3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You clarified my misunderstanding—that I need not concern myself with what the “roots” of the operator polynomial would be; all that is needed is the decomposition. $\endgroup$ – rorty Aug 29 '19 at 1:38
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.