Polynomial with operator as variable When I was reading my lectures of linear algebra I ran into the following question: Let $\lambda _1,\lambda_2\in \mathbb{C}$ such that $\lambda_1\neq \lambda_2$ and consider two polynomials $(t-\lambda_1)^m$ and $(t-\lambda_2)^m$. It is easy to see that they are coprime and it means that one  can find polynomials $p_1(t),p_2(t)\in \mathbb{C}[t]$ such that $p_1(t)(t-\lambda_1)^m+p_2(t)(t-\lambda_2)^m=1$. This is obvious to me but the following is not so clear.
Since $f:V\to V$ an operator and $V$ is vector space over $\mathbb{C}$ then we can plug in operator $f$ in the above polynomial and we get: $$p_1(f)(f-\lambda_1\cdot\text{id})^m+p_2(f)(f-\lambda_2\cdot \text{id})^m=\text{id}.$$
I was trying to understand it but I failed.
More precisely my question is this: if we have polynomial identity and if we plug in operator why do we still have identity?
Can anyone explain it to me in simpler way, please? Would be grateful for your help!
 A: Here is an attempt to explain this without appealing to any existing terminology or definitions.
First, let's say what we mean by $p(f)$ for a polynomial $p$.  If $p(t) = \sum_{k=0}^d a_k t^k$, then we define $p(f) = \sum_{k=0}^d a_k f^k$, where we define
$$
f^0 = \operatorname{id}, \quad f^k = f \circ f^{k-1} \quad k \geq 1.
$$
First, some strange notation: for a polynomial $p(t)$, I will write $[p(t)]_f$ to signify $p(f)$.  With that established, it suffices to make the following observation.

Claim: Take $p,q \in \Bbb C[t]$.  It holds that 
  $$
[p(t) + q(t)]_f = [p(t)]_f + [q(t)]_f = p(f) + q(f), \\
[p(t)q(t)]_f = [p(t)]_f[q(t)]_f = p(f)q(f).
$$

You might find proving this to be an informative exercise. Now, you have stated that we have
$$
p_1(t)(t-\lambda_1)^m+p_2(t)(t-\lambda_2)^m=1.
$$
It follows that 
\begin{align}
p_1(f)(f-\lambda_1 \operatorname{id})^m+p_2(f)(f-\lambda_2 \operatorname{id})^m &= 
[p_1(t)]_f{[(t-\lambda_1)]_f}^m+[p_2(t)]_f{[(t-\lambda_2)]_f}^m
\\ & = 
[p_1(t)]_f[(t-\lambda_1)^m]_f+[p_2(t)]_f[(t-\lambda_2)^m]_f
\\&= 
[p_1(t) (t-\lambda_1)^m]_f+[p_2(t) (t-\lambda_2)^m]_f
\\&= 
[p_1(t) (t-\lambda_1)^m + p_2(t) (t-\lambda_2)^m]_f = [1]_f = \operatorname{id}.
\end{align}
A: Consider the set $\operatorname{End}(V)$ of all endomorphisms of $V$ (i.e. $\Bbb C$-linear maps from $V$ to $V$).
We can define addition and multiplication on this set: given two endomorphisms $f, g$, we define $f + g$ as the endomorphism sending $v$ to $f(v) + g(v)$, and define $f \cdot g$ as the endomorphism sending $v$ to $f(g(v))$.
Furthermore, there are two special endomorphism: the "zero" endomorphism, denoted by $0$, which sends every vector $v$ to the zero vector; the "identity" endomorphism, denoted by $id$, which sends every vector $v$ to itself.
It turns out that $(\operatorname{End}(V), +, \cdot, 0, 1)$ is a (non-commutative) ring. We abbreviate the notation and simply call this ring $\operatorname{End}(V)$.
In fact, since $V$ is a $\Bbb C$-vector space, the ring $\operatorname{End}(V)$ has an extra structure: it contains $\Bbb C$ as a subring, if we identify any complex number $\lambda$ with the endomorphism sending $v$ to $\lambda v$.
In other words, $\operatorname{End}(V)$ is a $\Bbb C$-algebra.

What happens in your question is that we have a ring homomorphism $\phi$ from the ring of polynomials $\Bbb C[t]$ to the ring $\operatorname{End}(V)$, which sends $t$ to $f$. It is also a $\Bbb C$-algebra homomorphism, which means that for any $\lambda\in\Bbb C$, the image of $\lambda \in \Bbb C[t]$ is just $\lambda\in \operatorname{End}(V)$.
It then turns out that any polynomial $p(t)$ is sent to $p(f)$, and any identity in $\Bbb C[t]$ translates to an identity in $\operatorname{End}(V)$.
