Pythagorean Theorem: Relationships in Euclidean Space

In summary, the Pythagorean theorem can be generalized to higher dimensions by using an orthonormal basis for linear solids. This can be seen by projecting a linear solid onto the basis vectors and calculating its hypervolume using the sum of squares of projection coefficients. The determinant is a useful tool for this calculation, as shown in a referenced paper. The formula for the square of the p-volume in a parallelepiped is a special case of this generalization.
  • #1
lavinia
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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.
 
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  • #2
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.
 
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  • #3
I really don't know what you mean.

in higher dimensions
Length^2 = a^2 + b^2 + c^2 + d^2...length not volume
 
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  • #4
granpa said:
I really don't 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.
 
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  • #5
I just gave you the higher dimensional generalization of the Pythagorean theorem.

its length not volume.
 
  • #6
perhaps you can explain what you mean using a square as an example

[URL]http://202.38.126.65/navigate/math/history/Diagrams/PythagorasTheorem.gif[/URL]
 
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  • #7
BruceG said:
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?
 
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  • #8
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)
 
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  • #9
Here is a simple way to show your claim is correct. Take some parallelepiped and project it onto the basis vectors:

[tex]X_1 \wedge X_2 \wedge \ldots \wedge X_p = \sum_I a_I \; e^I[/tex]

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

[tex]g(X_1 \wedge X_2 \wedge \ldots \wedge X_p,X_1 \wedge X_2 \wedge \ldots \wedge X_p)[/tex]

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,

[tex]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)[/tex]

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

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

as desired.
 
  • #10
Ben Niehoff said:
Here is a simple way to show your claim is correct. Take some parallelepiped and project it onto the basis vectors:

[tex]X_1 \wedge X_2 \wedge \ldots \wedge X_p = \sum_I a_I \; e^I[/tex]

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

[tex]g(X_1 \wedge X_2 \wedge \ldots \wedge X_p,X_1 \wedge X_2 \wedge \ldots \wedge X_p)[/tex]

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,

[tex]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)[/tex]

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

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

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.
 
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Related to Pythagorean Theorem: Relationships in Euclidean Space

1. What is the Pythagorean Theorem?

The Pythagorean Theorem is a mathematical theorem that states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

2. Who discovered the Pythagorean Theorem?

The Pythagorean Theorem is named after the ancient Greek mathematician Pythagoras, although it is possible that it was known by other ancient civilizations as well.

3. How is the Pythagorean Theorem used in real life?

The Pythagorean Theorem is used in many real-life situations, such as in construction, architecture, and engineering. It can be used to calculate the length of a diagonal in a rectangular room, the height of a building, or the distance between two points on a map.

4. Is the Pythagorean Theorem only applicable to right triangles?

Yes, the Pythagorean Theorem only applies to right triangles. It will not work for other types of triangles, such as equilateral or isosceles triangles.

5. Are there any other versions of the Pythagorean Theorem?

Yes, there are several other versions of the Pythagorean Theorem, including the Law of Cosines and the Law of Sines. These versions can be used to solve for angles and sides in non-right triangles.

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