Writing $int_Omega nabla u^T M nabla v$ in terms of $H^1$ inner product of $u$ with another function

I am not quite sure my approach, but I do think that we need more restrictions on $M$. Since it would be quite long, instead of putting it on the comment, I write down my idea here.

Let $v in C_c^infty(Omega)$ for simplicity. Note that the $i$-th entry of $(Mnabla v)_i = sum_{j=1}^n a^{ij} v_{x_j}$. Therefore, $$ begin{align} int nabla u^T M nabla v & = int nabla u cdot (Mnabla v) \& = int sum_{i=1}^n u_{x_i} sum_{j=1}^n a^{ij} v_{x_j} \ & = int sum_{i,j=1}^n u_{x_i}a^{ij} v_{x_j} \ & = int sum_{i=1}^n u_{x_i}a^{ii} v_{x_i} + int sum_{ineq j} u_{x_i}a^{ij} v_{x_j}\ & = int sum_{i=1}^n u_{x_i}a^{ii} v_{x_i} + int u left(sum_{ineq j} a^{ij} v_{x_ix_j} + a_{x_i}^{ij}v_{x_j}right) end{align} $$ So in order to have the specific form you require, we need $a^{ii} = text{const}$ for $i= 1, cdots, n$, and $a_{x_i}^{ij} = 0$ for all $i neq j$, provided we have $v in C^infty_c$. However, if $v in H_0^1$ only, we may not be able to do integration by parts in the last equality, since we may not have the second derivative of $v$.