Hilbert-Schmidt and invertibility Using the Hilbert-Schmidt norm for matrices:
$$\|A\|^2 = \sum_{i,j}A_{ij}^2$$
show that if $\|H\|<1$, then $I-H$ is invertible.
 A: Let us prove the following lemma:
Lemma
$$
\min_{H \in \mathbb{C}^{n\times n}}\{||H||_2 : rank(I-H)<n\} = 1.
$$
which will prove the claim as we are given $||H||<1$.
Proof
If $rank(I-H)<n$, then there exists an $x \in \mathbb{C}^n$, with $||x||_2 = 1$ (possibly after scaling) such that
$$
(I-H)x = 0.
$$
Hence, $||Hx||_2 =||x||_2=1$, and finally since
$$
||Hx||_2 \leq ||H||_2||x||_2 = ||H||_2,
$$
we get $||H||_2 \geq 1$. For an example achieving the equality, consider the following $H$:
$$
\begin{bmatrix}
    1     & 0  & \dots & 0 \\
   0     & 0 & \dots & 0 \\
    \vdots & \vdots & \ddots & \vdots\\
   0      & 0 & \dots & 0
\end{bmatrix}.
$$
A: This works in any Banach space, finite or infinite-dimensional: 

if $\|H\|<1$, then $I-H$ is invertible. 

It is simply done by writing the inverse explicitly: let 
$$
T=\sum_{k=0}^\infty  H^n.
$$
This series converges due to the fact that $\|H\|<1$. And 
$$
T(I-H)=\lim_{n\to\infty}\sum_{k=0}^n H^k(I-H)=\lim_{n\to\infty}I-H^{n+1}=I.
$$
Similarly, one shows that $(I-H)T=I$. 
**edit: fixed an error
