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.