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I've read through several posts regarding this same proof in the same book (theorem $5.21,$ pg $145,$ Axler, $3$rd), but none of them addressed my particular issue.
For reference (This is the same proof but from an earlier edition)

I'm having trouble understanding the reasoning behind the very first step, i.e.
...$(v,Tv,\dots,T^nv)$...
Why are we starting with a list of powers of T? I understand the flow of the proof after this, but this first step just seems completely arbitrary to me.
Again, the theorem states: Every operator on a finite-dimensional, nonzero, complex vector space has an eigenvalue. ...and I just can't see where a list of powers of T, or even polynomials in general, fit into that.
I'm sorry this question is so vague, but I'm finding it difficult to express precisely in words. Hopefully this is just some ridiculously simple thing that I'm either overlooking or overthinking. I appreciate the help!

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  • $\begingroup$ Welcome to Maths SX! The reason behind this is probably (I don't have the book) that he considers the vector space as a finitely generated module over the ring of polynomials $\mathbf C[X]$, which operates on the vector space through $\; X\cdot v \stackrel{\text{def}}{=} Tv$, in order to use that the ring of polynomials over a field is a P.I.D. and its irreducible elements are known by D'Alembert-Gauß. $\endgroup$ – Bernard Jan 2 at 0:21
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    $\begingroup$ There are many connected ideas that the author of the proof already knows, even if you don't know them yet, that inspire looking at powers of $T$. (1) The fundamental theorem of algebra, that polynomials over $\mathbb{C}$ factor as a product of linear polynomials. (2) That to find values $r$ such that $(T-r)v=0$ for some non-zero $v$ it is necessary and sufficient that $\det(T-r)=0$. (3) that $\det(T-r)$ is a polynomial in $r$. (4) that polynomials are linear combinations of powers. (5) a method for finding eigenvalues is via Krylov subspaces $\endgroup$ – MoonLightSyzygy Jan 2 at 0:40
  • $\begingroup$ The latter being exactly a method for finding eigenvalues based on this proof, that consists in looking at linear dependencies of iterated applications of $T$ onto a non-zero vector. $\endgroup$ – MoonLightSyzygy Jan 2 at 0:44
  • $\begingroup$ he refers directly to the Fundamental Theorem, which is 4.13 on page 124, and factoring polynomials over the complexes, 4.14 on page 125. These are big theorems. The cyclic vector idea is used just to find a way to include them. $\endgroup$ – Will Jagy Jan 2 at 0:50
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    $\begingroup$ Meanwhile, the cyclic vector idea is fundamental for the Jordan Normal form, which he does, still without saying the word determinant. $\endgroup$ – Will Jagy Jan 2 at 1:00
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The first "trick" is that $\{v,Tv,\ldots, T^n v\}$ is a set of $n+1$ elements in a space of dimension $n$ and thus must be linearly dependendent. So there exist complex numbers $a_0,a_1,\ldots,a_n$ not all zero such that $$ a_0v + a_1Tv + \cdots + a_nT^nv = 0. $$ Write the $a_i$ as coefficients of a polynomial, i.e. $$ p(z) = a_0+a_1z+\cdots+a_nz^n, $$ with $0<m\leqslant n$ the largest index such that $a_m\ne 0$. Then we can factor $$ p(z) = a_0+a_1z+\cdots+a_mz^m = c(z-\lambda_1)\cdots(z-\lambda_m). $$ The second "trick" is that this polynomial also holds true for $z=T$ and $\lambda_i = \lambda_i I$. Then as $p(z)=0$ for all $z\in\mathbb C$ we have $$ 0 = p(T)v = c(T-\lambda_1I)\cdots(T-\lambda_mI)v, $$ which means that for some $\lambda_i$ we have $v\in\ker(T-\lambda_i)$, which means that $T-\lambda_i$ is not injective. In other words, $\lambda_i$ is an eigenvalue for $T$.

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    $\begingroup$ OK, so correct me if I'm wrong: He is basically taking advantage of the fact that you can express the 0 operator as a non-trivial linear combo of the powers of T from 0 to n, in order to obtain a polynomial, to which the fundamental theorem of algebra can be applied, and then showing that at least one of the subsequent roots is an eigenvalue? $\endgroup$ – David K Jan 2 at 1:35
  • $\begingroup$ That is the gist of the proof, yes. $\endgroup$ – Math1000 Jan 2 at 1:40
  • $\begingroup$ Thank you very much for your help. $\endgroup$ – David K Jan 2 at 1:41

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