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 x_{x} 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.