Property matrix elliptic operator Let $A(x)=(a_{i,j}(x))_{i,j=1}^n$ such that


*

*$A$ is bounded and

*$A$ is strictly elliptic, i.e. 
$$
\xi\cdot A\xi=\sum_{i,j=1}^n a_{i,j}(x)\xi_i\xi_j\ge \lambda |\xi|^2
$$
for a.e. $x\in\Omega$,$\forall\,\xi\in\mathbb{R}^n$.
Then I have to prove that $\exists\, m\in\mathbb{R}$ and a matrix $K(x)$ such that
$$
A(x)=m(I+K(x))
$$
where $\|K(x)\|_{op}\le k<1$ for a.e. $x\in\Omega$.


Idea:
let's show that $$
|\eta\cdot(I-\frac{A(x)}{m})\xi|\le k<1 \label{C}
$$
for every $\xi,\eta\in\mathbb{R}^n$ such that $\|\xi\|=\|\eta\|=1$.
Finally I will have to solve an optimization problem to find $m$.
Any help to solve \ref{C}?
 A: If $A\in\mathbb{R}^{n\times n}$ is symmetric, the strict ellipticity is equivalent to saying that $A$ is positive definite and hence has real and positive eigenvalues $\lambda_i$ such that $\lambda_{\max}\geq\lambda_i\geq\lambda=:\lambda_{\min}>0$ for all $i=1,\ldots,n$. Take $K:=I-\alpha A$, where 
$$\tag{1}0<\alpha<2/\lambda_{\max}$$
(so that $A=\frac{1}{\alpha}(I-K)$), and show that the eigenvalues $\mu_i$ of $K$ are contained in the open interval $(-1,1)$.
By using the spectral decomposition, we have that for each eigenvalue $\lambda_i$ there is a corresponding eigenvalue $\mu_i=1-\alpha\lambda_i$ of $K$. Therefore, the eigenvalues of $K$ are located in the interval
$$
[1-\alpha\lambda_{\max},1-\alpha\lambda_{\min}].
$$
From (1), we have
$$
1-\alpha\lambda_\max>1-\frac{2}{\lambda_\max}\lambda_\max=-1
$$
and
$$
1-\alpha\lambda_\min<1-0\cdot\lambda_\min=1.
$$
Since the 2-norm of a symmetric matrix is equal to its spectral radius and the eigenvalues of $K$ are located in $(-1,1)$, it follows that
$$
\rho(K)=\|K\|_2<1.
$$
