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> space as we know it is three dimensional because you can put three straight lines to meet at right angles to each other, and no more. To continue in the spir

> space as we know it is three dimensional because you can put three straight lines to meet at right angles to each other, and no more. To continue in the spirit of your pedantry, it’s not really important that the lines meet at right angles; to say that space is three dimensional is just to say that a point in space can be uniquely expressed by specifying three numbers: the ‘coordinates’ of the point. What you say is true, but dimensionality in general doesn’t require the notion of ‘angle’ to make sense. In vector space/linear algebra terms it’s to say that we live in three dimensional space because 3 is the size of a maximal linearly independent set of vectors, or alternatively the size of a minimal spanning set or vectors. They need not be orthogonal, but orthogonality does imply linear independence (it’s stronger).

lupire 3y ago
Can you give an example of any n-dimensional system where you can't make a basis of n perpendicular vectors? "Right angle" isn't any kind of angle, it's the angle that means "perpendicular" . You don't necessarily need any other angles besides 0 (parallel) and right (perpendicular) in a system.
- xanderlewis 3y ago
  > Can you give an example of any n-dimensional system where you can't make a basis of n perpendicular vectors? Well, yes. In a totally general vector space (in the mathematical sense of a vector space over a field; it’s the same notion as used in physics but perhaps more general) there’s no such thing as ‘perpendicular’ (aka ‘orthogonal’) — you need an inner product space for that. Of course, (at least in finite dimensions) you can always define such an inner product and you can certainly create orthogonal bases (see: Gram-Schmidt process), but my point was that the notion of dimension doesn’t require it to make sense. Indeed it’s not part of the usual definition; we only need to be able to talk about sets of vectors and their spans.
roywiggins 3y ago
Yeah, I had a bit in there at first on linear independence but I couldn't fit it in without feeling like I was digressing!

lupire 3y ago

Can you give an example of any n-dimensional system where you can't make a basis of n perpendicular vectors? "Right angle" isn't any kind of angle, it's the angle that means "perpendicular" . You don't necessarily need any other angles besides 0 (parallel) and right (perpendicular) in a system.

xanderlewis 3y ago
> Can you give an example of any n-dimensional system where you can't make a basis of n perpendicular vectors? Well, yes. In a totally general vector space (in the mathematical sense of a vector space over a field; it’s the same notion as used in physics but perhaps more general) there’s no such thing as ‘perpendicular’ (aka ‘orthogonal’) — you need an inner product space for that. Of course, (at least in finite dimensions) you can always define such an inner product and you can certainly create orthogonal bases (see: Gram-Schmidt process), but my point was that the notion of dimension doesn’t require it to make sense. Indeed it’s not part of the usual definition; we only need to be able to talk about sets of vectors and their spans.

roywiggins 3y ago

Yeah, I had a bit in there at first on linear independence but I couldn't fit it in without feeling like I was digressing!