, 1 and 2. We flip a coin to determine the state at time t=0. Thus, the probability of starting at state 1 is .5 as is the probability of starting in state 2. We represent these probabilities by the vector p0:

> p0:=vector([.5,.5]);

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A clock ticks every second, and at each tick, the state changes with probabilities of changing from 1 to 1, from 1 to 2 etc. given by the transition matrix T:

> T:=matrix(2,2,[.2,.8, .4,.6]);

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Remember what this means: The entry of the matrix gives the probability that the state changes at a click from i to j. That is, for example, it will change from 1 to 1 with the probability .2, from 1 to 2 with the probability .8 and so on. Clearly, the matrix product p1=(p0)T gives the probabilities that the device will be in state 1 or, respectively 2, after the first tick.

> p1:=evalm(p0&*T);

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Indeed, the probability that after the first tick the device is in state 1 is the probability that it started at 1, which is .5, times the probability that it will pass from1 to 1, which is .2, plus the probability that it started at 2, again .5, times the probability that it will pass from 2 to 1, which is .4. Thus the probability that the device is in state 1 after the first tick is .5*.2+.5*.4, and so on. To find the probabilities after the second tick, p2, we have to multiply again by the transition matrix T: p2=(p1)T=(p0)T^2. And so it goes.

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(a) Determine the probability vector for the states after 5 ticks and after 10 ticks.

(b) Suppose many hours have gone by. What are the probabilities of finding the device in state 1? In state 2?

(c) Suppose you don't flip a coin to start but just put the device in state 1 to start. p0 is now [1,0]. What are your answers to (a) and (b) now? What if you start at state 2?

(d) Is there an initial probability vector p0 that if you start by choosing the initial state according to this vector you will have the same probability vector p0 after the first, the second, in fact, after each tick? Discuss.

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