Exists $\delta > 0$ such that if $\|I-A\| < \delta$ then $A$ have a square root. Definition: Let $A \in \mathcal{M}_{n \times n}(\mathbb{R})$. $A$ has a square root when exists $B \in \mathcal{M}_{n \times n}(\mathbb{R})$ such that $B^2 = A$.
Proposition: Exists $\delta > 0$ such that if $\|I-A\| < \delta$ then $A$ has a square root.
Proof. Define $\varphi: \mathcal{M}_{n \times n}(\mathbb{R}) \to \mathcal{M}_{n \times n}(\mathbb{R})$ s.t. $\varphi(X) = X^2$. Note that $\varphi \in \mathcal{C}^1$ and $\varphi'(I) = T$, where $T(X) = 2X$, is invertible. By the Inverse Function Theorem exists $\delta > 0$ such that $\varphi$ is a diffeomorphism between $B_\delta(I)$ and $\varphi(B_\delta(I))$. It follows that if $\|I-A\| < \delta$, then exists $B = \varphi^{-1}(A)$, that is, $B^2 = A$.
Question: "How many" square roots has the matrix $\left[\begin{array}{cc}
1 & 0 \\
0 & 1 \end{array}\right]$?
Well, by my proof it has only one square root, but since the author chose a $2 \times 2$ matrix, I think it is wrong. Are my proof correct?
 A: If you want to check how many square roots $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$, we have to see how many solutions $(a,b,c,d)$ we have for
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
which can be rewritten as
\begin{align*}
a^2+bc&=1\tag{1}\\
ab+bd&=0\tag{2}\\
ac+cd&=0\tag{3}\\
bc+d^2&=1\tag{4}
\end{align*}
From (1) and (4) we obtain $a^2=d^2$, so $a=\pm d$, and in fact equation (4) is unnecessary.


*

*If $a=0$, then $d=0$. Equations (2) and (3) become trivial, and equations (1) and (4) are the same. In particular $bc=1$, so this gives us some solutions:
$$(0,t,\frac{1}{t},0),\qquad t\neq 0$$

*Now suppose $a\neq 0$, so $d\neq 0$. We have two cases
2.1. $a=d$.
In this case, equations (2) and (3) become $2ab=0$ and $2ac=0$. Since we are assuming $a\neq 0$, then $b=c=0$, and equation (1) becomes $a^2=1$, so $a=\pm 1$. This gives us one solution
$$(1,0,0,1)$$
2.2. $a=-d$.
In this case equations (2) and (3) become trivial, so we simply need to worry about equation (1). We again have two subcases.
2.2.1 $|a|=1$.
Then (1) becomes $bc=0$, so either $b=0$ or $c=0$, and the other variable is free, and we have solutions
$$(1,0,t,-1),\quad(1,t,0,-1),\quad(-1,0,t,1),\quad(-1,t,0,1),\qquad t\in\mathbb{R}$$
(note that for $t=0$ the first and second terms are equal, as are the third and fourth).
2.2.2 $|a|\neq 1$. In this case $bc=1-a^2\neq 0$, so $b$ and $c$ are nonzero and $c=\frac{1-a^2}{b}$. This gives us the last set of solutions
$$(s,t,\frac{1-s^2}{t},-s),\qquad s\neq0,\pm 1,\quad t\neq 0$$


So indeed we have lots of square roots for $I$ (uncountably many).
A: But see if your proposition is correct, then every matrix has a square root, else the quantity $||I−A||$ would be infinite for some $A$. What you mean to say is the following: 
There is a $δ>0$ such that $||I−A||<δ$ implies $A$ has a square root. Further, this does not imply it is unique, A may have some square root outside of the inverse image of your diffeomorphism. By assuming the square root is unique, you essentially assume $B_\delta(I)$ is the entire space.
