Why does $(a_0I +a_1 T + \dots + a_n T^n ) v = c(T - r_1 I) \dots (T - r_m I) v$ imply that $T - r_j I$ is not injective?

Question

I was reading down with determinants and it had the equation:

$$(a_0I +a_1 T + \dots + a_n T^n ) v = c(T - r_1 I) \dots (T - r_m I) v = 0$$

from which it it concluded that $T - r_j I$ is not injective i.e. non-injective meaning that it does NOT preserve distinctness in the mapping i.e. there is some collision in the mapping. I have a couple questions:

1. The first issue that I am having is that I don't even know what things we are mapping and what is the mapper. i.e. what are the sets we are relating, what is the non-injection over what what does $T,r_i, a_i, c, v$ all have to do with this injection?
2. Also, probably related to my confusion, what theorem is he using to make a relation of the polynomial $a_0 + a_1 z + \dots a_n z^n = c(z - r_1) \dots (x - r_m)$ to linear operators. Why is that equation applicable to doing $(a_0I +a_1 T + \dots + a_n T^n ) v = c(T - r_1 I) \dots (T - r_m I) v$ and what is its relation to this non-injectivness?

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For the sake making the answer self contained here is the theorem I am trying to understand:

Theorem 2.1 Every linear operator on a finite-dimensional complex vector space has an eigenvalue (Note: he defines one page before an eigenvalue as: A complex number $λ$ is called an eigenvalue of $T$ if $T −λI$ is not injective ).

Proof. To show that $T$ (our linear operator on V ) has an eigenvalue, fix any non- zero vector $v \in V$ . The vectors $v, T v, T^2v, \dots , T^nv$ cannot be linearly independent, because V has dimension $n$ and we have $n + 1$ vectors. Thus there exist complex numbers $a0,...,an$, not all $0$, such that $$a_0 v + a_1 T v + · · · + a_n T^n v = 0 .$$ Make the a’s the coefficients of a polynomial, which can be written in factored form as $$a_0 +a_1z+···+a_nz^n =c(z−r1)...(z−rm),$$ where $c$ is a non-zero complex number, each $r_j$ is complex, and the equation holds for all complex z. We then have $$ 0 = (a_0I +a_1T +···+a_nT^n)v = c(T −r_1I)...(T −r_mI)v ,$$ which means that $T − r_jI$ is not injective for at least one $j$. In other words, $T$ has an eigenvalue.

Answer 1

We're given

$$\left(a_0I+a_iT+\ldots+a_nT^n\right)v=c\left(T-r_1I\right)\cdot\ldots\cdot(T-r_mI)v$$

... for a non-zero vector $\;v\in V\;$ . Try to see it the following way: suppose $\;A_1,...,A_m\;$ are operators/square matrices s.t for a vector $\;0\neq v\in V\;$ we have

$$A_1\cdot\ldots\cdot A_m v=0\implies A_1\cdot\ldots\cdot A_m\;\;\text{ is a singular operator/matrix}\implies$$

at least one of $\;A_i\;$ is singular, as we know that the product of regular operators/matrices is again regular.

In our case, we get that at least one of $\;T-r_iI\;$ is singular $\;\iff r_i\;$ is an eigenvalue of $\;T\;$ , and we're done, and in finite dimension singular $\;\iff\;$ non-injective.