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Linear algebra question about subspaces

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Homework Statement



This is probably a very dumb question, but I just can't wrap my head around what I'm supposed to be doing.

The question is:

"Determine whether the set is a subspace of R3:
All vectors of the form (a,b,c) where a = 2b + 3c"

Homework Equations



u + v is an element of R3
ku is an element of R3

The Attempt at a Solution



My question is.. is R3 just the set of all vectors with 3 terms..(a,b,c), (d,e,f)...etc? OR does it mean that the vector I get after multiplying by the scalar has to be of the same form as u, such that a = 2b + 3c

I think I'm phrasing this wrong so I will give an example.

Let's say I am checking vector addition.

So u = (2b + 3c, b, c) and v = (2e + 3f, e, f)

When I add them I get (2b+3c+2e+2f, b+e, c+f)

Is that in R3 since it has 3 terms which in fact would make it a vector space? Or do I need to check if the first term (2b + 3c + 2e + 2f) = 2(b+e) + 3(c+f) ?
 

Answers and Replies

  • #2
CompuChip
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Is that in R3 since it has 3 terms which in fact would make it a vector space? Or do I need to check if the first term (2b + 3c + 2e + 2f) = 2(b+e) + 3(c+f) ?
Yes and yes.
It is in R³ for the reason you state. However, this only means that the given set is a subset of R³, but you are asked to show that it is a subspace, i.e. a subset which is a vector space in its own right. For that, it needs to satisfy some additional demands, for example that if you add two vectors in the subset, the result is again in the subset. Therefore, the answer to your second question is also yes.
 
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  • #3
HallsofIvy
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The basic vector space is R3, the space of vectors of the form (a, b, c) where a, b, and c can be any real numbers, vector addition is defined as (a, b, c)+ (d, e, f)= (a+ d, b+ e, c+ f) and scalar multiplication is defined by x(a, b, c)= (xa, xb, xc).

The subset is the set of vectors of the form (a, b, c) where a, b, and c are numbers satisfying a= 2b+ 3c. No, it is not enough just to check that the sum of two such things is in R3, you must show they are still in that subset: if (a, b, c) and (d, e, f) are in that set, then a= 2b+ 3c and d= 2e+ 3f. The sum of the two vectors is (a+ d, b+ e, c+ f) and you must check that a+ d= (2b+ 3c)+ (2e+ 3f) is, in fact, (2b+ 2e)+ (3c+ 3f)= 2(b+ e)+ 3(c+ f).

Don't forget that you must also check scalar multpliction. x(a, b, c)= (xa, xb, xc) and now you must check that xa= x(2b+ 3c)= 2(xb)+ 3(xc).
 
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  • #4
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That explains it very well! I think I was just getting a little confused. Thanks very much!
 

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