## Problem Set 6 Comments and Hints

Here are some relevant Maple commands:
with(LinearAlgebra):
A:=Matrix([[...
h,P:=Eigenvectors(A)
Determinant(P)
ReducedRowEchelonForm(Matrix([P,Vector([0,..,0])])
GaussianElimination(Matrix([P,Vector([u1,..,un])])
MatrixInverse(P).A.P
Eigenvalues(A)

Problem 1 7.1 #26 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 for x and then pull out the free variables. 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. In Part B, once you have values of theta that will work, you can plug them 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 ASULearn geometry of eigenvectors demo (M:=Matrix([[0,1],[-1,0]]); which is a rotation clockwise by 90 degrees) and try and generalize this reasoning here.

Diagonalizability Be very careful - 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. 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 - the 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.

Problem 6: 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.

For part C, 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 eigenvector ratio.