My Blog: Anh-Tourguide: Stereographic Projection, 3

In Cartesian coordinates $\xi=(\xi_1, \xi_2,...,\xi_{n+1})$ on the sphere $\mathbb S^n$ and $x=(x_1,x_2,...,x_n)$ on the plane, the projection and its inverse are given by the formulas

$\displaystyle\xi _i = \begin{cases} \dfrac{{2{x_i}}}{{1 + {{\left| x \right|}^2}}},&1 \leqslant i \leqslant n, \hfill \\ \dfrac{{{{\left| x \right|}^2} - 1}}{{1 + {{\left| x \right|}^2}}},&i = n + 1. \hfill \\ \end{cases}$

and

$\displaystyle {x_i} = \frac{{{\xi _i}}}{{1 - {\xi _{n + 1}}}}, \quad 1 \leqslant i \leqslant n$ .

It is not hard to see that

$\displaystyle\frac{{\partial ({\xi _1},...,{\xi _n})}}{{\partial ({x_1},...,{x_n})}} = \frac{2^n}{{{{\left( {1 + {{\left| x \right|}^2}} \right)}^{2n}}}}\det \left( {\begin{array}{*{20}{c}} {1 + {{\left| x \right|}^2} - 2x_1^2}&{ - 2{x_1}{x_2}}& \cdots &{ - 2{x_1}{x_n}} \\ { - 2{x_2}{x_1}}&{1 + {{\left| x \right|}^2} - 2x_2^2}& \cdots &{ - 2{x_2}{x_n}} \\ \vdots & \vdots & \ddots & \vdots \\ { - 2{x_n}{x_1}}&{ - 2{x_n}{x_2}}& \cdots &{1 + {{\left| x \right|}^2} - 2x_n^2} \end{array}} \right).$

Now we use the trick introduced in this entry. Let

$\displaystyle x = \left( {\begin{array}{*{20}{c}} {{x_1}} \\ \vdots \\ {{x_n}} \end{array}} \right)$

and let

$A=xx^T$ .

The determinant we are trying to compute is

$\displaystyle\det \left( {(1 + |x{|^2})I - 2A} \right) = {2^n}\det \left( {\frac{{1 + |x{|^2}}}{2}I - A} \right),$

which is the characteristic polynomial of $A$ evaluated at $\frac{1+|x|^2}{2}$ times $2^n$ .

Now, $A$ is certainly diagonalizable (which doesn’t even matter, but it makes it easier to think about), and we know its eigenvalues. Why do we know its eigenvalues? Because $A$ is a matrix of rank $1$ , hence nullity $n-1$ , hence $n-1$ of its $n$ eigenvalues are zero. What is the other eigenvalue? It’s the same as the sum of the eigenvalues, which is the trace of $A$ , which is $|x|^2$ . Put that information together, and we have that the characteristic polynomial of $A$ is

$\displaystyle \det(\lambda I-A)=\left(\lambda-|x|^2\right)\lambda^{n-1}=\lambda^n-|x|^2\lambda^{n-1}$ .

Substitute $\frac{1+|x|^2}{2}$ for $\lambda$ to get

$\displaystyle\det \left( {\frac{{1 + |x{|^2}}}{2}I - A} \right) = {\left( {\frac{{1 + |x{|^2}}}{2}} \right)^{n - 1}}\left( {\frac{{1 + |x{|^2}}}{2} - |x{|^2}} \right) = \frac{{1 - |x{|^2}}}{2}{\left( {\frac{{1 + |x{|^2}}}{2}} \right)^{n - 1}}$

which implies

$\displaystyle\det \left( {(1 + |x{|^2})I - 2A} \right) = (1 - |x{|^2}){(1 + |x{|^2})^{n - 1}}$ .

Thus

$\displaystyle\det \left( {\begin{array}{*{20}{c}} {1 + {{\left| x \right|}^2} - 2x_1^2} & { - {x_1}{x_2}} & \cdots & { - {x_1}{x_n}} \\ { - {x_2}{x_1}} & {1 + {{\left| x \right|}^2} - 2x_2^2} & \cdots & { - {x_2}{x_n}} \\ \vdots & \vdots & \ddots & \vdots \\ { - {x_n}{x_1}} & { - {x_n}{x_2}} & \cdots & {1 + {{\left| x \right|}^2} - 2x_n^2} \\ \end{array} } \right) = (1 - |x{|^2}){(1 + |x{|^2})^{n - 1}}$

which implies

$\displaystyle\frac{{\partial ({\xi _1},...,{\xi _n})}}{{\partial ({x_1},...,{x_n})}} = {\left( {\frac{2}{{1 + {{\left| x \right|}^2}}}} \right)^n}\frac{{1 - |x{|^2}}}{{1 + |x{|^2}}}.$

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Stereographic Projection, 3

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