Even if Maple does not provide a friendly answer: I guess you are
talking about the real case.

May be a basic geometric view on eigenvalues and eigenvectors helps:

Let f be linear with f(x) = lambda*x and lambda =/= 0, x =/= 0,
lambda in the field, x in the vector space. 

Then the line L = IR*x given by x is an invariant space under f
[i.e. f(L) = L as a set] and g = f restricted to the subspsace L
is simply a multiplication by lambda [i.e. g(y) = lambda*y for y
in L], just write it down by paper and pencil:

for y in L write y = alpha*x and you have f(y)= alpha*f(x) =
alpha*lambda*x = lambda*y, so it is in L.

So an eigenvector is not *one* vector, it *is* a 1-dim subspace
and for lambda=0 it is the whole kernel of f.

But a rotation in the plane always rotates any line and never leaves
them invariant - except the rotation is by 180° or 360°, i.e. your
theta is a multiple of Pi. Then your rotation equals +- identity.

More formally: you where already told that the eigenvalues are
complex, so in general do not exist in the Reals. Assuming that
theta is between +-Pi the command Eigenvalues(A) will give you
cos(theta)+-sqrt(1-cos(theta)^2)*I.

The imaginary part has to vanish: solve(1-cos(theta)^2=0, theta);

                                Pi, 0

However if you insist to work over the complex numbers it was
already said what has to done ...

Even if Maple does not provide a friendly answer: I guess you are
talking about the real case.

May be a basic geometric view on eigenvalues and eigenvectors helps:

Let f be linear with f(x) = lambda*x and lambda =/= 0, x =/= 0,
lambda in the field, x in the vector space. 

Then the line L = IR*x given by x is an invariant space under f
[i.e. f(L) = L as a set] and g = f restricted to the subspsace L
is simply a multiplication by lambda [i.e. g(y) = lambda*y for y
in L], just write it down by paper and pencil:

for y in L write y = alpha*x and you have f(y)= alpha*f(x) =
alpha*lambda*x = lambda*y, so it is in L.

So an eigenvector is not *one* vector, it *is* a 1-dim subspace
and for lambda=0 it is the whole kernel of f.

But a rotation in the plane always rotates any line and never leaves
them invariant - except the rotation is by 180° or 360°, i.e. your
theta is a multiple of Pi. Then your rotation equals +- identity.

More formally: you where already told that the eigenvalues are
complex, so in general do not exist in the Reals. Assuming that
theta is between +-Pi the command Eigenvalues(A) will give you
cos(theta)+-sqrt(1-cos(theta)^2)*I.

The imaginary part has to vanish: solve(1-cos(theta)^2=0, theta);

                                Pi, 0

However if you insist to work over the complex numbers it was
already said what has to done ...