# Prove operator has eigenvalue with fundamental theorem of algebra

Every operator on a finite-dimensional, nonzero, complex vector space has an eigenvalue.

Axler proves this as follows. Suppose $$T: V\rightarrow V$$ with $$\text{dim}(V) = n > 0$$. Choose $$v \in V$$ with $$v \neq 0$$. Then $$v, Tv, T^2v,\dots,T^nv$$ is not linearly independent because $$T$$ has dimension $$n$$ and there are $$n+1$$ vectors. Then there are $$a_0,\dots,a_n$$ not all $$0$$ such that $$0=a_0v+a_1Tv+\dots+a_nT^nv$$.

By the fundamental theorem of algebra, the polynomial $$a_0 + a_1z+\dots+a_nz^n = c(z-\lambda_1)\cdots(z-\lambda_m),$$ where $$m$$ is not necessarily equal to $$n$$ because $$a_n$$ could be zero.

Axler then uses this fact to say we have $$0=c(T-\lambda_1 I)\cdots(T-\lambda_m I)v$$, where $$T-\lambda_j$$ is not injective for at least one $$j$$.

Why is it that we can apply factorization/fundamental theorem of algebra to a polynomial of operators?

You know that $$\sum_k a_k T^k v=0$$ for some $$a_k$$. Define the polynomial $$p(z)\equiv\sum_k a_k z^k$$. You know that there are $$\lambda_k$$ and $$c$$ such that $$p(z)=c\prod_k (z-\lambda_k).$$ This also means that $$p(T)=c\prod_k (T-\lambda_k I)$$, by definition of $$p(T)$$.
I think what's important to note here is that you don't need $$z$$ to be a "number" for this decomposition to hold. If $$z$$ is, say, in $$\mathbb C$$, then you can think of the $$\lambda_k\in\mathbb C$$ as roots of the polynomial. If $$z=T$$ is an operator, then the $$\lambda_k$$ are not the roots of the polynomial anymore, in the sense that you don't have $$p(T)=0$$. However, you can still decompose the polynomial in the same way.
Our starting point, $$\sum_k a_k T^k v=0$$, now implies that $$p(T)v\equiv\sum_k a_k T^k v=0$$, that is, $$c\prod_k (T-\lambda_k)v=0$$.
The only way for this to happen is that $$(T-\lambda_j)$$ has a non-empty kernel for some $$j$$, which is equivalent to it being not injective.
In genearl we can't because, in general, operators do not commut. But here the omly operators ar $$\operatorname{Id},T^2,\ldots,T^n$$, which do commute. Now if, for instance, you know that$$x^2-3x+2=(x-1)(x-2),$$you can deduce that$$T^2-3T+2\operatorname{Id}=(T-\operatorname{Id})(T-2\operatorname{Id}),$$since\begin{align}(T-\operatorname{Id})(T-2\operatorname{Id})&=T^2-T-2T+2(T-\operatorname{Id})(T-2\operatorname{Id})\\&=T^2-3T+2\operatorname{Id}.\end{align}And the same argument applies to any polynomial decomposition.