Proving Vector Spaces: Get Expert Help Now!

[Chadlee88]

i'm really really confused abt vector spaces and how to prove
if something is a vector space

Could som1 please help!

Example: why is V = {(1 2)} not a vector space??

 

[quasar987]

The set V is a vector space if it satisfies all the properties* of vector space. If anyone of those property is not met, then V is automatically not a vector space. For your exemple V={(1,2)}, i.e. the set containing only the vector (1,2), the property 3 is not satisfied because, the only element of V is (1,2), and (1,2)+(1,2) = (2,4) $\neq (1,2)$. For 3 to be satisfied, V should contain the null vector (0,0) also. *the 8 properties of a vector space: http://en.wikipedia.org/wiki/Vector_space#Formal_definition

 

[Chadlee88]

I think i get how to solve them now but for this example I'm not too sure if I'm right, could u please tell me if I'm wrong:

Question: Determine whether the subsets of R^3 given by all vectors of the following form are vector spaces Note: elem = elements of

a) {a,0,b)|a,b elem R}

Let v = (a1,0,b1) where v elem V

v+0 = v Therefore it passes on the 3rd property

Let w = (a2,0, b2)

v + w = (a1,0,b1) + (a2,0,b2) = ((a1+a2) + 0 + (b1+b2)) #

Since (a1+a2) elem R, 0 elem R and (b1+b2) elem R # is of the form
(a,0,b) where a,b elem R, the vector space is closed under addition.

If m elem R then
m x v = (a1m, 0, b1m) where (a1m,0,b1m) is of the form (a,0,b) where a,b elem R. the vector space is closed under scalar multiplication.

Therefore all vectors of the form {a,0,b)|a,b elem R} are vector spaces.

 

[HallsofIvy]

I was tempted to be very hard nosed and answer that a set cannot be a vector space! A vector space is a set of objects (called "vectors") together with two operations: addition of vectors and scalar multiplication. Of course, I understand that you meant "with the usual operations on R², but you should understand the difference.
In addition to the fact that there is no 0 vector, the set containing the single vector (1, 2) is not a vector space because adding vectors (the only possible addition is (1, 2)+ (1, 2)) gives (2, 4) which is not in the set. Even more simply, scalar multiplication of vectors:
2(1, 2)= (2, 4), 3(1, 2)= (3, 6), -1.5(1, 2)= (-1.5, -3) gives vectors not in the set.
On the other hand, {(a, 2a)}, the set of multiples of (1, 2) where a can be any real number is a vector space- a one-dimensional subspace of R².

a) {a,0,b)|a,b elem R}

Let v = (a1,0,b1) where v elem V

v+0 = v Therefore it passes on the 3rd property

No, that's not the "3rd property" (existence of an additive identity; the zero vector). The question is whether or not the 0 vector is of the form (a, 0, b). The answer is still "yes" because a and b can be any real numbers-including 0. (0, 0, 0)= (a, 0, b) with a=0, b=0.

Let w = (a2,0, b2)

v + w = (a1,0,b1) + (a2,0,b2) = ((a1+a2) + 0 + (b1+b2)) #

Typo here: you mean ((a1+a2, 0, (b1+ b2)), not "+".

Since (a1+a2) elem R, 0 elem R and (b1+b2) elem R # is of the form
(a,0,b) where a,b elem R, the vector space is closed under addition.

If m elem R then
m x v = (a1m, 0, b1m) where (a1m,0,b1m) is of the form (a,0,b) where a,b elem R. the vector space is closed under scalar multiplication.

Therefore all vectors of the form {a,0,b)|a,b elem R} are vector spaces.

Yes, that is correct. Strictly speaking, the operations must also satisfy such things as the associative law, commutative law, etc. but since the operations are just the standard operations on R³ we already know they are satisfied. What you have done is show that this set, with the usual operations, is a (2-dimensional) subspace of R³.
To prove a "subspace", you really only need to show that addition and scalar multiplication are closed- that the result is still in the set. It isn't necessary to show that the 0 vector is in the set separately- that follows from closure of the operations. If v is a vector in the set, then, since scalar multiplication is closed, (-1)v= -v is in the set and, since addition is closed, v+ (-v)= 0 is in the set.

Now, what if the problem had been {(a, 1, b)} where a, b are real numbers. Can you show that this is not a vector space?