For given a subset of the set of all $m\times n$ real matrices can we find a Lipschitz function $s.t.$ a given condition hold? Let $m,n \in \mathbb N$ and let $$\mathbb R^{m\times n}:=\left\{A : A \   is  \ (m\times n) \ real  \  \ matrix \right\}$$
Let  $\ E \subseteq \mathbb R^{m\times n}$  be given and let $\Omega \subseteq \mathbb R^n $  be open.
Does there exist a Lipschitz function $u: \Omega \to \mathbb R^m$ $s.t.$ $\nabla u$ $\in E \ a.e. $ in  $\Omega$.  Here  $\nabla u$ denotes differentiation of $u$. 
Here it is also given that $u$ should not be of the form $u(x)=Ax$ for some $A\in E$.
Can anyone please help me how can I start to solve it...
 A: In general $u$ does not exist. If $E$ has only two elements $A$ and $B$ and if $u$ is Lipschitz, not affine,  and $\nabla u$ takes values $A$ and $B$, then necessarily $A-B=a\otimes n$. 
Moreover the gradient jumps from $A$ to $B$ on hyperplanes perpendicular to $n$. This is called the Hadamard jump condition. You can find a proof in a paper of Ball and James "Fine Phase Mixtures as Minimizers of Energy", Arch. Ration. Mech. Anal. 100 (1987) 13–52. 
If $E$ has more than two elements, and two of them satisfy the Hadamard jump condition, then you can construct $u$. If none of the matrices of $E$ satisfy the Hadamard jump condition, then the situation is much more complicated. 
Edit Let $z\left(  x\right)  :=u\left(  x\right)  -Bx$, and let $C:=A-B$. Then
$\nabla z\left(  x\right)  =\chi_{F}\left(  x\right)  C$ for $\mathcal{L}^{N}$
a.e. $x\in\Omega$ and for some set $F$. Since $\chi_{F}$ is not constant, we may find $\varphi\in C_{c}^{\infty}\left(  \Omega\right)
$ such that$$
\int_{F}\nabla\varphi\,dx\neq0.
$$
Define
$$
n:=\frac{\int_{F}\nabla\varphi\,dx}{\left\vert \int_{F}\nabla\varphi
\,dx\right\vert }.
$$
We have
\begin{align*}
0  &  =\frac{1}{\left\vert \int_{F}\nabla\varphi\,dx\right\vert }\int_{\Omega
}\left(  \frac{\partial z_{i}}{\partial x_{j}}\frac{\partial\varphi}{\partial
x_{k}}-\frac{\partial z_{i}}{\partial x_{k}}\frac{\partial\varphi}{\partial
x_{j}}\right)  dx\\
&  =C_{ij}n_{k}-C_{ik}n_{j},
\end{align*}
and so the vectors $C_{i\cdot}:=\left(  C_{i1},\ldots,C_{iN}\right)  $ are
paralell to $n$, i.e., there exists $a=\left(  a_{1},\ldots,a_{d}\right)
\in\mathbb{R}^{d}$ such that $C_{i\cdot}=a_{i}n$, and this proves $A-B=a\otimes n$.
