Minimal polynomial and Fredholm operator.

Question

Let $X$ be a Banach space and let $m_{T}$ be the minimal polynomial of $T \in\mathcal{L}(X)$. We have $$\sigma_{e}(T)=\{\lambda \in \mathbb{C}, \lambda-T \not \in \Phi(X)\}.$$ Where, $\Phi(X)$ is the set of Fredholm operators on $X$. Do we have this inclusion : $$\sigma_{e}(T)\subset \{\lambda \in \mathbb{C}: m_{T}(\lambda)=0 \}.$$

Answer 1

The answer is affirmative, but please see the comments below. To prove it let us first show the following result.

Lemma. Let $A$ be a unital complex algebra and let $a\in A$. Suppose that $p$ is a polynomial in $\mathbb C[X]$ such that $p(a)=0$. Then the spectrum of $A$, namely $$ \sigma (a) = \{\lambda \in \mathbb C: a-\lambda \text{ is not invertible}\}, $$ is contained in the set of roots of $p$.

Proof. Given $\lambda $ in $\sigma (a)$, suppose by contradiction that $p(\lambda )\neq 0$. Observig that the polynomial $p(\lambda )-p(z)$ vanishes on $\lambda $, we may write $$ p(\lambda )-p(z) = (z-\lambda )q(z), $$ for some polynomial $q$. Substituting $z$ for $a$ we get $$ p(\lambda ) = (a-\lambda )q(a), $$ and hence also $$ 1 = (a-\lambda )\frac{q(a)}{p(\lambda )}, $$ which implies that $a-\lambda $ is invertible, contradicting the fact that $\lambda $ is in $\sigma (a)$. QED

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The next question is how to define the minimal polynomial of an operator $T$ acting on an infinite dimensional Banach space. Mimicking the finite matrix definition I'd say the most natural one is to let $m_T$ be the generator of the ideal formed by all polynomials $p$ in $\mathbb C[X]$ such that $p(T)=0$.

Of course the main trouble is that it is impossible to define the characteristic polynomial in infinite dimensions and hence there is no chance of even stating the Cayley-Hamilton Theorem. In other words, more often than not, only the zero polynomial will vanish on $T$, in which case $m_T=0$. If so, the question of whether or not $$ \sigma_{e}(T)\subseteq \{\lambda \in \mathbb{C}: m_{T}(\lambda)=0 \} $$ has an uninteresting, trivial affirmative answer.

In the rare instances that some nonzero polynomial vanishes on $T$, meaning that $m_T\neq 0$, then $m_T$ also vanishes on the image of $T$ under the quotient map $$ \pi:\mathcal L(H) \to \mathcal L(H)/ \mathcal K(H), $$ where $\mathcal K(H)$ is the ideal of compact operators.

By the lemma we then have that $$ \sigma(\pi (T))\subseteq \{\lambda \in \mathbb{C}: m_{T}(\lambda)=0 \}. $$ Combining this with the fact that $\sigma_e(T) =\sigma(\pi (T))$, we have arrived at an affirmative answer.