<s>[INST] Showing that $\mathbf{X}^{2} + \mathbf{X} = \mathbf{A}$ has a solution 
Show that there exists some $\epsilon >0$ s.t. for all $\mathbf{A}\in \mathbb{R}^{2\times 2} $
with $|( \mathbf{A})_{i, j}| < \epsilon $ for all $i, j$ (let the space of all such matrices be $E$) the equation
$$
\begin{align*}
\mathbf{X}^{2} + \mathbf{X} = \mathbf{A}
\end{align*}
$$
has some solution $\mathbf{X}\in \mathbb{R}^{2\times 2} $.

My approach would be to use the Banach fixed point theorem:
Consider the mapping
$$
\begin{align*}
  \Phi \colon E \to \mathbb{R}^{2, 2}, \quad \Phi ( \mathbf{X})
  = \mathbf{A} - \mathbf{X}^{2}
.\end{align*}
$$
So first we need to find an $\epsilon > 0$ s.t. $\Phi [ E]\subseteq E$. Since
$$
\begin{align*}
\begin{bmatrix}
  a & b \\ c & d  
\end{bmatrix} ^{2} = \begin{bmatrix}
a^{2} + cb & ab + db \\ ac + c d &  b c + d ^{2} 
\end{bmatrix} 
\implies \forall i, j \in \{ 1, 2\}\colon ( \mathbf{X}^{2})_{i, j} < 2 \epsilon^{2}
\end{align*}
$$
choosing any $\epsilon < 1 / 2$ will assure that $\Phi ( \mathbf{X}^{2}) \in E$ for $\mathbf{X} \in E$.
Next we have to determine when $\Phi $ becomes a contraction. One has
$$
\begin{align*}
\left\| \Phi ( \mathbf{X}) - \Phi ( \mathbf{Y})\right\|_{\infty }  
&= 
\left\| \mathbf{X}^{2} - \mathbf{Y}^{2}\right\|_{\infty} 
\\ 
&=  \max_{i \in \{ 1, 2\}} \sum_{j = 1}^{2} \left|   ( \mathbf{X}^{2})_{i, j}- ( \mathbf{Y}^{2})_{i, j} \right|
\\
&\leqslant \max_{i \in \{ 1, 2\}} \sum_{j = 1}^{2}\!\left(  \left|   ( \mathbf{X}^{2})_{i, j}\right|+ \left|  ( \mathbf{Y}^{2})_{i, j}  \right|
\right)
\\
& < 2 \!\left( 2 \epsilon ^{2} + 2 \epsilon ^{2}\right) = 8 \epsilon ^{2}
\end{align*}
$$
and
$\left\| \mathbf{X} - \mathbf{Y}\right\|_{\infty} < 4\epsilon $ meaning we simply have to choose some
$\epsilon $ satisfying
$$
\begin{align*}
2 \epsilon ^{2} < \epsilon  \iff \epsilon  < \frac{1}{2}
.\end{align*}
$$
First of all, is my approach correct? Secondly, are there more elegant solutions (this question was in the
context of an analysis course, meaning I didn't think about any elegant linear algebra solutions).
 [/INST] Lemma: There exists $\epsilon>0$ such that if $B$ is a $2\times 2$ matrix with $\|B-\frac{1}{4}I \|_\infty<\epsilon$, then the equation $Y^2=B$ has a solution $Y$.
Proof: Since the eigenvalues of $S$ are given by
$$\lambda_{1,2}(S)=\frac{1}{2} \left(S_{11}+S_{22}\pm \sqrt{(S_{11}+S_{22})^2-4(S_{11}S_{22}-S_{12}S_{21})} \right ),$$ and $\frac{1}{4}I$ has positive eigenvalues, there exists a neighborhood of $\frac{1}{4}I$ such that every matrix in that neighborhood whose eigenvalues are real must have positive eigenvalues as well. In other words, there exists $\epsilon>0$ such that for every matrix $B$ with $\|B-\frac{1}{4}I\|_\infty<\epsilon$, either $B$ has non-real eigenvalues or $B$ has real positive eigenvalues.
Next, by Jordan decomposition, there exists a matrix $P$ such that $P^{-1}BP=\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}$ with $\lambda_1,\lambda_2>0$ (if $B$ has two positive eigenvalues) or $P^{-1}BP=t\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$ for some $t>0$ and angle $\alpha$ (if $B$ has non-real eigenvalues). In either case, it is easy to see that there exists $Z$ with $Z^2=P^{-1}BP$. To be precise, in the former case, let $Z=\begin{bmatrix} \sqrt{\lambda_1} & 0 \\ 0 & \sqrt{\lambda_2}\end{bmatrix}$ and in the latter case, let $Z=\sqrt{t}\begin{bmatrix} \cos (\alpha/2) & -\sin (\alpha/2) \\ \sin (\alpha/2) & \cos (\alpha/2) \end{bmatrix}$. We let $Y=PZP^{-1}$ to get $Y^2=B$.
Using this lemma, now let $B=A+\frac{1}{4}I$. Hence there exists $Y$ with $Y^2=B$. Finally, let $X=Y-\frac{1}{2}I$ to get $X^2+X=A$.
</s>