An exercise on Linear algebra in PDE I'm struggling to show some exercise given in the PDE book of Krylov(Lecture s on Elliptic and Parabolic Equations in Sobolev Spaces.

Exercise 1.4.7. Let $A=(a^{ij})$ and $U=(u_{ij})$  be $2\times 2$ symmetric matrices. Assume that 
  $$ \mu |\xi|^2 \leq \xi^T A \xi \le \nu |\xi|^2$$ 
  for all  $\xi \in\mathbb{R}^2$, where $\mu,\nu>0$ are some constants. Prove that 
  $$ \frac{1}{2\mu^2} \left(\sum_{i,j=1}^2 a^{ij} u_{ij}\right)^2 \geq \frac{\mu^2}{2\nu^2} \left (\sum_{i,j=1}^2 u^2_{ij} \right)+\det U.$$

 A: Indeed you can use $\mathrm{tr}(AU)=\left(\sum_{i,j=1}^2 a^{ij} u_{ij}\right)$. You may have found this in the hints to the exercises at the end of the chapter,  note that in the hint for this exercise, it is also given that the trace is invariant under orthogonal transformation. The first orthogonal transformation associated with symmetric matrices is the one that is associated with the diagonalisation of the matrix.
I'll not give away everything, but these steps should guide you to the answer:
1) Write $A$ as $A = P^T D P$.
2) Observe that $\mathrm{tr}(AU) = \mathrm{tr}(PAP^T P U P^T) = \mathrm{tr}(DPUP^T)$.
3) Make an estimate for $\mathrm{tr}(DP U P^T)$ by estimating the eigenvalues of $A$ (and thus the elements of $D$).
4) Now transform $U$ back using $P$ again. You should now have $\mathrm{tr}(AU) \geq \mu  \mathrm{tr}(U)$.
5) Use the diagonalization of $U$ to get the sum of the eigenvalues as the trace.
6) Now square on both sides and write out using the transformation of $U$ again you can show that the expression you on the right hand side of your squared inequality equals $\sum_{i,j=1}^2 u^2_{ij} +2 \det U$.
7) Now you can obtain the desired inequality by dividing by 2 and using $\frac{\mu}{\nu} \leq 1$.
I hope this helps you, if you have any questions about the details, please ask.
