Degenerate Linear Operator Plus Identity is Fredholm Operator

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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.


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

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
\(\ds \) \(<\) \(\ds + \infty\) Definition of Degenerate Linear Transformation

$\Box$


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

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
\(\ds \) \(<\) \(\ds + \infty\) Definition of Degenerate Linear Transformation

$\blacksquare$


Sources

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