MHB Definition of SO(n): O(n) & SL(n, \mathbb{R}) Intersection

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The discussion clarifies the definitions of various matrix groups, specifically O(n), GL(n, \mathbb{R}), and SL(n, \mathbb{R}), leading to the conclusion that SO(n) can be defined as the intersection of SL(n, \mathbb{R}) and O(n). The participant initially questioned whether defining SO(n) as the set of orthogonal matrices with determinant 1 would be more logical, but recognized that both definitions are indeed equivalent. The conversation emphasizes the importance of understanding these definitions in the context of linear algebra and matrix theory. Additionally, it highlights how mathematical terminology can convey multiple properties simultaneously, similar to descriptive language in English. Overall, the definitions and their relationships are crucial for grasping the structure of these mathematical groups.
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Four definitions:
1) Define [math]M_n( \mathbb{R} )[/math] as the set of all n x n matrices over [math]\mathbb{R}[/math].

2) Define [math]O(n) = \{ A \in M_n ( \mathbb{R} ) | A A^T = I \}[/math]

3) Define [math]GL(n, \mathbb{R} ) = \{ A \in M_n( \mathbb{R} ) | \det A \neq 0 \}[/math]

4) Define [math]SL(n, \mathbb{R} ) = \{ A \in GL(n, \mathbb{R} ) | \det A = 1 \}[/math]

(I presume these are all standard, but the way the definitions are made plays into my question.)

My source now defines [math]SO(n) = SL(n, \mathbb{R} ) \cap O(n)[/math]

As typical I was able to sort this out as I typed it. My question was that wouldn't it make more sense to define [math]SO(n) = \{ A \in O(n) | \det A = 1 \}[/math] but now I see that both definitions are equivalent.

So no worries!

-Dan
 
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topsquark said:
Four definitions:
1) Define [math]M_n( \mathbb{R} )[/math] as the set of all n x n matrices over [math]\mathbb{R}[/math].

2) Define [math]O(n) = \{ A \in M_n ( \mathbb{R} ) | A A^T = I \}[/math]

3) Define [math]GL(n, \mathbb{R} ) = \{ A \in M_n( \mathbb{R} ) | \det A \neq 0 \}[/math]

4) Define [math]SL(n, \mathbb{R} ) = \{ A \in GL(n, \mathbb{R} ) | \det A = 1 \}[/math]

(I presume these are all standard, but the way the definitions are made plays into my question.)

My source now defines [math]SO(n) = SL(n, \mathbb{R} ) \cap O(n)[/math]

As typical I was able to sort this out as I typed it. My question was that wouldn't it make more sense to define [math]SO(n) = \{ A \in O(n) | \det A = 1 \}[/math] but now I see that both definitions are equivalent.

So no worries!

-Dan

This takes advantage of a special construction of English: the juxtaposition of two adjectives means both apply. That is saying "the tall English dude" is the same as saying:

The dude is English, and the dude is tall. In Mathese:

$\text{Dude} \in \{\text{tall dudes}\} \cap \{\text{English dudes}\} \iff \{\text{Dude} \in \{\text{English dudes}\}: \text{Dude} \in \{\text{tall dudes}\}\}$

In the case at hand: "special" means having determinant 1, and "orthogonal" means having the transpose as the inverse. The corresponding noun to "dude" in my example above is "linear (group)" (the group of UNITS of the ring $M_n(\Bbb R)$).

This is quite common, for example one may specify the set of all real numbers which are positive integers. While it is natural to interpret this as: first restrict to integers, and then restrict to positive integers, it is also possible to take the intersection of the positive reals with the integers.
 
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