Using induction (specifically not Cayley-Hamilton), prove the existence of a nonzero polynomial $p$ such that $p(T) = 0$ for $T \in \mathcal{L}(V)$ This question aims to prove that there exists a nonzero polynomial $p \in \mathcal{P}_m(\mathbb{F})$ such that $p(T) = 0$, for $T \in \mathcal{L}(V)$ ($\dim V = m$), without using the Cayley-Hamilton theorem or matrices. There's multiple parts, but it essentially proceeds by induction; I will try to be brief in where I am stuck and what I have tried in my attempts. 
Question premise: We first begin with a base case, where $m = 0$, and a polynomial such as $p(x) = 1$ works since $p(T) = $Identiy operator, which is zero on the trivial vector space. By strong induction, we can assume the conclusion is true for dimensions $0, 1, ..., m-1$. Now, the question proceeds as such: we pick a nonzero $v \in V$ and let $n$ be smallest positive integer in which $v, Tv,  ..., T^nv$ is linearly dependent. Then, we let $W := span(v, Tv, ..., T^{n-1}v)$ with dimension $n$. 
Part (a) asks us to show $W$ is an invariant subspace for $T$. 
Part (b) sets up $T_w \in \mathcal{L}(W)$ which is restriction of $T$ to $W$, such that $T_w(w) := Tw$. It wants us to show there exists a nonzero $q \in \mathcal{P}_n(\mathbb{F})$ for which $q(T_w) = 0$.
Part (c) sets up $T_{V/W} \in \mathcal{L}(V/W)$ defined as $T_{V/W}(v+W) = Tv + W$. It then asks us why a nonzero $h\in \mathcal{P}_{m-n}(\mathbb{F})$ exists such that $h(T_{V/W}) = 0$.
Part (d) brings it all together and defines a $p = qh$ from the previous parts, with nonzero $p \in \mathcal{P}_m(\mathbb{F})$ . It asks for us to show $p(T) = 0$ to complete the inductive step of the entire proof.
My attempt at a solution:
a) (Omitted, I have solved this part with a high degree of certainty and let us assume that it is true, i.e. $W$ is an invariant subspace for $T$)
b) We first note that $v, ..., T^{n-1}v$ is a basis for $W$, as is $v, T_wv, ..., T^{n-1}_wv$ since $v \in W$ and $T_w(v) = Tv$ for $v \in W$. I will use a construction that was hinted that I should use from a previous problem I did: construct $T^n_wv + a_{n-1}T^{n-1}_wv + ... + a_0v = 0$ for unique $a_{n-1}, ..., a_0 \in \mathbb{F}$ (since $v, ..., T^{n-1}_w$ is linearly independent). Now, let $\mu_{T_w,v} (x) := x^n + a_{n-1}x^{n-1} + ... + a_0$ so $\mu_{T_w,v}(T_w)v = 0$. Similarly, we can create this for all basis vectors {$v, T_wv, ..., T^{n-1}_wv$}: $$ T^n_wT^j_wv + a_{n-1}T^{n-1}_wT^j_wv + ... + a_0T^j_wv = 0$$ for all $j = 0, ..., n-1$. This is such that $$\mu_{T_w, T_w^jv}(T_w)T^j_wv = 0$$ for all $j = 0, ..., n-1$. We can create a polynomial $q(x)$ that is the least common multiple of $\mu_{T_w, T_w^jv}$ for all $j = 0, ..., n-1$. Thus, $q(x)$ is divisible by $\mu_{T_w, T_w^jv}(x)$ for all $j = 0, ..., n-1$ and that means, from $q(x) = h(x)\mu_{T_w, T_w^jv}(x) + 0$ for some $h(x)$, we can say $q(T_w)T^j_wv = 0$ for all $j = 0, ..., n-1$. 
Any $w \in W$ can be written as a linear combination of basis vectors, so $$q(T_w)w = q(T_w)(c_1v + ... + c_nT^{n-1}_wv)$$ for some constants $c_1, ..., c_n \in \mathbb{F}$. This turns into $c_1q(T_w)v + ... + c_nq(T_w)T^{n-1}_wv$, but each $q(T_w)T^j_wv = 0$ for all $j = 0, ..., n-1$ so $q(T_w)w = 0$ for all $w \in W$, and so $q(T_w) = 0$, as desired.
My issue: If this is a valid approach at all, how can I show that $q \in \mathcal{P}_n(\mathbb{F})$? (i.e. How should I be constructing $q$ so that it has degree $n$?)
c) For this one I am a bit stuck; I was thinking of using the inductive hypothesis that there exists a $p \in \mathcal{P}_{m-n}(\mathbb{F})$ such that $p(T) = 0$, but I'm not sure how to make this true of the new operator $T_{V/W}$.
d) I figured part (d) would follow from (b) and (c), but just thinking about it, I'm not sure how to figure out how $p(T) = q(T)h(T) = 0$ given that $q(T_w) = 0$ only for $T_w$ and $h(T_{V/W}) = 0$ only for $T_{V/W}$. Any sort of hint would be appreciated for this one, or perhaps I'll have a better idea once I work through the answers to (b) and (c).
In summary, my questions:


*

*For part (b), if my approach is valid, how can I ensure $q$ has degree $n$?

*For part (c), how do I prove the existence of $h(T_{V/W}) = 0$ and is using the inductive hypothesis along the right track (I feel like I never really used it)?

*Any hints for part (d), particularly how to interpret $q(T)$ and $h(T)$?

 A: For (b), you don't need to construct $\mu_{T_w,T^j_w v}$ for each $j$ because $\mu_{T_w,v}$ already works. Let's simplify our notation and write $\mu$ instead of $\mu_{T_w,v}$. Notice that $\mu(T_w)T^j_w v=(x^j\mu)(T_w)v=[T_w^j\circ\mu(T_w)]v=T^j_w0=0$, for any $j$. This implies that $\mu(T_w)$ is the zero map.
Part (c) follows from the induction hypothesis because the dimension of $W$ is at least $1$, so the dimension of $V/W$ is at most $m-1$, so the hypothesis applies. I think you have got the wrong idea about the question and got confused. In the question, the operator $T$ is not fixed, instead the question asks you to show that "Let $V$ be an $m$ dimensional vector space, then for any linear map from $V$ to itself, there exists some polynomial $p$ of degree $m$ so that $p(T)=0$." In other words, the statement holds for any vector spaces and any linear map on it.
For part (d), you could try to show that $q(T)h(T)$ is the zero map by first proving that $q(T)h(T)$ is zero on $W$ (which follows from part (b) directly). Then the hint is to think about what $h(T_{V/W})=0$ means, for example what does the image of $h(T)$ lie inside? One last remark: $q(T)h(T)$ shall be understood as a composition of maps $q(T)\circ h(T)$, and by properties of polynomials this is commutative $(q\cdot h)(T)=q(T)\circ h(T)=h(T)\circ q(T)$.
