Confused on why this subspace spans R^3, matrices fun

[mr_coffee]

I'm studying for my exam on monday and i saw ap roblme that says:
Determine if the following is a subspace of R^3.
For it to be a subspace it has to pass
1) must have the zero vector
2) closure under addition
3) closure under multiplaication

here is my work:
http://img206.imageshack.us/img206/8067/lastscan1qn.jpg


now it fails because it r1+r1-2 right? but why? because why wouldn't
3(s1+s2) also fail? because ur getting a different number if u like say, let s1 and s2 be 1, or r1 and 2 be 1, so did they both fail? Also if there was say

What if there were only 2 variables, only r and s, and U = [r s r], r & s are in R, now would this automaticallya not be a subpsace in R^3?

 

[LeonhardEuler]

I'm studying for my exam on monday and i saw ap roblme that says:
Determine if the following is a subspace of R^3.
For it to be a subspace it has to pass
1) must have the zero vector
2) closure under addition
3) closure under multiplaication
here is my work:
http://img206.imageshack.us/img206/8067/lastscan1qn.jpg
now it fails because it r1+r1-2 right? but why? because why wouldn't
3(s1+s2) also fail? because ur getting a different number if u like say, let s1 and s2 be 1, or r1 and 2 be 1, so did they both fail?

Actually, it doesn't fail. The problem said that r,s, and t were real numbers. So, if r1 and r2 are real numbers, then (r1+r2-1) is a real number. Call it R. Then r1+r2-2=R-1, which means that this component of the vector is ok.

Also if there was say
What if there were only 2 variables, only r and s, and U = [r s r], r & s are in R, now would this automaticallya not be a subpsace in R^3?

No, it would still be a subspace. Being a subspace of a vector space means being a subset that is also a vector space (with the same rules of addition and scalar multiplication as the original vector space). So the x-y plane is a subspace of R^3, as is the x-axis.

 

[mr_coffee]

LeonhardEuler, thanks for the reply! My professor did this problem,and we are having an exam on monday, and he told us it fails, because r1+r2-1 isn't equal to r1+r2-2, he said u cna't get the same number so it fails, is he wrong?

 

[LeonhardEuler]

Well, before I go saying that the professor is wrong, I would need to know the exact wording of the problem. As I understand it, r is allowed to be any real number and vectors in the subspace are allowed to have the first component equal to (r-1). Is that the case?

 

[mr_coffee]

He really didn't expand on what it said, all it says is:
Determine if the follwing is a subpsace of R^3. then the problem says
U = {
r-1
3s
t,

r,s,t in R }

So they are all real numbers.

 

[matt grime]

I think the question is badly phrased (either by you or your teacher). As a set the vectors given are all of R^3, and so that subset is certainly a subspace of R^3. Have you failed to define something properly?

In any case, what he's trying to get at is that whilst it is a vector space it is not a subspace since we have to define a different addition on the first component, that is the addition of x+y is different if done in R^3 or done in the set itself. Ie the operation u@v=u+v-1 on the underyling set R is an isomorphism of additive groups but does not identify R with a subgroup of itself under addition.

 

[mr_coffee]

How is it suppose to be phrased if you want it to fail in the case of r1-r2 -2, because that's how he thought it failed, so what would u have to add to the directions for that case to occur? (failure under closure of addition)

 

[matt grime]

Unless you define the operations on the vectors it fails to be well phrased since as a set that is clearly all of R^3! You need to explain how the description (parametrization) of the vectors in terms of r,s,t interacts with the operations.

It is sort of clear what he's getting at...

 

[HallsofIvy]

How is it suppose to be phrased if you want it to fail in the case of r1-r2 -2, because that's how he thought it failed, so what would u have to add to the directions for that case to occur? (failure under closure of addition)

If, for example your vectors were of the form (r-1, s, r) where the first component was always 1 less than the third, then it would not be a subspace because (1, 1, 2) is in that set and so is (2, 5, 3). But their sum, (3, 6, 5) is not in the set: 3= 5-2 not 5-1. Of course, here the zero vector (0, 0, 0) is also not in the set.

In the example you first gave, (r-1, 3s, t), if r, s, and t can be any numbers, let x= r-1, y= 3s, z= t and you have (x, y, z) where x, y, and z can be any numbers: that's all of R³. The set you gave spans all of R³, which, interestingly, is what you title this thread.
If the problem was to show that this set was NOT a subspace, why did you title the thread "confused on why this set spans R³"?

 

[mr_coffee]

Thanks Ivey, cleary my professor is wrong, becuase no the problem wasn't to show that this set was NOT a subspace the question says:
Determine if the following is a subspace of R^3. and yes r,x, and t can be any real number. Because the problem says: r,s,t in R. Also he said the set of vectors (r-1,3s,t) does contain the zero vector shows that he didn't set any pre-conditions that says, the first component is less than the thrid. So as a general rule, can you say, as long as the addition makes sense, it passes under closure of addition? like as you showed,
3 = 5-2 not 5-1.

Note: the reason i asked, "so what would u have to add to the directions for that case to occur? (failure under closure of addition)" was becuase i was trying to see if my professor just forgot to tell us somthing, but that wasn't the case.

 

[HallsofIvy]

"can you say, as long as the addition makes sense, it passes under closure of addition?"

I'm not sure what you mean by "the addition makes sense". In the example I gave, (r-1, s, r), didn't the addition "make sense"?

By the where, where did you get "Ivey"?

 

[mr_coffee]

hah Sorry, I don't know why I put an 'e' in there. When i said, "the adddition make sense" i ment, like 5-2 = 3, but 5-1 is not equal to 3, so if u made 2 vector sets like I did above to prove closure under addition works, how does this prove it?
(r1-1, 3s1, t1) + (r2-1,3s2,t2);
(r1+r2-2 ,3(s1+s2), t1+t2);

So since its in the same form as
(r-1,3s,t) it works? but how is it in the same form, if its not longer
r-1, but insteed, R-2? if R = r1+r2?

THe more i look into this the more i get confused!

Like.
let r = 0, u get -1 for the case where (r-1,3s,t).
for case 2:
r1+r2-2
r1 = 0, r2 = 0, u get -2
not -1!