Q:

If $I+B$ is invertible in $l^2(\mathbb{Z})$ where $B$ is a compact opreator, is $I+B^N$ invertible where $B^N$ is a truncated version of $B$? 

Let $I$ be the identity operator in $l^2(\mathbb{Z})$ and let $B$ be a compact operator in $l^2(\mathbb{Z})$ and suppose $I+B$ is invertible.  $B$ can be written as a bi-infinite matrix with elements $B_{i,j}$ for $i,j \in \mathbb{Z}$:
$$
\begin{bmatrix}
\ddots & \vdots & \vdots &  \vdots &  \\
\dots & B_{-1,-1} & B_{0,-1} & B_{1,-1} & \dots \\
\dots & B_{-1,0} & B_{0,0} & B_{1,0} & \dots \\
\dots & B_{-1,1} & B_{0,1} & B_{1,1} & \dots \\
 & \vdots & \vdots & \vdots & \ddots\\
\end{bmatrix}
$$
Define $B^N$ as a truncated version of $B$ such that its elements are the same as $B$ when both $i$ and $j$ are less than $N$, and equal to zero when at least one of $i$ or $j$ are greater than or equal to $N$.
Is $I+B^N$ invertible? 
As $N \to \infty$, we have that $B^N \to B$ so it seems that by taking $N$ large enough we can make $B^N$ as close to $B$ as we want so it 'should' become invertible at some stage.  But even if this is the case, this explanation is a bit hand-wavey, how could we state it more rigorously? 

A:

Here is a counterexample: Define the entries of $B$ by
$$
b_{ij} = \begin{cases} 1/i & \text{ if } j=i, \ i\ge 1,\\
-1/i & \text{ if } j=i-1, \ i \ge 1,\\
0 & \text{otherwise}
\end{cases}
$$
So $B$ is equal to $D(I-R)$, where $D$ is diagonal and compact, $R$ is right-shift.  So $B$ is compact. 
Then the non-trivial rows/columns of $(I+B^n)x=0$ form a linear system of $n$ equations
in the unknowns $x_0 \dots x_n$, which has non-trivial solutions. 
So $I+B^N$ is not injective for all $N$.
Let me argue that $I+B$ is injective (and hence invertible by compactness of $B$): Assume $(I+B)x=0$. Then $x_i=0$ for all $i\le0$, and
$$
-i^{-1} x_{i-1} + (1+i^{-1})x_i=0
$$
for all $i\ge1$. Since $x_0=0$ this implies $x_1=0$, and by induction $x=0$.

The claim is true if $\|B^n - B \|_{\mathcal L(l^2)}\to0$. Assume not, then every for $n$ there is $x_n$ with $\|x_n\|=1$ and $(I+B^n)x_n=0$. By reflexivity of $l^2$, $x_n \rightharpoonup x$. Then
$$
(I+B)x = (I+B_n-B_n+B)x = (I+B_n)(x-x_n) + (I+B_n)x_n + (B-B_n)x. 
$$ 
The right-hand side converges to zero.  Hence $x=0$ by injectivity of $I+B$. In addition,
$x_n = -B_nx_n \rightharpoonup -x$ and
$$
\|B_nx_n-Bx\|\le \|(B_n-B)x_n\| + \|B(x_n-x)\| \to 0. 
$$
Hence the convergence $-x_n =B_nx_n \to Bx $ is strong, which implies $\|Bx\|=1$, a contradiction to $x=0$.

