An Equation in Matrices

Let $\displaystyle a=\frac{1}{\sqrt{1-x^2}}$ and $\displaystyle b=\frac{x}{\sqrt{1-x^2}}.$ It is left as an exercise to prove by mathematical induction that

$\displaystyle \left( \begin{array}{ccc} a & 0 & b\\ 0 & 1 & 0\\ b & 0 & a \end{array} \right)^{n}=\left(\begin{array} a_{n} & 0 & b_{n}\\ 0 & 1 & 0\\ b_{n} & 0 & a_{n} \end{array}\right),$

where $\displaystyle a_{n}=\frac{1}{2}[(a+b)^n+(a-b)^n]$ and $\displaystyle b_{n}=\frac{1}{2}[(a+b)^n-(a-b)^n].$ In our case,

$\displaystyle a_{n}=\frac{1}{2}\left[\sqrt{\left(\frac{1+x}{1-x}\right)^n}+\sqrt{\left(\frac{1-x}{1+x}\right)^n}\right]$ and $\displaystyle b_{n}=\frac{1}{2}\left[\sqrt{\left(\frac{1-x}{1+x}\right)^n}+\sqrt{\left(\frac{1+x}{1-x}\right)^n}\right].$

We need to solve the equations $\displaystyle a_{n}=\frac{2}{\sqrt{3}}$ and $\displaystyle b_{n}=\frac{1}{\sqrt{3}}.$ Define $\displaystyle t=\sqrt{\left(\frac{1-x}{1+x}\right)^n}\gt 0.$

The first equation gives $\displaystyle t\in\left\{\frac{1}{\sqrt{3}},\sqrt{3}\right\}$ whereas from the second equation we get $\displaystyle t\in\left\{\frac{1}{\sqrt{3}},-\sqrt{3}\right\}.$ It follows that $\displaystyle t=\frac{1}{\sqrt{3}},$ implying $\displaystyle x=\frac{\sqrt[n]{3}-1}{\sqrt[n]{3}+1}.$

First of all note that matrix $M$ acts in the $x,z$-plane of $\mathbb{R}^3$ with coordinates $x,y,z.$ We thus can reduce the problem to investigating a $2\times 2$ matrix

$\displaystyle M=\left( \begin{array}{cc} \frac{1}{\sqrt{1-x^2}} & \frac{x}{\sqrt{1-x^2}}\\ \frac{x}{\sqrt{1-x^2}} & \frac{1}{\sqrt{1-x^2}} \end{array} \right).$

For convenience rewrite the matrix as

$\displaystyle M=\frac{1}{\sqrt{1-x^2}}\left( \begin{array}{cc} 1 & x\\ x & 1 \end{array} \right).$

We may observe that matrix $M$ has two orthogonal eigenvectors, with two eigenvalues whose product is $1:$

$\displaystyle M\left(\begin{array}0 1\\1\end{array}\right)=\frac{1+x}{\sqrt{1-x^2}}\left(\begin{array}0 1\\1\end{array}\right)$ and $\displaystyle M\left(\begin{array}{r}1\\-1\end{array}\right)=\frac{1-x}{\sqrt{1-x^2}}\left(\begin{array}{r}1\\-1\end{array}\right).$

This means that

$\displaystyle M^{n}\left(\begin{array}0 1\\1\end{array}\right)=\left(\frac{1+x}{\sqrt{1-x^2}}\right)^{n}\left(\begin{array}0 1\\1\end{array}\right)$ and $\displaystyle M^{n}\left(\begin{array}{r}1\\-1\end{array}\right)=\left(\frac{1-x}{\sqrt{1-x^2}}\right)^{n}\left(\begin{array}{r}1\\-1\end{array}\right).$

For $\displaystyle x=\frac{1}{2},$ $\displaystyle M=\left( \begin{array}{cc} \frac{2}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{array} \right).$ We are thus looking for an $x$ that satisfy

$\displaystyle \left(\frac{1+x}{\sqrt{1-x^2}}\right)^{n}=\sqrt{3}$ and $\displaystyle \left(\frac{1-x}{\sqrt{1-x^2}}\right)^{n}=\frac{1}{\sqrt{3}}.$

The two equations are equivalent. Squaring the first and taking the $n-th$ root gives a quadratic equation:

$(1+\sqrt[n]{3})x^2+2x+(1-\sqrt[n]{3})=0,$

with two roots $\displaystyle x=\frac{-1\pm\sqrt[n]{3}}{1+\sqrt[n]{3}}.$ Obviously $x=-1$ is unsuitable, leaving $\displaystyle x=\frac{-1+\sqrt[n]{3}}{1+\sqrt[n]{3}}.$

The problem, due to Dan Sitaru, has been posted by Leo Giugiuc at the CutTheKnotMath facebook page along with his own solutionn (Solution 1 above).