Let's suppose we have two variables $(\alpha, \beta)$ on two functions $k_1, k_2$ that can be defined in terms of matrix relations:

$$ k_1 = \left| \xi^T \mathcal{F}^T \mathcal{F} \xi \right| $$

$$ k_2 = \left| \xi^T \mathcal{G}^T \mathcal{G} \xi \right| $$


$$ \mathcal{F}= \left( \begin{matrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{matrix} \right) $$

$$ \mathcal{G}= \left( \begin{matrix} d_{11} & d_{12} \\ d_{21} & d_{22} \end{matrix} \right) $$

$$ \xi = \left( \begin{matrix} \alpha \\ \beta \end{matrix} \right) $$

Suppose that the determinants of $\mathcal{F}$ and $\mathcal{G}$ are non-zero. It's possible to invert the relation such that:

$$ \alpha = f(k_1, k_2) $$ $$ \beta = g(k_1, k_2) $$

However the multiple solutions return by the CAS (mathematica) are extremely complicated. Is there a compact way to express the functions $f$ and $g$ supposing we are only interested in positive values for $k_1$ and $k_2$?


Write $F$ and $G$ for $\mathcal F$ and $\mathcal G$, respectively. Since $M:=F^TF$ and $N:=G^TG$ are symmetric positive definite matrices, there is a $2\times2$ nonsingular matrix $A$ such that $C:=A^TMA$ and $D:=A^TNA$ are diagonal positive definite matrices, say with the diagonal entries $c_1,c_2$ and $d_1,d_2$, respectively. Moreover, without loss of generality $c_1=c_2=1$. (See details on this below.)

Also, $k_1=\xi^TM\xi$ and $k_2=\xi^TN\xi$ (taking the absolute values is not needed, since $M$ and $N$ are positive definite). So, $k_1=\eta^TC\eta=\eta_1^2+\eta_2^2$ and $k_2=\eta^TD\eta=d_1\eta_1^2+d_2\eta_2^2$, where $[\eta_1,\eta_2]^T=\eta:=A^{-1}\xi$.

Solving now the equations $k_1=\eta_1^2+\eta_2^2$ and $k_2=d_1\eta_1^2+d_2\eta_2^2$ for $\eta_1$ and $\eta_2$, one finds the expressions of $\xi=A\eta$ in terms of $k_1$ and $k_2$.

Details on the simultaneous diagonalization of $M$ and $N$. Let $R:=(F^{-1})^TNF^{-1}$, so that $N=F^TRF$. Since $R$ is symmetric positive definite, we have $R=Q^TDQ$ for some orthogonal matrix $Q$ and some diagonal positive definite matrix $D$. Thus, letting $A:=(QF)^{-1}$, we have $N=F^TRF=F^TQ^TDQF=(A^{-1})^TDA^{-1}$ and $M=F^TF=F^TQ^TQF=(A^{-1})^TIA^{-1}$, whence indeed $A^TMA=I$ (the identity matrix) and $A^TNA=D$ are diagonal positive definite matrices.

  • 1
    $\begingroup$ I have added details on the simultaneous diagonalization. $\endgroup$ – Iosif Pinelis Apr 1 '16 at 14:58

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