A nonlinear matrix equation
Author
Bùi Quang Tú
Last Updated
9 anni fa
License
Creative Commons CC BY 4.0
Abstract
A explicit solution to a nonlinear matrix equation in n=3 dimension.
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Problem
Let $\theta,\beta$ be $3\times 3$ skew-symmetric matrices and $\sigma$ be a $3\times 3$ matrix. Find symmetric $S,T$ such that:
$$(S-\theta)(T-\beta)=\sigma$$
Solution
Assuming nonsingularity whenever necessary
From $(S-\theta)(T-\beta)=\sigma$ we have $S=\theta+\sigma (T-\beta)^{-1}$
S is hermitian if and only if $$S=\theta+\sigma (T-\beta)^{-1}=\theta^\dagger+((T-\beta)^{-1})^\dagger\sigma^\dagger$$ or equivalently: $$2\theta=(T+\beta)^{-1}\sigma^\dagger-\sigma(T-\beta)^{-1}$$ $$2(T+\beta)\theta(T-\beta)=\sigma^\dagger(T-\beta)-(T+\beta)\sigma$$ $$2T\theta T +(2\beta\theta-\sigma^\dagger)T+T(\sigma-2\theta\beta)-2\beta\theta\beta+\sigma^\dagger\beta+\beta\sigma=0$$
We denote $R=2\theta;A=(2\theta\beta-\sigma);Q=-2\beta\theta\beta+\sigma^\dagger\beta+\beta\sigma$
then $A^\dagger=(2\beta\theta-\sigma^\dagger)$
The equation becomes $$TRT+A^\dagger T-TA+Q=0$$ with given $A$ and skew-hermitian $R,Q$.
To find 1 solution we assume that $T$ is diagonal.
Denote R = $ \left( \begin{array}{ccc}
0 & c & b \\
-c & 0 & a \\
-b & -a & 0 \end{array} \right)$, A = $ \left( \begin{array}{ccc}
m_1 & m_2 & m_3 \\
m_4 & m_5 & m_6 \\
m_7 & m_8 & m_9 \end{array} \right)$, Q = $ \left( \begin{array}{ccc}
0 & q_1 & q_2 \\
-q_1 & 0 & q_3 \\
-q_2 & -q_3 & 0 \end{array} \right)$,
T = $ \left( \begin{array}{ccc}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z \end{array} \right)$
Then we have $TRT$ = $ \left( \begin{array}{ccc}
0 & cxy & bxz \\
-cxy & 0 & ayz \\
-bxz & -ayz & 0 \end{array} \right)$,
$A^\dagger X$ = $ \left( \begin{array}{ccc}
0 & m_4y & m_7z \\
m_2x & 0 & m_8z \\
m_3x & m_6y & 0 \end{array} \right)$
Then we have the explicit form of the equation:
$$cxy+m_4y-m_2x=q_1(1)$$ $$bxz+m_7z-m_3x=q_2(2)$$ $$ayz+m_8z-m_6y=q_3(3)$$
This system of equations is solved by eliminate $z$ (by $(2)$ and $(3)$) then calculate $y$ from $x$ (by the identity of $xy$). Then we are left with a quadratic equation of $x$.
Have $x$ we can solve $y,z$.
The explicit solution is obtainable but not worth calculated by hand.
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