Why does polynomial factorization generalize to matrices

I'm reading about linear algebra and I came across with the following theorem where I have a problem convincing myself:

Theorem 2.1 $\,$ Every linear operator on a finite-dimensional complex vector space has an eigenvalue.

Proof:

To show that $T$ (our linear operator on $V$) has an eigenvalue, fix any nonzero vector $v \in V$. The vectors $v, Tv, T^2v,..., T^nv$ cannot be linearly independent, because $V$ has dimension $n$ and we have $n + 1$ vectors. Thus there exist complex numbers $a_0,...,a_n$, not all $0$, such that

$$a_0v + a_1Tv + ··· + a_nT^nv = 0.$$ Make the $a$’s the coefficients of a polynomial, which can be written in factored form as $$a_0 + a_1z + ··· + a_nz^n = c(z − r_1)\cdots(z − r_m),$$ where $c$ is a non-zero complex number, each $r_j$ is complex, and the equation holds for all complex $z$. We then have

$${\color{red}{ 0=(a_0I + a_1T + ··· + a_nT^n)v= c(T − r_1I)\cdots(T − r_mI)v}},$$ which means that $T − r_j$ I is not injective for at least one $j$. In other words, $T$ has an eigenvalue.$\;\blacksquare$

I'm having trouble with the factorized form of the matrix polynomial (in red). I understood that the factorization holds for a polynomial by the fundamental theorem of algebra but why does this also hold for matrices?

In other words, why is the the part I highlighted true? Does such an factorization always exist?

Could I have some help to see this? Thank you =)

P.S. here is my reference (page 3).

UPDATE:

Someone else also has asked the same question before it seems.

What matters is that the matrices involved, namely powers of $T$, commute with each other. With that in mind, the legitimacy of the factorisation should be clear: just think about expanding the brackets using the associative and distributive properties of matrix multiplication. A more sophisticated argument can be obtained by viewing the equation in terms of an action of the polynomial ring $\mathbb{C}[x]$, in which factorisation is more familiar.

This follows simply from the universal property of polynomial rings, which implies that any polynomial equation in $\, R[x]\,$ will persist when "evaluated" into any ring where the images of the constants commute with the image of $x$ (which is precisely the condition necessary for such a map to be a ring homomorphism).

Indeed, a polynomial ring is designed precisely to have this property, i.e. it is the most general ("free") ring that contains $\,R\,$ and a new element $\,x\,$ that commutes with all elements of $R$. Because we use only the ring axioms and constant commutativity when proving polynomial equations, such proofs persist in said ring images where constant commutativity persists.

This is true in your example because constants $\,r\,$ map to a constant matrices $\,rI\,$ which commute with $\,T = $ image of $x$.

This implies that all familiar polynomial equations (e.g. Binomial Theorem and difference of squares factorization) persist to be true when evaluated into any ring where the constants commute with the indeterminates. Ditto for many other ubiquitous polynomial equations, e.g. cyclotomic polynomial factorizations, polynomial Bezout identities for the gcd, resultants, etc. Therefore such equations represent universal laws (identities), modulo said constant commutativity.

These ideas are brought to the fore when one studies $(R-)$algebras, which are rings containing a central image of $R$, i.e. where the images of elements of $R$ commute with everything. Any polynomial equation that holds true in $\,R[x_1,\ldots,x_n]\,$ will persist to be true when evaluated into any $R$-algebra, i.e. it is an identity (law) of $R$ algebras. In fact it is easy to show that an equation holds true in $\,R[x_1,\ldots,x_n]\,$ iff it is true in all all $R$ algebras. Hence the equations that hold true $\,R[x_1,\ldots,x_n]\,$ are precisely the identities (universal laws) of $R$-algebras.