The linear map that sends $p \in \mathcal{P}_{N^2}(\mathbb{C})$ to $p(T) \in \mathcal{L}(V)$ (Ex. 17 Sec 5B in Axler)? [Preface: All references are to the 3rd Edition of LADR by Axler]
Exercise: The exercise asks to rewrite a proof for statement 5.21: "every operator on a finite-dimensional, nonzero, complex vector space has an eigenvalue", but "using the linear map that sends $p \in \mathcal{P}_{N^2}(\mathbb{C})$ to $p(T) \in \mathcal{L}(V)$."
Proof / Solution: If I understand the proof correctly, we use the fact that dim $\mathcal{P}_{N^2}(\mathbb{C}) = n^2+1 > $ dim $\mathcal{L}(V) = n^2$ to show that the function is not injective. Define $S \in \mathcal{L}(\mathcal{P}_{N^2}(\mathbb{C}) \to \mathcal{L}(V))$ by $S(p) = p(T)$. It follows that $S$ is not injective, meaning $\exists p\in \mathcal{P}_{N^2}(\mathbb{C}), p \neq 0$ such that $S(p) = p(T) = 0$. We know that $p(z) = c(z-\lambda_1)\dots(z-\lambda_m)$ for all $c, \lambda_1,\dots,\lambda_m \in \mathbb{C}$, so we rewrite $p(T) = c(T-\lambda_1)\ldots(T-\lambda_m) = 0$. At least one of these $(T-\lambda_j) = 0$ for $j = {1,\dots,m}$ so $\lambda_j$ is an eigenvalue.
Definition of $p(T)$ (if helpful): Suppose $T \in \mathcal{L}(V)$ and $p \in \mathcal{P}(F)$ is a polynomial given by $p(z) = a_0 + a_1z + a_2z^2 + \dots + a_mz^m$ for $z \in F$. Then $p(T)$ is the operator defined by $p(T) = a_0I + a_1T + a_2T^2 + \dots + a_mT^m$.
My Questions (from most to least important):

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*I'm having trouble wrapping my head around what this linear map $S$ actually looks like. I know $p(T) \in \mathcal{L}(V)$ can be represented as an $n \times n$ matrix, so can I think of transformation $S$ as taking all the coefficients of $\mathcal{P}_{N^2}(\mathbb{C})$ and placing them in this $n \times n$ matrix? And if so what order should I put them in and what do I do with the extra coefficient?

*Is this proof suggesting that $p(T)=0$ for all $v \in V$? This seems to contradict 4.7 from the book that states "If a polynomial is the zero function, then all coefficients are 0". Unless this only applies to $p(z)$ for $z \in F$ but not $p(T)$ for $T \in \mathcal{L}(V)$?

*In some questions, the concept of "building-your-own-function" comes up. Do I need to check that the function $S$ is i) well-defined and ii) linear? What other things should I check in proofs where I can arbitrarily "build-my-own-function"?

*[Optional] The previous exercise in the book is essential the same except it asks you to use "the linear map that sends $p \in \mathcal{P}(\mathbb{C})$ to $(p(T))v \in V$". Combining these two results, does this suggest that I can use any arbitrary higher-dimensional vector space to prove the result (so long as the map is linear)?

 A: Before I address your questions at the end, let me point some missing steps / errors in the proof you gave.

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*You'll want to rule out the case that $p$ is a non-zero constant polynomial, so that you can actually apply the fundamental theorem of algebra to factor $p$.

*You have that $$p(T) = c(T - \lambda_1 I) \cdots (T - \lambda_m I)$$ is identically zero on $V$. It does not follow that one of the $(T - \lambda_kI)$'s is identically zero. What you need to do is conclude that one of these merely fails to be injective, and show that this implies that $T$ has an eigenvalue.

