Is there a way to express elements of a symmetric non-singular $2\times2$ matrix through its eigenvalues?
Physics behind: I have an expression $f({\bf I},{\bf II})$ depending on the first and second fundamental forms, more particularly, expressed in terms of their elements ${\bf I}_{11}$, ${\bf I}_{12}$ and ${\bf I}_{22}$, and ${\bf II}_{11}$, ${\bf II}_{12}$ and ${\bf II}_{22}$, respectively. The eigenvalues of ${\bf I}$ are the squares of the principal stretches $\lambda_1^2$ and $\lambda_2^2$, and the eigenvalues of ${\bf II}$ are the principal curvatures $\kappa_1$ and $\kappa_2$. I have to express $f$ as a function $\tilde{f}(\lambda_1,\lambda_2,\kappa_1,\kappa_2)$.
More information: More precisely, I have to express $$ f({\bf I},{\bf II})=\frac{1}{\det\left({\bf I}\right)}\left[{\bf I}_{11}\left({\bf II}_{11}^2+2{\bf II} _{12}^2+3{\bf II}_{22}^2-2{\bf II}_{11}{\bf II}_{22}\right)-8{\bf I}_{12}{\bf II}_{11}{\bf II}_{12}+2{\bf I}_{22}\left({\bf II}_{11}^2+3{\bf II}_{12}^2\right)\right] $$ in terms of $\lambda_1$, $\lambda_2$, $\kappa_1$, $\kappa_2$.
Idea: Since ${\rm tr}A={\rm e}_1+{\rm e}_2$ and $\det A={\rm e}_1\cdot{\rm e}_2$, it would be a solution to express $f({\bf I},{\bf II})$ in terms of ${\rm tr}{\bf I}$, ${\rm tr}{\bf II}$ and $\det{\bf I}$, $\det{\bf II}$.
Edit: The problem is reduced to representation of $\langle a,{\bf I}a\rangle$ and $\langle b,{\bf II}b\rangle$ in terms of $\lambda_1$, $\lambda_2$ and $\kappa_1$, $\kappa_2$ for some constant $a, b\in\mathbb{R}^2$ such that $|a|^2=|b|^2=1$. In fact $$ f({\bf I},{\bf II})=\langle a_1,{\bf I}a_1\rangle\left(12\langle b_1,{\bf II}b_1\rangle-{\rm tr}{\bf II}\right)^2+\langle a_2,{\bf I}a_2\rangle\left(12\langle b_2,{\bf II}b_2\rangle-{\rm tr}{\bf II}\right)^2+\langle a_3,{\bf I}a_3\rangle\left(12\langle b_3,{\bf II}b_3\rangle-{\rm tr}{\bf II}\right)^2 $$ I thought about $$ \langle a,{\bf I}a\rangle=\langle a,{\bf I}a\rangle-\lambda\langle a,a\rangle+\lambda=\langle a,\left({\bf I}-\lambda E\right)a\rangle+\lambda, $$ because if $\lambda$ is an eigenvalue of ${\bf I}$ then $$ \langle a,{\bf I}a\rangle =\lambda, $$ but can`t use it to transform $f$ into $\tilde{f}$.