Let $E$ be a Banach space. Let $\mathcal K(E)$ be the space of all compact (bounded linear) operators from $E$ to $E$. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. Let $I:E \to E$ be the identity map.


Below is the Fredholm alternative in Brezis' *Functional Analysis*.



> 
> **Theorem 6.6** Let $T \in \mathcal K(E)$. Then
> 
> 
> (a) $\dim N(I-T) < \infty$,
> 
> 
> (b) $R(I-T)$ is closed, and more precisely $R(I-T) = N(I-T^\*)^\perp$,
> 
> 
> (c) $N(I-T) = \{0\} \iff R(I-T) = E$,
> 
> 
> (d) $\dim N(I-T)=\dim N(I-T^\*)$.
> 
> 
> 


The [proof of (d)](https://math.stackexchange.com/questions/4703547/how-a-contradiction-is-obtained-in-a-proof-of-the-fredholm-alternative) by the author is by contradiction. My goal is to give a more direct proof of (d).



> 
> Could you elaborate if the map $I-S$ in my below attempt is actually injective?
> 
> 
> 




---


**My attempt** By (a), $\dim N(I-T) < \infty$. Then $N(I-T)$ has a (closed) complement subspace $G$ in $E$, i.e., $E= N(I-T) \oplus G$. Let $\pi\_1 : E \to N(I-T)$ be the (continuous linear) projection map. Then $\pi\_1$ is surjective with $N (\pi\_1) = G$.


We have $T \in \mathcal K(E) \iff T^\* \in \mathcal K(E^{\*})$. By (a), $\dim N(I-T^\*) < \infty$. By (b), $\operatorname{codim} R(I-T) < \infty$. Then $R(I-T)$ has a (closed) complement subspace $L$ in $E$, i.e., $E= R(I-T) \oplus L$. Let $\pi\_2 : E \to L$ be the (continuous linear) projection map. Then $\pi\_2$ is surjective with $N(\pi\_2) = R(I-T)$.


Let $\Lambda:= \pi\_2 \circ \pi\_1:E \to L$. Then $\Lambda$ is linear continuous. We have $\dim L = \dim N(I-T^\*) < \infty$. Then $\Lambda$ has finite rank and thus $\Lambda \in \mathcal K(E)$.


Let's prove that $\Lambda$ is surjective. Let $S := T-\Lambda$. Then $S \in \mathcal K(E)$. First, we'll show that $I-S$ is injective. Indeed, if $(I-S)u=0$ then $(I-T)u + \Lambda u =0$. Because $E= R(I-T) \oplus L$ and $R(\Lambda) \subset L$, we get $(I-T)u=\Lambda u=0$. Then $u \in N(I-T) \cap N(\Lambda)$. We want to prove $u=0$.



Is $I-S$ in my attempt of Fredholm alternative injective?
No, in general $I-S$ will not be injective.


Indeed, suppose e.g. that $E=\mathbb R^2$ and
$$T=\begin{bmatrix}2&1\\ -1&0 \end{bmatrix};$$
here we will identify the linear operators with their matrices in the standard basis of $\mathbb R^2$.


Then (assuming $G$ and $L$ are the orthogonal complements of $N(I-T)$ and $R(I-T)$ respectively) we have
$$\pi\_1=\frac12\begin{bmatrix}1&-1\\ -1&1 \end{bmatrix};$$
$$\pi\_2=\frac12\begin{bmatrix}1&1\\ 1&1 \end{bmatrix};$$
$$\Lambda=\begin{bmatrix}0&0\\ 0&0 \end{bmatrix};$$
$$I-S=I-T=\begin{bmatrix}-1&-1\\ 1&1 \end{bmatrix}.$$
So, $I-S$ is not injective.