To try and address your first three (non-optional) questions at the end:

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*If you want to be very explicit, you could write the map $S$ as $$S(a_0 + a_1z + \cdots + a_{n^2+1}z^{n^2+1})v = a_0v + a_1Tv + \cdots + a_{n^2+1}T^{n^2+1}v,$$ where $v \in V$. Since $S$ is a linear operator from an $(n^2+1)$-dimensional space to an $n^2$-dimensional space, its matrix (with respect to a choice of basis from the domain and from the codomain) will have the dimensions $n^2 \times (n^2+1)$. The space $P_{n^2}(\mathbb C)$ has a natural choice of basis, but $\mathcal L(V)$ might not, so, to avoid making arbitrary choices, I think it'd be best to view the map $S$ by just the above expression (or the one you gave).

*Yes, $p(T)$ will be identically zero on $V$. This doesn't contradict 4.7 since 4.7 applies to polynomials $p(z) \in P_{n^2}(\mathbb C)$, and $p(T) \in \mathcal L(V)$ is not a polynomial.

*In this case, it's clear that $S$ is well-defined and linear. If you have any doubts in your mind that it's linear, I think it's worth writing out a detailed proof once, just to make sure. As for well-definedness, you aren't making any arbitrary choices that depend on $p$ when you define $S(p)$, so there's nothing to worry about there.

A: The proof/solution you gave is not quite correct, and given your questions, some of the things you say in the proof, while correct, it appears you do not yet have a full grasp of yet. Here is the proof/solution redone.
Proof. Fix $T\in \mathcal L(V)$ and define a map $S\colon \mathcal P_{n^2}(\mathbb C)\to \mathcal L(V)$ by $S(p) = p(T)$. This map is well-defined and linear (in principle this requires proof, but let's say we agree).
Because $\dim \mathcal P_{n^2}(\mathbb C) = n^2 + 1 > n^2 = \dim \mathcal L(V)$, the rank-nullity theorem tells us that there is a non-zero polynomial $p\in \mathcal P_{n^2}(\mathbb C)$ such that $p(T)$ is the zero linear transformation. Also, by the fundamental theorem of algebra, we know that $p$ factorizes into a product of linear factors, so $p(T) = c(T-\lambda_1)\dots(T-\lambda_m)$ for some constants $c\ne 0,\lambda_1,\dots,\lambda_m$, where $m = \deg p$. The operator $c(T-\lambda_1)\dots(T-\lambda_m)$ is the zero map, so at least one of the factors, say $T-\lambda_1$ is not injective (it is not necessarily true that $T-\lambda_1 = 0$!). Therefore, $T$ has an eigenvector with eigenvalue $\lambda_1$. $\square$
Now for the questions you had.

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*The map $S$ literally takes a polynomial, such as $p(x) = x^2 - 1$, and sends it to the operator obtained by evaluating $p$ on $T$: $p(T) = T^2 - I$ ($I$ is the identity operator). I think this is the simplest and most useful way to understand the map $S$. It is possible in principle to express $S$ as an $n^2\times(n^2+1)$ matrix by choosing a basis for the domain and codomain, but it would not be as useful of a description.


*The rank-nullity theorem guarantees that $p$ is in the kernel of the map $S$. By the definition of $S$, and the definition of kernel, this means that $p(T) = 0$. If we use the example $p(x) = x^2-1$ again, this is saying $T^2 - I = 0$, as an operator, so $(T^2-I)v = 0$ for all $v\in V$. This is not a contradiction because we are not saying the polynomial $p(x)$ is $0$, just that the operator $p(T)$ is $0$.


*In principle, yes you need to prove that the function you give is well-defined and linear, though in practice, it is often obvious that the map you give is well-defined and linear. The time you really need to pay attention to when your map is well-defined is when you are defining a map whose domain is a quotient space, because we often want to define a map by prescribing what happens to an arbitrary member of an equivalence class, and we may get ourselves in trouble if we do not check that we get the same result when an equivalent element is used. For example, consider the map
$$T\colon \mathbb Z/12\mathbb Z\to \mathbb Z$$
defined by $T(n) = n+2$. Seems harmless enough, but this map is not well-defined because $n = n + 12$ in $\mathbb Z/12\mathbb Z$, so by my definition, $T(3) = 3$, and also $T(3) = 15$. A function must be singly-valued, so this is not a well-defined map.
