For the first three, either you'll get a spectral radius of 1 with exactly one eigenvalue having absolute value 1 (converges to a steady state) OR you'll get a spectral radius of 1 with multiple eigenvalues having absolute value 1 (oscillates) OR you'll get a spectral radius greater than 1 (diverges) OR you'll get a spectral radius less than 1 (converges to 0). In the first case, to compute the steady state, start with some 3-vector with entries summing to 1 (like (1/3, 1/3, 1/3)) and multiply by A a bunch of times (10-20) until you see the vector settle into a steady state.
For the two population problems, the analysis is the same, except that you use the special matrix described in the "Leslie model" section (with s's and f's). The asymptotic growth rate is just the spectral radius. For your own curiosity, if the AGR is 1.04, that means that the following year's population is 104% that of the current year, i.e., 4% growth. If it is 0.95, that means 5% loss.
For #3, it will be handy to note that A^2=A. This means that A solves x^2-x....
For #4, you are supposed to prove something, and the point is to use induction to prove it. Induction works a lot like recursion: You prove that it works for a base case and then show how to get from one step (k) to the next one (k+1). That's adequate as a proof because it gives an argument for any given level N - the statement at level N is true because the statement at level N-1 is true, which is true because the statement at level N-2 is true, which is true because...the statement at level 1 (the base case) is true.
OK, so what's the point? Check the formula for one (or even a couple) small values of n (the smallest), i.e., at least the case n=1. To get a feel for how it is working, you could also compute n=2 and/or n=3, but that's really overkill from the perspective of proof. Now assume that the statement is true for n=k, i.e., 1+...+k^3 = (k(k+1)/2)^2. Assuming that, prove the formula for the case n=k+1, i.e., prove that 1+...+k^3+(k+1)^3=((k+1)k/2)^2. Here's a hint: Start with 1+...+k^3+(k+1)^3. You already have a formula (from the assumption) for 1+...+k^3. Use it. Now try to combine your two terms, possible by factoring out a couple (k+1)'s....
Just to drive the point home, you have now shown (directly) that the formula works for n=1. Given the other (inductive) part of your proof, you also now know that it is true for n=2. Using that part of the proof again, you know that it is true for n=3 (and so on, ad infinitum). Cool, huh?
Hopefully this assignment is a bit more reasonable. For #5, please note that a closed walk is just a walk that begins and ends at the same place. Sorry that I didn't have the time to define this in class, but at least it was in the book....
This assignment is actually a bit more challenging than I had first realized. The first couple are just plug and chug problems. After that, things get a bit harder.
For number 3, you need to factor out (x-r) where r is the real root that you can spot graphically (or just by inspection, for that matter). After that, it should be straightforward (but don't forget about the other parts, namely finding the modulus and sketching....).
I gave you a formula at the end of class to help with #4. Note that there will be three solutions to this problem, so you should let k=0, k=1, and k=2 in the formula that I gave you.
For #5, the idea is to think about what that modulus means. The modulus of a vector is the distance of the end of the vector from the origin. Think of |x| as |x-0|, so that |x-0| is the distance between x and 0. Similarly, |x-1| indicates the distance between two points (which two?). Given that, what does |x-1|=1 mean?
#7 is actually significantly harder than I first thought. I will talk you through the first part. The second part is your problem. For the first part (about cos(theta)), you want to write down two versions of Euler's formula and add them. The key here is to note that cos(-theta)+i*sin(-theta)=e^(-i*theta), so (by some simple trig identities) cos(theta)-i*sin(theta)=e^(-i*theta). Adding this to the original version of Euler's formula should give you something very close to what you are trying to prove....
For #3, I failed to tell you in class last time what a standard list of BSAHS's is (although it is in the book). It's just the list of basic solutions (BSAHS's) in the order specified by the variable order. For example, if you have BSAHS's cooresponding to x1 and x2, you would list the one corresponding to x1 first, then the one corresponding to x2 (instead of the reverse order). No big deal.
For #4, please note that the vectors are not necessarily either distinguished solutions or basic solutions - they could be neither (and some are!).
There are two proofs on this homework set - #6 and #7. For #6, try to argue using the fact that I gave you regarding the distribution of 0s and 1s among the basic solutions (and also discussed at the top of page 36). The definition of linear independence says that vectors are linearly independent if the only linear combination of them that yields the zero vector is the trivial linear combination (with all coefficients zero). So, given a linear combination of these basic solutions that yields zero, why must the coefficients all be 0?
For #7, I suggest starting by assuming that the statement is false and then coming up with a contradiction. Proof by contradiction is a standard proof technique - if the statement being false leads to some contradiction, then the statement must be true.
12. This one is a little confusing. Try writing down an augmented matrix (smaller ones make less work!) in RREF and which has at least one nonpivot (free) column, using the regular variable order x1, x2, x3. Now pretend that your variable order is x1, x3, x2 and tweak the augmented matrix appropriately. Put this new matrix into RREF. Now what was your original free variable and what is the new one?
I ran short on time in class today, so please check out these notes for more details about RREF for nonnumerical matrices and also generalized augmented matrices.
Tips for 1.1
7. (i) It is tempting to multiply through by xy. However, "simplifying" (which is all you are allowed to do here) does not include multiplying by extraneous quantities. Instead, consider combining everything with a common denominator on one side of the equation (with 0 on the other side)....
GENERAL TIP: Some partial solutions are in the back of the book. Be careful, though - the answer in the book may not be complete. In particular, none of them have justification, which some problems require.