Pythagorean theorem

[lavinia]

The Pythagorean theorem relates the length of a vector to its projection onto an orthonormal basis for Euclidean space.

Does it also work in the same way for parallograms, and higher dimensional linear solids such as paralleopipeds? I take an n dimensional linear solid and project it onto an orthonormal basis for the space of n dimensional solids and then compute its hypervolume from the sum of squares of the projection coefficients.

[BruceG]

Here is a nice paper on this

http://www.jyi.org/volumes/volume2/issue1/articles/barth.html

Basically the determinant is the general tool for calculation of a k-volume in k-space, but this paper explains doing k-volume in n-space when k<n.

[granpa]

I really dont know what you mean.

in higher dimensions
Length^2 = a^2 + b^2 + c^2 + d^2...


length not volume

[lavinia]

I really dont know what you mean.

in higher dimensions
Length^2 = a^2 + b^2 + c^2 + d^2...


length not volume

If i have an orthonormal basis for a vector space then I can project linear solids onto elementary linear solids formed from the basis vectors. The projection coefficients are related to the volume of the original solid by some formula. The formula generalizes the Pythagorean theorem - and must logically be a consequence of it.

[granpa]

I just gave you the higher dimensional generalization of the Pythagorean theorem.

its length not volume.

[granpa]

perhaps you can explain what you mean using a square as an example

http://202.38.126.65/navigate/math/history/Diagrams/PythagorasTheorem.gif

[lavinia]

Here is a nice paper on this

http://www.jyi.org/volumes/volume2/issue1/articles/barth.html

Basically the determinant is the general tool for calculation of a k-volume in k-space, but this paper explains doing k-volume in n-space when k<n.

Thanks for this paper. It was helpful. It does not deal with the problem of projections onto an orthonormal basis for linear solids. Do you have another reference?

[lavinia]

Ok here is a special case.

take an orthonormal basis E1 ... E4 in R^4 and two orthonormal vectors

X and Y.

X and Y span a square of area 1 and can be written as linear combinations of the squares spanned by the basis E1 ... E4.

X^Y = z12 E1^E2 + ... z34E3^E4

An ugly but straight forward computation shows that sum zij^2 = 1. ( I think)

[Ben Niehoff]

Here is a simple way to show your claim is correct. Take some parallelepiped and project it onto the basis vectors:

X_1 \wedge X_2 \wedge \ldots \wedge X_p = \sum_I a_I \; e^I

where I is some ordered multi-index, and e^I stands for the wedge product of all the e^i in I. The square of the p-volume in this parallelepiped is given by

g(X_1 \wedge X_2 \wedge \ldots \wedge X_p,X_1 \wedge X_2 \wedge \ldots \wedge X_p)

where g is the natural extension of the inner product to the p-th exterior power of our vector space. Expanding in the orthonormal basis,

g(X_1 \wedge X_2 \wedge \ldots \wedge X_p,X_1 \wedge X_2 \wedge \ldots \wedge X_p) = \sum_I \sum_J a_I a_J \; g(e^I, e^J)

The terms in the sum on the right vanish unless the multi-indices I and J are identical (this is because the basis is orthonormal; e.g. the projection of e1^e2^e3 on e1^e2^e4 is zero, etc.). Therefore only the diagonal part of the sum contributes, giving

\mathrm{Area}(X_1 \wedge X_2 \wedge \ldots \wedge X_p)^2 = \sum_I a_I^2

as desired.

[lavinia]

Here is a simple way to show your claim is correct. Take some parallelepiped and project it onto the basis vectors:

X_1 \wedge X_2 \wedge \ldots \wedge X_p = \sum_I a_I \; e^I

where I is some ordered multi-index, and e^I stands for the wedge product of all the e^i in I. The square of the p-volume in this parallelepiped is given by

g(X_1 \wedge X_2 \wedge \ldots \wedge X_p,X_1 \wedge X_2 \wedge \ldots \wedge X_p)

where g is the natural extension of the inner product to the p-th exterior power of our vector space. Expanding in the orthonormal basis,

g(X_1 \wedge X_2 \wedge \ldots \wedge X_p,X_1 \wedge X_2 \wedge \ldots \wedge X_p) = \sum_I \sum_J a_I a_J \; g(e^I, e^J)

The terms in the sum on the right vanish unless the multi-indices I and J are identical (this is because the basis is orthonormal; e.g. the projection of e1^e2^e3 on e1^e2^e4 is zero, etc.). Therefore only the diagonal part of the sum contributes, giving

\mathrm{Area}(X_1 \wedge X_2 \wedge \ldots \wedge X_p)^2 = \sum_I a_I^2

as desired.

thanks. That's very cool. I guess I was trying to prove this formula for the square of the p-volume. Will keep trying. It makes it a lot easier knowing that it is true.