= Solution
A function $f:\mathbb R^m\to\mathbb R^r$ is <differentiable function>[differentiable] at $x$ if there is a linear map $L:\mathbb R^m\to\mathbb R^r$ such that
$$
f(x+h)=f(x)+Lh+o(\lVert h\rVert)
\quad\text{as }h\to0.
$$
The map $L$, which is unique, is the <derivative> $f'(x)$.
Expanding in a matrix increment $H$ gives
$$
(A+H)^3=A^3+A^2H+AHA+HA^2+O(\lVert H\rVert^2).
$$
Therefore $p$ is differentiable everywhere and
$$
\boxed{p'(A)H=A^2H+AHA+HA^2-3H}.
$$
The <inverse function theorem> states that if $f$ is <continuously differentiable function>[continuously differentiable] near $a$ and $f'(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivative
$$
(f^{-1})'(y)=\bigl[f'(f^{-1}(y))\bigr]^{-1}.
$$
In the normalized case $f(0)=0$ and $f'(0)=I$, continuity of $f'$ gives a closed ball $\overline B_r(0)$ on which
$$
\lVert I-f'(x)\rVert\le q<1.
$$
For small $y$, define $T_y(x)=x-f(x)+y$. The <mean value inequality> makes $T_y$ a <contraction mapping> with constant $q$. If $\lVert y\rVert\le(1-q)r$, then
$$
\lVert T_y(x)\rVert\le q\lVert x\rVert+\lVert y\rVert\le r,
$$
so $T_y$ maps the ball into itself. The <contraction mapping theorem> gives a unique <fixed point> $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,
$$
\lVert x_y-x_z\rVert
\le q\lVert x_y-x_z\rVert+\lVert y-z\rVert,
$$
so $\lVert x_y-x_z\rVert\le\lVert y-z\rVert/(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.
For the polynomial map in the question,
$$
p(2I)=I,
\qquad
p'(2I)H=(4+4+4-3)H=9H.
$$
The derivative is invertible, so the inverse function theorem supplies some $\epsilon>0$ and a continuously differentiable $q:D_\epsilon(I)\to\mathcal M_n$ such that
$$
\boxed{p\circ q=\operatorname{id}|_{D_\epsilon(I)}}.
$$
There can be no inverse on all of $\mathcal M_n$, because
$$
p(2I)=I=p(-I),
$$
so $p$ is not <injective function>[injective].
Back to article page
