Prove operator has eigenvalue with fundamental theorem of algebra

Question

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?

Answer 1

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.

Answer 2

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.