Degenerate Linear Operator Plus Identity is Fredholm Operator

Theorem

Let $U$ be a vector space.

Let $T : U \to U$ be a degenerate linear operator.

Let $I_U : U \to U$ be the identity operator.

Then:

$T + I_U$

is a Fredholm operator.

Proof

We need to show that both:

$\map \ker {T + I_U}$

and:

$U / {\Img {T + I_U} }$

are finite-dimensional.

Recall that:

$(1):\quad \map \dim {\Img T} < +\infty$

since $T$ is degenerate.

This article needs to be tidied. In particular: Make proper headings so they look like headings Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code.

$\map \dim {\map \ker {T + I_U} } < +\infty$

Let $u \in \map \ker {T + I_U}$.

That is:

$Tu + u = 0$

Thus:

$u = \map T {-u} \in \Img T$

Hence:

$\map \ker {T + I_U} \subseteq \Img T$

Therefore:

\(\ds \map \dim {\map \ker {T + I_U} }\)

\(\le\)

\(\ds \map \dim {\Img T}\)

Dimension of Proper Subspace is Less Than its Superspace

\(\ds \)

\(<\)

\(\ds + \infty\)

by $(1)$

$\Box$

$\map \dim {U / {\Img {T + I_U} } } < +\infty$

Observe that:

$\map \ker T \subseteq \Img {T + I_U}$

since if $u \in \map \ker T$, then:

$u = T u + u \in \Img {T + I_U}$

Thus:

\(\ds \map \dim {U / {\Img {T + I_U} } }\)

\(=\)

\(\ds \map {\mathrm {codim} } {\Img {T + I_U} }\)

Definition of Codimension of Vector Subspace

\(\ds \)

\(\le\)

\(\ds \map {\mathrm {codim} } {\map \ker T}\)

Codimension of Proper Subspace is Greater

\(\ds \)

\(=\)

\(\ds \map \dim {U / \map \ker T}\)

Definition of Codimension of Vector Subspace

\(\ds \)

\(=\)

\(\ds \map \dim {\Img T}\)

$U / \map \ker T \cong \Img T$ by First Isomorphism Theorem

\(\ds \)

\(<\)

\(\ds + \infty\)

by $(1)$

$\blacksquare$

Sources

• 2002: Peter D. Lax: Functional Analysis: Chapter $27$: Index Theory