Showing that any square matrix in $\mathbb{R}^{n \times n}$ has a "square root" Prove that there exists $\delta > 0$ so that for all square matrices $A\in \mathbb{R}^{n\times n}$ with $\|A-I\| < \delta$ (where $I$ denotes the identity matrix) there exists $B\in \mathbb{R}^{n\times n}$ so that $B^2=A$.

My attempt so far:
$$A-I= \begin{bmatrix}
a_{11}-1 & a_{12} & ... & a_{1n} \\
a_{21} & a_{22}-1 & ... & a_{2n} \\
\vdots \\
a_{n1} & a_{n2} & ... & a_{nn}-1 \\
\end{bmatrix} $$
Taking $x=(1,0,...,0)\in \mathbb{R}^n$, we have that
$$\|A-I\|_{op}=\underset{\|x\|=1}{\sup}\|(A-I)x\|= \sqrt{(a_{11}-1)^2+...+a_{n1}^2}<\delta$$
Intuitively, this seems to suggest that for each basis vector of $\mathbb{R}^n$ means that $A-I$ can be made close to the zero matrix, and hence, close to being a diagonal matrix. But I am having trouble going from here. Anyone have any hints?
 A: Hint:
$$
\sqrt{x}=\sqrt{1-(1-x)}=\sum_{k=0}^\infty {1/2\choose k} (x-1)^k
$$
By the binomial series. Can you use the above taking $x=A$ and proving it is a norm convergent series for a square root of $A$ with your bound?
A: $x^{T}Ax\geq x^{T}x+x^{T}(A-I)x \geq (1-\delta) x^{T}x \geq 0$. Hence $A$ (being non -negative definite) is digonalizable with non-negative diagonal entries.  Replace diagonal entries by their square roots to get $B$. 
A: Let 
$I \subset U \subset \Bbb R^{n \times n}, \tag 1$
where $U$ is open.  Then we may define the matrix function
$F(B) = B^2:U \to \Bbb R^{n \times n}, \tag 2$
and we note that
$F(I) = I^2 = I, \tag 3$
that is, $I$ is in the range of $F(B)$.
$F(B) = B^2$ is in fact differentiable; we have, for $\Delta \in \Bbb R^{n \times n}$, 
$F(B + \Delta) = (B + \Delta)^2 = (B + \Delta)(B + \Delta) = B^2 + B\Delta + \Delta B + \Delta^2, \tag 4$
$F(B + \Delta) - F(B) - (B \Delta + \Delta B) = (B + \Delta)^2 - B^2 - (B\Delta + \Delta B)  = \Delta^2; \tag 5$
$\Vert F(B + \Delta) - F(B) - (B \Delta + \Delta B) \Vert = \Vert (B + \Delta)^2 - B^2 - (B\Delta + \Delta B) \Vert = \Vert \Delta^2 \Vert \le \Vert \Delta \Vert^2; \tag 6$
since
$\dfrac{\Vert \Delta \Vert^2}{\Vert \Delta \Vert} = \Vert \Delta \Vert \to 0 \; \text{as} \; \Delta \to 0, \tag 7$
we see that $F(B)$ is differentiable and that its derivative is the linear map
$DF(B)(\Delta) = B\Delta + \Delta B; \tag 8$
thus
$DF(I)(\Delta) = I \Delta + \Delta I = 2 \Delta, \tag 9$
which is clearly non-singular.  It follows from the inverse function theorem that there is some neighborhood $V$ of $I$ and a function
$R:V \to U \subset R^{n \times n}, \: R(I) = I, \tag{10}$
such that for $A \in V$
$F(R(A)) = (R(A))^2 = A; \tag{11}$
if we now choose $\delta > 0$ sufficiently small we may ensure that the set
$B(I, \delta) = \{A \in \Bbb R^{n \times n}, \; \Vert I - A \Vert < \delta \} \subset V, \tag{12}$
and thus for $A \in B(I, \delta)$ we may set
$B = R(A), \tag{13}$
and we have
$B^2 = (R(A))^2 = F(R(A)) = A, \tag{14}$
as desired.  $OE\Delta$
