Stuck on this Algebra Problem

[AKG]

T be a linear operator (I think they mean "Let T be a linear...") on a finite-dimensional vector space V, and let W_1 be a T-invariant subspace of V. Let x \in V such that x \notin W_1. Prove the following results: There exists a unique monic polynomial g_1(t) of least positive degree such that g_1(T)(x) \in W_1.
If h(t) is a polynomial for which h(T)(x) \in W_1, then g_1(t) divides h(t).
g_1(t) divides the minimal and characteristic polynomials of T.
Let W_2 be a T-invariant subspace of V such that W_2 \subseteq W_1, and g_2(t) be the unique monic polynomial of least degree such that g_2(T)(x) \in W_2. Then g_2(t) divides g_2(t).
If \beta is a basis of V, and \beta _{W_1} is a basis of W_1 such that \beta _{W_1} \subseteq \beta, then define W_1 \prime = Span(\beta - \beta _{W_1}). V = W_1 \oplus W_1 \prime. \forall v \in V, \exists w_1 \in W_1, w_1 \prime \in W_1 \prime such that v = w_1 + w_1 \prime.

g_1(T)(x) = g_1(T)(w_1 + w_1 \prime) = g_1(T)(w_1) + g_1(T)(w_1 \prime)

Now, if the restriction of T to W_1 \prime were an operator on W_1 \prime, then there would be a unique monic polynomial of least degree such that g_1(T)(w_1 \prime) = 0, namely the minimal polynomial of T restricted to W_1 \prime. Then, if g_1(t) is a polynomial over the same field that underlies W_1, I can assert that g_1(T)(w_1) \in W_1, and thus the result is proved. Can I prove these "if"s? If not, is there another way to prove the result? I haven't looked at the rest of it yet.

[AKG]

Well, assuming that it's true that W_1' is T-invariant, more can be proven:

(b) If h(T)(x) \in W_1, h(T)(w_1') = 0\ \forall w_1' \in W_1', so if T_{W_1'} is the restriction of T to W_1', then h(T_{W_1'}) = T_0. My book proves that if g_1(t) is the minimal polynomial of a linear operator T_{W_1'} on a finite-dimensional vector space W_1', then for any polynomial h(t), if h(T_{W_1'}) = T_0, g_1(t) divides h(t). Of course, insted of "W_1'", my book proves it for "V", and instead of the minimal polynomial being "g_1(t)" it calls it "p(t)" in the proof, but the result still holds.

(c) It suffices to show that g_1(t) divides the minimal polynomial of T, since the minimal polynomial of T already divides the characteristic polynomial of T (a corollary of the theorem from my book that I just mentioned). Let p(t) be the minimal polynomial of T. Then,

\forall v \in V,\ v = w_1 + w_1',\ w_1 \in W_1,\ w_1' \in W_1':

p(T)(v) = p(T)(w_1) + p(T)(w_1') = 0

Since W_1 and (supposedly) W_1' are T-invariant, and since (supposedly) p(t) takes coefficients from the same field that underlies W_1 and W_1', we have the above equality holding if and only if:

p(T)(w_1) = 0,\ p(T)(w_1') = 0

Since this must be true for all w_1', p(T_{W_1'}) = T_0, so, from (b), g_1(t) | p(t).

(d) Define W_2' in a way similar to how W_1' was defined. Let g_2(t) be the minimal polynomial of T restricted to W_2'. Then it is the unique polynomial which satisfies the hypotheses, and since W_2 \subseteq W_1, g_1(t) | g_2(t) by (b).

All of this is pretty simple, but it all rests on two (huge) assumptions, mainly the assumption that if W_1 is T-invariant, then so is W_1'.

[AKG]

I think I'm on to something I can use, but too tired to figure out exactly what to do with it. Anyways, define S to be the projection of V on W_1, and U to be the projection of V on W_1 \prime. If we define g_1(t) to be the minimal polynomial of UT, I think we'll have something.

T = ST + UT
g_1(T) = g_1(ST + UT)
g_1(T) = a_n(ST + UT)^n + \dots a_0I

Now if any power of ST is composed with UT, we will get zero since SU = 0, so:

g_1(T)= g_1(ST) + g_1(UT) = g_1(ST).

Now UT really is an operator on the vector space W_1 \prime, that is, W_1 \prime really is UT-invariant. I believe the rest should hold with a few adjustments. Now, is it problematic that I assume that p(t) takes coefficients from the same field as the one that underlies the subspaces?

[shmoe]

Can I prove these "if"s? If not, is there another way to prove the result? I haven't looked at the rest of it yet.[/list]

No you can't. V any vector space, W_1 any non-trivial subspace, T the projection map on W_1. You can't pick a W_2 in the manner you describe to be T-invariant.

Think quotient space.