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There are many examples on the Web of the use of Cavalieri's principle in determining areas and volumes of 2-D and 3-D geometrical figures. The Wikipedia link uses the principle as both a proof and geometric interpretation of Cramer's rule with explicit diagrams in 2-D.

For the Vandermonde matrix in 2-D

$$V_2(x_1,x_2)=\left[ \begin{array}{} 1 & 1 \\ x_1 & x_2 \end{array} \right],$$

Cramer's rule gives

$$\bar{x}_1\left[ \begin{array}{} 1 \\ x_1\end{array} \right]+\bar{x}_2\left[ \begin{array}{} 1 \\ x_2\end{array} \right]=\left[ \begin{array}{} 1 \\ 0\end{array} \right],$$

or equivalently,

$$\bar{x}_1\left[ \begin{array}{} 1 \\ x_1\end{array} \right]=\left[ \begin{array}{} 1 \\ 0\end{array} \right]-\bar{x}_2\left[ \begin{array}{} 1 \\ x_2\end{array} \right],$$

where $$\bar{x}_1=\frac{|\left[ \begin{array}{} 1 & 1 \\ 0 & x_2 \end{array} \right]|}{|\left[ \begin{array}{} 1 & 1 \\ x_1 & x_2 \end{array} \right]|},$$

so

$$|\left[ \begin{array}{} 1 & 1 \\ 0 & x_2 \end{array} \right]|=\bar{x}_1|\left[ \begin{array}{} 1 & 1 \\ x_1 & x_2 \end{array} \right]|=det(\bar{x}_1\left[ \begin{array}{} 1 \\ x_1\end{array} \right],\left[ \begin{array}{} 1 \\ x_2\end{array} \right] )=det(\left[ \begin{array}{} 1 \\ 0\end{array} \right]-\bar{x}_2\left[ \begin{array}{} 1 \\ x_2\end{array} \right],\left[ \begin{array}{} 1 \\ x_2\end{array} \right] ).$$

$$$$ Choose $x_2>x_1>0$ to illustrate the geometry. The first and last determinants are the areas of two parallelograms $P_1$ and $P_2$ that can easily be drawn. Both are framed on the left side by the vector $\left[ \begin{array}{} 1 \\ x_2\end{array} \right]$ from the origin to the point $(1,x_2)$. $P_1$ is framed on the adjoining side at the origin by the vector $\left[ \begin{array}{} 1 \\ 0\end{array} \right]$ from the origin to $(1,0)$, while $P_2$ is framed on the adjoining side by the vector $\left[ \begin{array}{} 1 \\ 0\end{array} \right]-\bar{x}_2\left[ \begin{array}{} 1 \\ x_2\end{array} \right]$ extending from the origin to the point $(1-\bar{x}_2,-\bar{x}_2 x_2)$. So, the left sides of both parallelograms are the same, and the right sides lie on the line through $(1,0)$ that is parallel to the vector $\left[ \begin{array}{} 1 \\ x_2\end{array} \right]$ (the line traced by the parametrized position vector $\left[ \begin{array}{} 1 \\ 0\end{array} \right]-t\left[ \begin{array}{} 1 \\ x_2\end{array} \right]$); therefore, since the two figures lie on the left and right between the same parallels and their cross-sections on any line parallel to these sides are the same (just $|\left[ \begin{array}{} 1 \\ x_2\end{array} \right]|$ or $0$), I can invoke Cavalieri's principle to claim they have the same area.

For the 3-D case, the parallel lines are replaced by parallel planes and areas by volumes, and the argument to invoke Cavalieri's priciple proceeds analogously from examining the parallelepipeds of the relation

$$|\left[ \begin{array}{} 1 & 1 & 1\\ 0 & x_2 & x_3\\0 & x_2^2 & x_3^2\end{array} \right]|=det(\left[ \begin{array}{} 1 \\0\\0\end{array} \right]-\bar{x}_2 \left[ \begin{array}{} 1 \\x_2\\x_2^2\end{array} \right]-\bar{x}_3 \left[ \begin{array}{} 1 \\x_3\\x_3^2\end{array} \right],\left[ \begin{array}{} 1 \\x_2\\x_2^2\end{array} \right],\left[ \begin{array}{} 1 \\x_3\\x_3^2\end{array} \right]).$$

Questions:

1) Is there any reason that invoking Cavalieri's principle at higher dimensions for a geometric proof of the inversion of the Vandermonde matrix would fail? (Inversion through Cramer's rule is certainly valid for any dimension.)

2) Are there any other applications in the literature of Cavalieri's principle for dimensions above three?

Edit: A 4-D application

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  • $\begingroup$ According to en.wikipedia.org/wiki/… the Cavalieri principle has to do with area-preserving shears in the plane. This is true regardless of which matrix you are trying to invert. The inverse of the Vandermonde matrix is a complicated mess of symmetric polynomials. Hopefully the geometric route gives more insight! $\endgroup$ – john mangual Jun 2 '14 at 12:40
  • $\begingroup$ John, see the MO question "Geometric proof of the Vandermonde determinant?" for a quite elegant expression for the inverse of $V$ in terms of the elementary symmetric polynomials. I use V as a specific example because of my general interest in its association to standard simplices and some operational formulas. One user questioned the validity of invoking Cavallieri for a geometric proof of Cramer's rule for higher dimensions, ergo the question with the shearing presented in vector detail. $\endgroup$ – Tom Copeland Jun 2 '14 at 16:59
  • $\begingroup$ Cavalieri's principle, of course, applies to much more general transformations in 2- and 3-D, hence the second question. $\endgroup$ – Tom Copeland Jun 2 '14 at 17:06
  • $\begingroup$ I deleted my MO answer containing the inverse matrix for the Vandermonde matrix, but will present it again in some notes at my website/petite-arxiv on the Vandermonde matrix in construction. $\endgroup$ – Tom Copeland Jul 8 '14 at 6:10
  • $\begingroup$ See Roy Smith's mathoverflow.net/questions/28268/do-you-read-the-masters/… $\endgroup$ – Tom Copeland Jan 29 '15 at 0:10

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