OpenStudy (anonymous):

prove: If {v1, v2, ...,vn} in V then also {c1v1, c2v2,..cnvn} in V for some scalar c where c1,c2.. not = 0

2 years ago
OpenStudy (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.

2 years ago
OpenStudy (anonymous):

Thank U.. I have seen them, but here we have different scalars.

2 years ago
OpenStudy (mathmate):

That is where the closure under addition comes in.

2 years ago
OpenStudy (anonymous):

you mean this one c · (u + v) = c · u + c · v?

2 years ago
OpenStudy (mathmate):

If \(u\in V\) and \(v\in V\), then \(u+v\in V\)

2 years ago
OpenStudy (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.

2 years ago
OpenStudy (anonymous):

you mean that I have to combine between addition and scalar multiplication?

2 years ago
OpenStudy (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.

2 years ago
OpenStudy (anonymous):

now I got it... Thank U very much.. I will try it.

2 years ago
OpenStudy (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.

2 years ago
OpenStudy (anonymous):

I think it is a vector space.. I am not sure.

2 years ago
OpenStudy (zzr0ck3r):

There is not enough information to answer this.

2 years ago
OpenStudy (thomas5267):

Unless V is given as a vector space then it is trivial to prove it.

2 years ago
OpenStudy (zzr0ck3r):

right, and thus it seems like that is not what we are to assume.

2 years ago