Show there is a unique solution I came across this problem.
For $X=(x_1,x_2,x_3,x_4)$, define $A(X)=(x_1^2-x_2^2-x_3^2-x_4^2,2x_1x_2,2x_1x_3,2x_1x_4)$. For any $Z\in\mathbb{R}^4$ with $|Z|\leq 1/16$ prove that the equation $X-A(X)=Z$ has a unique solution such that $|X|\leq 1/8$. $|X|$ is the Euclidean norm of $X$.
My thought is to use the inverse function theorem. Let $I$ be the identity on $\mathbb{R}^4$. the derivative of $I-A$ is $\begin{bmatrix}1-2x_1 & -2x_2 & -2x_3 & -2x_4\\ -2x_2 & 1-2x_1 & 0 & 0\\ -2x_3 & 0 & 1-2x_1 & 0\\ -2x_4 & 0 & 0 & 1-2x_1 \end{bmatrix}$. The determinant is 
$(1-2x_1)^2((1-2x_1)^2-4x_2^2-4x_3^2-4x_4)$. So this quantity needs to be nonzero for $|X|\leq 1/8$? And if this is true, $I-A$ is bijective from the ball of radius $1/8$ onto its image. So I'd then have to show that the image of this ball is contains the ball of radius $1/16$? Is this the idea?
 A: Write $X=(t,{\bf x})$ and $Z=(s,{\bf z})$ with ${\bf x}, {\bf z}\in{\mathbb R}^3$. Then
$$A(X)=(t^2-|{\bf x}|^2, 2t{\bf x})\ ,$$
and the equation $X-A(X)=Z$ expands to
$$\bigl(t-t^2+|{\bf x}|^2, (1-2t){\bf x}\bigr)=(s,{\bf z})\ .$$
It follows that necessarily
$${\bf x}={1\over1-2t}{\bf z}\ ,$$
which is unproblematic as long as $|t|<{1\over2}$. In addition we now have the following scalar equation for $t$:
$$t-t^2+{z^2\over(1-2t)^2}=s\ ,\tag{1}$$
which we have to solve for given $Z$ with $|s|\leq{1\over16}$, $\>z:=|{\bf z}|\leq{1\over16}$. Put
$$f(t,s,z):= t-t^2+{z^2\over(1-2t)^2}-s\ .$$
Then $f(0,0,0)=0$ and
$$f_t(0,0,0)=1+4z^2\ne0\ .$$
The implicit function theorem now allows the following conclusion: There is a neighborhood $U$ of $(s,z)=(0,0)$ and a neighborhood $V$ of $t=0$ such that for all $(s,z)\in U$ there is exactly one solution $t=\psi(s,z)\in V$ of $(1)$, whereby $\psi$ is a real analytic function of $s$ and $z$. 
Of course, for exact estimates concerning the sizes of $U$ and $V$ more work is necessary. The following figure shows the graph of the LHS of $(1)$ for various values of $z$. It shows that the claim made in your question holds true.

A: Let $F(X) = X - A(X)$ and $X_0 \in \mathbb{R}^4$. By the inverse function theorem we are able to conclude that if $\det F'(X_0) \neq 0$, then there exists an open set $U_{X_0} \subset \mathbb{R}^4$ containing $X_0$ such that the function $F_0 : U_{X_0} \to F(U_{X_0}) : X \mapsto F(X)$ is bijective.
Let $\overline{B}(r) = \{ X \in \mathbb{R}^4 : |X| \le r \}$.
So given $Z \in \overline{B} \left(\frac{1}{16}\right)$, we need to find $X_0 \in \mathbb{R}^4$ such that $\det F'(X_0) \neq 0$ and $Z \in F(U_{X_0})$, such that $|\underbrace{F_0^{-1}(Z)}_{X_Z}| \le \frac{1}{8}$, and after that, we still need to show that among all points in $B \left(\frac{1}{8} \right)$, $X_Z$ is the unique point such that $F(X) = Z$.
How would we choose $X_0$ to guarantee that $U_{X_0} \cap B \left(\frac{1}{8} \right) \neq \emptyset$? Because the theorem gives us no control over $U_{X_0}$ we will need to inspect it by hand. This seems no easier than a direct computation approach.
Perhaps there is an alternative (and desirably easier) approach? Well... notice that
$$ X-A(X)=Z \iff A(X) + Z = X$$
This means that $X$ is a solution of the equation $X-A(X)=Z$ if and only if it is a fixed point of the function $G(X) = A(X) + Z$. Thus the statement is equivalent to showing that for every $Z \in \overline{B} \left(\frac{1}{16}\right)$, the function $G : \overline{B} \left(\frac{1}{8}\right) \to \overline{B} \left(\frac{1}{8}\right) : X \mapsto A(X) + Z$ is a contraction, whence it follows that it has an unique fixed point. Having in mind that $|A(X)| = |X|^2$ we see that the function $G$ is well-defined because for all $X \in \overline{B} \left(\frac{1}{8}\right)$
$$|G(X)| \le |A(X)| + |Z| = |X|^2 + |Z| \le \left( \frac{1}{8} \right)^2 + \frac{1}{16} \le \frac{1}{8} $$
Now to show that $G$ is a contraction it is enough to show that the absolute value of $G'(X)$'s eigenvalues are all less than $1$ for all $X \in \overline{B} \left(\frac{1}{8}\right)$. For that consider $\rho(A)$ the maximum absolute value of $A$'s eigenvalues and the matrix norm
$$ \|A\| = \max \{ |Av| : v \in \mathbb{R}^4, |v| = 1 \} $$
Therefore, if $X \in \overline{B} \left(\frac{1}{8}\right)$, then
$$ \rho(G'(X)) \le \|G'(X)\| = \sqrt{\rho(G'(X)^T G'(X))} = 2 |X| \le \frac{1}{4} $$
Hence the result.
(Moreover, note that the hypothesis we were working in $4$ dimensions were not used, meaning that the exactly same proof holds for the general case of $\mathbb{R}^n$.)
