Fredholm alternative interpretation Let $T \in \mathcal{K}(E)$, that is, $T$ is a compact operator in the Banach space $E$. Consider $N(T)=\{x \in E: Tx=0\}$ and $R(T)=\{y \in E:y=Tx 
 ~(\exists x \in E)\}$.
In Brézis book we have the following result:
Theorem 6.6 (Fredholm Alternative) Let $T \in \mathcal{K}(E).$ Then
a) $N(I-T)$ is finite dimensional,
b) $R(I-T)$ is closed, and more precisely $R(I-T)=N(I-T^*)^\perp$,
c) $N(I-T)=\{0\} \Leftrightarrow R(I-T)=E$,
d) $\dim N(I-T)=\dim N(I-T^*)$.
After the theorem the author says:
Remark 4: The Fredholm alternative deals with the solvability of the equation
$u − T u = f$. It says that
$\bullet $ either for every $f \in E$ the equation $u − T u = f$ has a unique solution,
$\bullet $ or the homogeneous equation $u−T u = 0$ admits $n$ linearly independent solutions,
and in this case, the inhomogeneous equation $u−T u = f$ is solvable if and only
if $f$ satisfies $n$ orthogonality conditions, i.e., $N(I-T^*)^\perp$.
My question: I would like to know how to interpret the second part of the Remark 4.
For example, in the first part, if for every $f \in E$ the equation $u − T u = f$ has a unique solution, then the only solution for $u-Tu=0$ is $u=0$. Therefore, $N(I-T)=\{0\}$. If this is not the case, then there exists $f \in E$ such that $u-Tu=f$ has more than one solution, for example, $u_1\neq  u_2$. Therefore, $u_1-u_2 \neq 0$ and $u_1-u_2 \in N(I-T)$, that is, $N(I-T) \neq \{0\}$.
I still can't understand how this second part is justified (especially the bold part):
or the homogeneous equation $u−T u = 0$ admits $n$ linearly independent solutions,
and in this case, the inhomogeneous equation $u−T u = f$ is solvable if and only
if $f$ satisfies $n$ orthogonality conditions, i.e., $N(I-T^*)^\perp$.
 A: A way to understand what mean these conditions is to consider the finite dimensional case, ie you take a square matrix $A$ of size $n$ for $I-T$ and try to solve $AX=F$. Two cases:

*

*either $AX=B$ has an unique solution,

*or $A$ has rank $m<n$ thus solving $AX=0$ will give $m-n$ independent vectors. Now as the range of $A$ is not all of $\mathbb{R}^n$ if you consider a $B\in \mathbb{R}^n$ there will be a solution only if it is in the orthogonal of the space span by the $m-n$ Vectors.
A simple example is to solve for
$$\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}.$$
$A$ has rank $1$ and $AX=0$ is spanned by $e_1=(1,-1,0)$ and $e_2=(1,0,-1)$,for instance. $AX=B$ has a solution if $B\cdot e_i=0$ for $i=1,2$ which gives $b1=b2=b3$. Hope that helps!

A: If $N(I-T)\ne\{0\}$, then by the theorem you know that $n=\dim(I-T)<\infty$. So there are $n$ linearly independent solutions $g_1,\ldots,g_n$ to the homogeneous problem.
And $f$ is a solution of the full problem if and only if $f\in R(I-T)=N(I-T^*)^\perp$. That is, $f$ is a solution if and only if $f$ is orthogonal to each of $g_1,\ldots,g_n$; these are the "$n$ orthogonality conditions".
