## Problem Set 6 Comments and Hints

Here are some relevant Maple commands:
with(LinearAlgebra):
A:=Matrix([[...
evalf(Eigenvectors(A))
h,P:=Eigenvectors(A)
Try:
li := <P|Vector([0,0])> (or Vector([0,0,0])) or similar
span:=<P|Vector([x,y])> (or Vector([x,y,z])) or similar
Diagonalizability:
MatrixInverse(P).A.P
Eigenvalues(A)

Problem 1 by hand and Maple For the by-hand portion, first form det(lamdaI-A)=0 and solve for lambda. For each value of lambda, solve (lamdaI-A)x=0. Remember to compare with Maple and resolve any apparent conflicts or differences.

Problem 2 Rotation matrices In Part A, you want to make the number under the square root positive or 0 - solve for the ALL the values of theta that allow this. Once you have any value of theta that will work, you can plug it in and use the Eigenvectors command [the non-zero eigenvectors in the columns of P that Maple produces are a basis]. For help with Part C, look at the second example in the Geometry of Eigenvectors (M:=Matrix([[0,1],[-1,0]]); which is a rotation clockwise by 90 degrees) and try and generalize this reasoning here. If you need to express an angle like 90 degrees, use expressions like Pi/2 in Maple.

Problem 3: Foxes and Rabbits (Predator-prey model) For similar class work, see the ASULearn for Foxes and Rabbits demo and the Dynamical Systems and Eigenvectors demo as well as the rural and urban population problem and the eigenvector decomposition in Maple. In part C, be sure to test out what happens to the fox and rabbit populations with different constants in the limit and to fill in the blanks, like we did in class examples. You can use evalf(Eigenvectors(A)) to obtain the decimal approximation of the eigenvalues and eigenvectors.

For part D, you want to make 1.05 unknown, say a k variable, and solve to force an eigenvalue of 1. If you input into Maple, Maple will output a pretty messy eigenvalue involving a square root. So, in this case, it is better to solve by hand so that we can make things easier by using the knowledge that we want lamda equal to 1 for stability: - plug in lambda=1 into the Eigenvalue equation: determinant (lambda I - A) = determinant(I-A)=0. The only unknown in this equation is the value of k so you can solve. Then you can plug that k value back in to Maple to determine the ratio via the eigenvectors.
Caution: each term in xx contributes to the foxes and rabbits - the tops of each column vector add to make up the foxes, and the bottoms of each column vector add to make up therabbits. It is incorrect to say that one eigenvector represents rabbits, while the other represents foxes.

Diagonalizability Be very careful - Caution: if Maple outputs distinct eigenvalues, then you can quote the theorem that states that there will be n linearly independent eigenvectors and hence the original matrix will be diagonalizable, however, if not, then this theorem does not apply (we saw a counterexample to the converse in class - that had repeated eigenvalues but was diagonalizable). You must check directly whether there are n linearly independent eigenvectors as the columns of P by setting up Px=0 and seeing if it has only the trivial solution (definition of linear independence).

For example, see class notes - one 3x3 matrix A we worked with is diagonalizable because the columns of P are linearly independent - (det(P) is not 0, so Px=0 has only the trivial solution) even though A has repeated eigenvalues.