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About LinkBacksWhich of the sollowing subsets of R^3 are subspaces of R^3? The set of all vectors of the form:
a. (a,b,c) where a=c=0
b. (a,b,c) where a=-c
c. (a,b,c) where b=2a+1
If u and v are vectors, then u+v must be in R^3 and if r is any real number then ru is also in R^3. I'm not sure what else there is. According to the two I do know, I think the first two are subspaces and the last is not because r(2a_1+1)=(2ra_1+r), which would no longer satisfy b=2a+1.
Oh the last condition is that the set must be nonempty.. but usually that's so trivial to check.
Anyway,
This is for the proofs
just take two arbitrary vectors and add them and see if you get something of the same form... then take an arbitrary vector and an arbitrary real number and do scalar multiplication.. and see if you get something of the same form..
so for the disproofs..
Like for #3
you have (a, 2a+1, c) where a and c are real numbers.
So i'd take some vector that I know is of the form (a, 2a+1, c)...
hmm let's pick.. a = 1, c = 2.
We get (1, 3, 2).
Now, if this thing is a subspace, we should be able to multiply this by any number and get something is in the set (it'd be closed under scalar multiplication).
However, multiply it by 0:
0 * (1,3,2) = (0, 0, 0). Here, a = 0, and c = 0. the middle term is 0, so that mean 2a+1 = 0.. if we solve this, a = -1/2. But we just said a = 0. We know a can't have two values. Therefore, we get a contradiction, and we know that (a, 2a+1, c) is not closed under scalar multiplication. Therefore it is not a subspace of R^3.
BTW: Later on, if you haven't learned it already. (It's been awhile, so I think this is true, but you should check me on it). I think that to be a subspace, the smaller thing has to have the same zero vector as the bigger thing. Therefore, if something is a subspace of R^3, it must contain the zero vector from R^3 (0,0,0). So, in the one I just disproved, it'd be easy to just say.. (0,0,0) is not in the set. Therefore, it's not a subspace.
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