Question

Explain whether the set is a subspace of R^3 if it is not indicate which closure properites failed: d) All vectors (a,b,c) for which a+b+c=1

Answer 1

I don't understand subspaces that well, do not know why this is not a subspace could someone explain

Answer 2

or would it be a subspace if all vetors of form (a, -a, c)

Answer 3

a subspace of a vector space needs
1.) non-empty (show it contains the 0 vector)
2.) closure under vector addition
3.)closure under scaler multiplication

Answer 4

the set of all vectors (a,b,c) such that a+b+c = 1 would not be a subspace because
1.) (0,0,0) is not in the set
2.) if (a,b,c) and (d,e,f) are in the set, a+b+c = 1 and d+e+f = 1 but (a+d, b+e, c+f) would be a+d+b+e+c+f = 2

Answer 5

Ask more questions like this, Asllsmiles. Fun math.

Answer 6

I am quite confused sorry

Answer 7

So are you saying that a b or c must be 0?

Answer 8

in order for it to contain a zero vector?

Answer 9

they would have to be for (0,0,0) to be in the set, but then a+b+c would not equal 1, so it doesn't have the right form.

Answer 10

Okay but what were you talking about with d e and f?

Answer 11

And if in R^3 a b and c have to be 0,0,0 is that the only possible solution that could be in R^3?

Answer 12

suppose you have two vectors in the set, (a,b,c) and (d,e,f). Then
a+b+c = 1 and d+e+f=1
but when you add the two vectors you get
(a,b,c)+(d,e,f) = (a+d, b+e, c+f).

You need for this new vetor to have the same form as the other two, namely
a+d + b + e +c+f =1
but we can see that
a+d + b + e +c+f =2

So the new vector is not in the set. Thus the set is not closed under addition.

Answer 13

as for the zero vector... A subspace has to be a vector space itself, so the zero vector has to be in the subspace.

Answer 14

This concept's hard to understand. I sort of get what you're saying but I don't see how you just got d e f? I see how it's equal to 2. When you said closed under addition so when you add those vectors up would they have to be 0

Answer 15

So that you include the zero vector or am I just getting this completely wrong

Answer 16

wait... let's start from the beginning.
the set of vectors you are interested in is
\[\{(a,b,c) \in \mathbb{R}^3 | a+b+c = 1\}\]

Answer 17

For this to be a subspace or R^3 you need to show that
1.) (0,0,0) is in the set
2.) adding any two vectors in the set gives you another vector that is also in the set
3.)multiplying any vector in the set by a number gives you a vector that is also in the set.

Answer 18

Okay that make sense so far

Answer 19

ok...

(0,0,0) can't be in the set since 0 + 0 + 0 not = 1

and yes, the only way to get (0,0,0) if for a=0, b=0, and c=0

Answer 20

Okay but could you have other values for a b c and still maintain a zero vector? So like if I have a=1 b=-1 c=0?

Answer 21

No if a=1, b=-1, c=0 then the vector would be (1,-1,0) which is not the zero vector.

Answer 22

zero vector for R^3 = (0,0,0)
uniquely

Answer 23

give me two examples of vectors that are in the set
\[\{(a,b,c) \in \mathbb{R}^3 | a+b+c = 1\}\]

Answer 24

1 0 0? I just don't see any other solutions because you said that you MUST have a (0,0,0)?

Answer 25

or else it wouldn't be a subspace as it's not non empty?

Answer 26

Sorry for the confusion

Answer 27

It's ok.... for the above set to be a subspace or R^3 it MUST have (0,0,0) in it.... but it doesn't .... so it is NOT a subspace of R^3 :)

Answer 28

Oh okay! But is that the ONLY solution. (0,0,0) to be a subspace of R^3?

Answer 29

Okay so this would work b=a+c because you can get 0

Answer 30

(0,0,0) is a subspace of R^3. But I don't think you meant that.

For any set of vectors to be a subspace of R^3, you must be able to show that (0,0,0) would be in it.

--------------------------------

Ok.... any set of vectors of the form (a,a+c,c)

Is (0,0,0) of that form??? Yes. So (0,0,0) is in the set.

But we haven't show that it's a subspace yet, just that the first condition is met.

Answer 31

Okay I think I understand now, thanks for all your help!!!