Linear Transformation is Fredholm Operator iff Pseudoinverse exists
Theorem
Let $U, V$ be vector spaces.
Let $T: U \to V$ be a linear transformation.
Then $T$ is a Fredholm operator if and only if $T$ has a pseudoinverse.
Proof
Recall the definitions:
$S$ and $T$ are pseudoinverse to each other if and only if:
- $T \circ S - I_U$ is degenerate
and:
- $S \circ T - I_V$ is degenerate.
$T$ is a Fredholm operator if and only if:
- $(1): \quad \map \ker T$ is finite-dimensional
- $(2): \quad$ the quotient space $V / \Img T$ is finite-dimensional.
Sufficient Condition
Let $T$ have a pseudoinverse $S : V \to U$.
That is, both:
- $D_U := S \circ T - I_U$
and:
- $D_V := T \circ S - I_V$
are degenerate.
In particular:
- $\map \ker T \subseteq \map \ker {S \circ T} = \map \ker {D_U + I_U}$
and:
- $\Img {D_V + I_V} = \Img {T \circ S} \subseteq \Img T$
By Degenerate Linear Operator Plus Identity is Fredholm Operator:
- $\map \dim {\map \ker T} \le \map \dim {\map \ker {D_U + I_U} } < +\infty$
and:
- $\map \dim {V / \Img T} \le \map \dim { V / \Img {D_V + I_V} } < +\infty$
$\Box$
Necessary Condition
Let $T$ be a Fredholm operator.
By Existence of Complementary Subspace, there are subspaces $U' \subseteq U$ and $V' \subseteq V$ such that:
- $U = \map \ker T \oplus U'$
and:
- $V = \Img T \oplus V'$
Consider the projections on direct summands:
- $P : U \to \map \ker T$
and:
- $Q : V \to V'$
Observe:
- $\Img P = \map \ker T$
and:
- $\Img Q = V' \cong V / \Img T$
where the last since the isomorphism follow from First Isomorphism Theorem.
Thus by hypothesis, these are finite-dimensional.
In particular, $P$ and $Q$ are degenerate.
- $U' \cong U / \map \ker T \cong \Img T$
In particular:
- $T \restriction_{U'} : U' \to \Img T$
is an isomorphism.
Define $S : V \to U$ by:
- $S u = \begin{cases} \paren {T \restriction_{U'} }^{-1} u &: u \in \Img T \\ 0 &: u \in V' \end{cases}$
Then:
- $S \circ T = I_U - P$
- $T \circ S = I_V - Q$
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Sources
- 2002: Peter D. Lax: Functional Analysis: $2.2$: Index of a Linear Map