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 March 15th, 2018, 04:51 AM #1 Newbie   Joined: Aug 2017 From: Zimbabwe Posts: 8 Thanks: 0 Question on vector spaces Could we have two vector spaces, each with its own set of basis vectors, but these basis vectors are related according to the following way? A particular set of vectors in the first vector space may exist "all over the place", but when you represent the same information in the second vector space, the discrete vectors in the first space can still be made out in the second space, but line up end to end to form one composite vector in it. Last edited by skipjack; March 15th, 2018 at 09:03 AM.
 March 15th, 2018, 04:52 AM #2 Newbie   Joined: Aug 2017 From: Zimbabwe Posts: 8 Thanks: 0 I was thinking the first set of basis vectors have to be factors of the second.
 March 24th, 2018, 03:00 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 You seem to have only a vague idea of what a "vector space" is. The only operations in a vector space are vector addition and scalar multiplication. You cannot talk about "factors" of vectors nor does it make sense to talk about vectors "end to end"..
May 17th, 2018, 08:23 AM   #4
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Quote:
 Originally Posted by Country Boy You seem to have only a vague idea of what a "vector space" is. The only operations in a vector space are vector addition and scalar multiplication. You cannot talk about "factors" of vectors nor does it make sense to talk about vectors "end to end"..

I went back and tried to study a bit more ...but with youtube videos, so forgive me if I am still naive...

I have a few questions that I would appreciate answers for

If I have a vector space of the following form. There is a multidimensional space that these vectors live on, and a particular matrix formed from some vectors has a determinant of zero.

Now are we able to apply curvature to the vector space in order to increase the value of the determinant for zero to something positive? And how would we do this?

Last edited by skipjack; June 18th, 2018 at 08:40 PM.

 June 18th, 2018, 11:25 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 "Curvature" is not defined for general vector spaces. The determinant formed by taking n vectors in $\displaystyle R^n$ is zero if and only if the vectors are linearly dependent. There is no way to make that determinant non-zero. Last edited by skipjack; June 18th, 2018 at 08:41 PM.

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