<s>[INST] $X^2 + X =A$ with $X, A\in \text{Mat}_{2,2} (\mathbb{R})$ . Show that there exists a solution $X$ for a given $A$ Prove the following statement:
There exists an $\epsilon > 0$ such that the following holds: 
If $A = (a_{ij}) \in \text{Mat}_{2,2} (\mathbb{R})$ a matrix with $(|a_{ij}|) < \epsilon$ for $i,j \in \{1,2\}$, then the following equation
$$X^2 + X = A$$
has a solution $X \in \text{Mat}_{2,2} (\mathbb{R})$
My Idea on how to solve this:
Let $X =  \begin{bmatrix}
v& w \\ x & y 
\end{bmatrix}$.
Therefore $X^2 + X = \begin{bmatrix}
v^2 + v + w x& v w + w y + w\\ v x + x y + x & w x + y^2 + y
\end{bmatrix} = \begin{bmatrix}
a_0 & a_1 \\ a_2 & a_3\end{bmatrix}$
Lets now define the function $$ 0=h(v,w,y,x,a_0,a_1,a_2,a_3) =\begin{cases} v^2 + v + w x - a_0 \\ v w + w y + w - a_1\\v x + x y + x -a_2
\\w x + y^2 + y-a_3
\end{cases} $$
We can now calculate the derivative of $h$:
$$dh = \begin{bmatrix}
 2v + 1 &  x & 0 & w & -1&0&0&0\\ 
w& v+y+1& w& 0& 0&-1&0&0\\
x & 0&x&v +1 & 0&0&-1&0
\\0&x&2y+1&w& 0&0&0&-1
\end{bmatrix}$$
The idea now would be to apply the implicit function theorem and show that there exists an $X$ which solves this equation. I am not sure though if this approach is correct.
Last but not least.. this question comes from an analysis sheet, so I assume one should use the methods of analysis to solve it.
Is my approach the correct way? And how does one proceed from here?
Feel free to use another approach.
Thank you for your time.
 [/INST] My way of attacking this problem uses both some algebra, and some analysis; also, there is nothing in the following which requires we restrict ourselves to $M_2(\Bbb R)$, the set of $2 \times 2$ real matrices; everything we say applies equally to $M_n(\Bbb R)$.
First, some algebra:  the equation
$X^2 + X = A \tag 1$
has a solution $X$ for some $A$ if and only if the equation
$X^2 + X + \dfrac{1}{4} I = \dfrac{1}{4}I + A \tag 2$
has the same solution $X$ as well; also, since
$(X + \dfrac{1}{2}I)^2 = X^2 + X + \dfrac{1}{4}I, \tag 3$
we may write (2) as
$(X + \dfrac{1}{2}I)^2 = \dfrac{1}{4} I + A, \tag 4$
which has a solution precisely when there exists a matrix $C$ such that
$C^2 = \dfrac{1}{4}I + A, \tag 5$
in which event we may take
$X + \dfrac{1}{2} I = C, \tag 6$
or
$X = C - \dfrac{1}{2}I. \tag 7$
The reader informed in the machinations of high-school algebra will recognize the above process as completing the square of the polynomial $X^2 + X$.
The preceding discussion indicates that we need to resolve the question of the existence of $C$ satisfying (5) when the entries of $A$ are sufficiently small.  To this end we turn to the tools of analysis.  Consider the mapping
$B \mapsto B^2 \tag 8$
defined in some neighborhood of $\frac{1}{2}I$; for any linear map $H$ we have
$(B + H)^2 = B^2 + HB + BH + H^2, \tag 9$
whence
$(B + H)^2 - B^2 - (HB + BH) = H^2, \tag{10}$
whence
$\Vert (B + H)^2 - B^2 - (HB + BH) \Vert = \Vert H^2 \Vert \le \Vert H \Vert^2; \tag{11}$
since the right-hand side of (11) is of the from $\Vert H \Vert o(\Vert H \Vert)$, we find that the derivative $DB^2$ at any $B$ is the linear map
$DB^2(H) = HB + BH; \tag{12}$
with
$B = \dfrac{1}{2}I, \tag{13}$
(12) becomes
$DB^2(H) = H \left (\dfrac{1}{2}I \right ) + \left (\dfrac{1}{2}I \right ) H = H, \tag{14}$
that is, $DB^2$ is the identity mapping when (13) binds.  We may now invoke that powerful tool of analysis, the inverse function theorem, to infer that there is a neighborhood $U$ of $\frac{1}{4}I$ and a continuously differentiable function
$S:U \to M_n(\Bbb R), \; S \left ( \dfrac{1}{4} \right ) = \dfrac{1}{2}, \tag{15}$
satisfying, for every $E \in U$,
$(S(E))^2 = E; \tag{16}$
that is, $S$ is an effective square-root on $U$.
The hypothesis placed upon $A$, that $\vert a_{ij} \vert < \epsilon$ for all $i, j$, allows us to conclude that for $\epsilon$ sufficiently small
$\dfrac{1}{4} I + A \in U; \tag{17}$
thus we may take
$C = S \left (\dfrac{1}{4}I + A \right ); \tag{18}$
then
$C^2 = \dfrac{1}{4}I + A, \tag{19}$
and with $X$ as in (7) we have a solution to (1).
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