I have a 2nd rank tensor $M(x,y)$ (basically a 2 by 2 matrix) and I want to find the expression of a new "rotated" tensor $N(x,y)$ with the following informations:
I have the 2D vector fields $a(x,y)$ and $b(x,y)$ that follows: $$b(x,y) = M(x,y)\,a(x,y)$$
I want to find the tensor N(x,y) "rotated" by an angle $\theta$ that satisfies: $$\,b(x',y')=N(x,y)\,a(x',y')$$
with $$R = \begin{bmatrix}\cos\theta & -\sin\theta \\\sin\theta & \cos\theta \end{bmatrix},\qquad \begin{pmatrix}x'\\y'\end{pmatrix}=R\begin{pmatrix}x\\y\end{pmatrix}$$
Is that even possible or do I need more information?