I am trying to show that $curl\,curl\,(\mathcal E)=0$, where $\mathcal E$ represents the symmetric gradient, i.e., for vector-valued function $u(x_1,x_2)=(u_1(x_1,x_2),u_2(x_1,x_2))$ $$ \mathcal E(u)= \begin{bmatrix} \partial_1 u_1 & \frac12(\partial_2 u_1+\partial_1 u_2)\\ \frac12(\partial_2 u_1+\partial_1 u_2) & \partial_2 u_2 \end{bmatrix} $$ I know there is a formula sayinng that $$ curl\,curl\,= \nabla div - \Delta $$ and I am tryinng to apply this formula. However, I got confused that, say $\mathcal E(u)$ is a 2 by 2 matrix, so $\Delta \mathcal E(u)$ should be a 2 by 1 vector. But $div(\mathcal E(u))$ is a 2 by 1 vector and $\nabla div \mathcal E(u)$ is then a 2 by 2 matrix... so their dimension does not match... where I am wrong? Also, I tried to explicitly write down the computation but they did not cancel to 0...
update: for a reference I found, they write $\nabla div - \Delta$ as a 2 by 1 vector, but no explicit formula can be found there...
Any help is really welcome!