# Stuck on this Algebra Problem

1. Nov 1, 2004

### 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.

Last edited: Nov 1, 2004
2. Nov 1, 2004

### 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 $W_1'$.

Last edited: Nov 1, 2004
3. Nov 1, 2004

### 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?

4. Nov 1, 2004

### shmoe

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.