anonymous
  • anonymous
prove: If {v1, v2, ...,vn} in V then also {c1v1, c2v2,..cnvn} in V for some scalar c where c1,c2.. not = 0
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmate
  • mathmate
Hint: http://www.math.niu.edu/~beachy/courses/240/06spring/vectorspace.html read especially closure properties under addition and multiplication by a constant c.
anonymous
  • anonymous
Thank U.. I have seen them, but here we have different scalars.
mathmate
  • mathmate
That is where the closure under addition comes in.

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More answers

anonymous
  • anonymous
you mean this one c · (u + v) = c · u + c · v?
mathmate
  • mathmate
If \(u\in V\) and \(v\in V\), then \(u+v\in V\)
mathmate
  • mathmate
Here what you are trying to prove is closure, i.e. a transformed vector still belong to V. There are two closure properties in the link that you need to focus on.
anonymous
  • anonymous
you mean that I have to combine between addition and scalar multiplication?
mathmate
  • mathmate
Closure: If u and v are any vectors in V, then the sum u + v belongs to V. Closure: If v in any vector in V, and c is any real number, then the product c · v belongs to V. See if you can spot them in the linked article. You need these properties to do your proof.
anonymous
  • anonymous
now I got it... Thank U very much.. I will try it.
thomas5267
  • thomas5267
What are you exactly trying to prove? If V is a finite-dimensional real/complex vector space, then it is rather easy to show that it is true. If V is something else, then it depends on what V is.
anonymous
  • anonymous
I think it is a vector space.. I am not sure.
zzr0ck3r
  • zzr0ck3r
There is not enough information to answer this.
thomas5267
  • thomas5267
Unless V is given as a vector space then it is trivial to prove it.
zzr0ck3r
  • zzr0ck3r
right, and thus it seems like that is not what we are to assume.

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