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RichE
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#21
Report 14 years ago
#21
(Original post by evariste)
the space generated by the zero vector 0 satifies the conditions for a vector space with the proviso we ignore the usual scalar identity element being "1".
since 1.v=v.1=v for all v in a vector space V is the definition of scalar identity we can replace 1 by 0 in the 0-space as we have 0.0=0.0=0.
ie for all v in 0-space 0.v=v.0=v
(since only vector in 0 space is the zero space.)
The set {0} is a subspace of any vector space - irrespective of any provisos.

You seem to be talking the fact that {0} may be considered a field if you're willing to allow for the possibility that 0=1.
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evariste
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#22
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#22
(Original post by RichE)
The set {0} is a subspace of any vector space - irrespective of any provisos.

You seem to be talking the fact that {0} may be considered a field if you're willing to allow for the possibility that 0=1.
no confusion between field and vector space.confusion arose because for a moment there i forgot that any scalar multiplying the zero vector is the zero vector and not just 0
of course no proviso is needed since
k.0=0.k=0 for any scalar not just the scalar 0.
im a fool
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Gaz031
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#23
Report Thread starter 14 years ago
#23
(Original post by RichE)
Yes, so that they have the same orientation as i,j,k in that order, rather than i,k,j which is left-handed.
Does this just refer to the order in which i evaluate the vector product?

I guess you've realised there are two unit vectors normal to a and b. Well if you have the formula

axb = |a| |b| [email protected] n^

then the angle @ is measured anti-clockwise from a to b as viewed from the top of n^. (If you can picture that.)
The unit vectors are just in opposite directions?
If i end up doing it the wrong way around can i simply add or subtract 360 degrees to obtain the acute angle?
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