$X,Y$ are Banach spaces and $A\in B(X,Y)$ is a Fredholm operator (that is, the dimensions of ker($A$) and coker($A$) are both finite), then are closed linear subspaces ker($A$) and Im($A$) complemented? (A closed linear subspace $H$ in a Banach space $Z$ is called complemented iff there is a closed linear subspace $G$ such that $H+G=Z$ and $H \cap G=0$)

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I think you mean complemented rather than complementary. – Jonas Meyer May 18 '12 at 16:06
They are both complemented: $\ker(A)$ is finite-dimensional, hence complemented, and $\text{im}(A)$ is finite-codimensional, so any algebraic complement is finite-dimensonal, hence closed. – Scott LaLonde May 18 '12 at 17:17
@Scott L:why is finite-dimensional subspace complemented? For example we can consider the kernel of an unbounded non-zero linear functional and its algebraic complement. – Hezudao May 19 '12 at 1:11
@Adterram the kernel of a unbounded linear functional is never closed. It's a fact of life that a linear functional is bounded if and only if its kernel is closed. – user12014 May 19 '12 at 1:18

Yes.

1. A finite-dimensional subspace $E$ of a Banach space $X$ is closed. Choose a basis $e_{1},\ldots,e_{n}$ of $E$, use Hahn–Banach to extend the dual functionals $\varphi_{i} :E \to \mathbb{R}$ determined by $\varphi_i(e_j) = \delta_{ij}$ to continuous linear functionals $e_{i}^\ast: X \to \mathbb{R}$ and check that $P : X \to X$ given by $Px = \sum_{i=1}^n e_{i}^\ast(x)\cdot e_i$ is a projection of norm $\|P\| \leq n$ and range $E$.

Note that every $x \in X$ can be uniquely written as $x = Px + (1-P)x$ and that $F = \ker{P} = \operatorname{im}(1-P)$ so that $E \oplus F \to X$ given by $(e,f) \mapsto e+f$ is a continuous linear map with inverse $(Px,(1-P)x)$. In other words, every finite-dimensional subspace of a Banach space is complemented.

2. Every closed subspace $E$ of finite co-dimension in a Banach space $X$ is complemented.

To see this, choose a basis $e_{1},\ldots,e_{n}$ of the Banach space $X/E$, choose pre-images $f_{1},\ldots,f_n$ of $e_1,\ldots,e_n$ under the canonical projection $X \to X/E$ and note that the linear span $F$ of $f_1,\ldots,f_n$ is a complement of $E$.

3. If the range $R = T(E)$ of a bounded linear operator $T: E \to F$ between Banach space has finite codimension in $F$ then $R$ is closed.

To see this, consider the Banach space $\bar{E} = E/\ker{T} = \operatorname{coim}{T}$ and factor $T$ over the injective bounded operator $\bar{T}: \bar{E} \to F$. Note that $R = T(E) = \bar{T}(\bar{E})$. Choose a basis $f_{1},\ldots,f_{n}$ of an algebraic complement of $R$ and consider the operator $$S: \bar{E} \oplus \mathbb{R}^n \to F,\quad S(\bar{e},x) \mapsto \bar{T}(\bar{e}) + \sum_{i=1}^n x_i f_i.$$ Now observe that $S$ is a continuous bijection, hence a homeomorphism by the open mapping theorem and conclude by observing that $R = S(\bar{E} \oplus 0)$ is the image of a closed subspace of $\bar{E} \oplus \mathbb{R}^n$, hence $R=T(E)$ is closed (of course, this argument also shows that $R = T(E)$ is complemented in $F$).

Now let $T:E \to F$ be a bounded operator with finite-dimensional kernel and whose range has finite co-dimension in $F$. By point 1. $K = \ker{T}$ is closed and complemented, by point 3. $T(E)$ is closed and therefore by 2. or otherwise $T(E)$ is complemented in $F$.

This implies that every Fredholm operator $T: E \to F$ is isomorphic to the operator $$K \oplus \bar{E} \longrightarrow T(E) \oplus C, \quad (k,\bar{e}) \longmapsto (\bar{T}\bar{e},0)$$ where $\bar{E} = E/\ker{T}$, $C = F/T(E)$ and $\bar{T}:\bar{E} \to T(E)$ is an isomorphism of Banach spaces. This is often called the canonical factorization of a Fredholm operator.

In more algebraic terms, a Fredholm operator induces two split short exact sequences $$\ker{T} \rightarrowtail E \twoheadrightarrow \operatorname{coim}T\quad\text{ and } \quad\operatorname{im}T \rightarrowtail F \twoheadrightarrow \operatorname{coker}{T},$$ and, also contrary to general operators, the induced map $\bar{T}:\operatorname{coim}T \to \operatorname{im}T$ is an isomorphism of Banach spaces. These split exact sequences and the isomorphism $\bar{T}$ are ultimately the reason for index theory to work the way it works and why “Fredholm operators behave very much like matrices”.

One of the best sources on basic Fredholm theory I know is in one of the first chapters of:

a slightly expanded form of which can be found in in Hirzebruch–Scharlau's Einführung in die Funktionalanalysis if you read German.

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