Question

Given vector x,
where x=<1,0,1>+t_1<2,1,2>+t_2<1,1,1>

Determine whether or not this set is a subspace of R^3

Answer 1

x is not a set, it's a vector, so the question seems screwy
...or is x supposed to be a set?

Answer 2

I think the question is telling us that vector x, is the set we are dealing with and from there, we are trying to determine if it's a subspace of R^3

Answer 3

how can a single vector be a subspace? they must mean that S is the set of all vectors x such that
x=<1,0,1>+t_1<2,1,2>+t_2<1,1,1>

right away we can ask ourselves "does this set contain the zero vector?"

Answer 4

i.e. are there any t_1 and t_2 such that \(\vec x=\vec 0\) ?

Answer 5

Ahh yes, I think you're right. The question must be asking us that if all the vectors x in the set, and if that would be a subspace

Answer 6

you will get a quick answer by proving that the set does not conatin \(\vec x=\vec0\)

Answer 7

Ok, so what I just did is solve for t_1 and t_2 to see if they existed if it contained the zero vector, and I found that t_1=-1 and t_2=1 when I set the vector equal to the zero vector

Answer 8

oh you are right, I somehow thought that the zero vector was not in this

so now it's on to proving closure under addition and scalar multiplication

Answer 9

that, I think, will show that this is not a subspace

Answer 10

x_1= <1,0,1>+t_1<2,1,2>+t_2<1,1,1>
x_2= <1,0,1>+t_3<2,1,2>+t_4<1,1,1>
is x_1+x_2 in S ?

Answer 11

I'm not sure...

Answer 12

well what is the sum of
x_1= <1,0,1>+a<2,1,2>+b<1,1,1>
x_2= <1,0,1>+c<2,1,2>+d<1,1,1>

?

Answer 13

<1+2a+b+1+2c+d,a+b+c+d,1+2a+b+1+2c+d> is that right?

Answer 14

no, just add each vector...

Answer 15

x_1= <1,0,1>+a<2,1,2>+b<1,1,1>
x_2= <1,0,1>+c<2,1,2>+d<1,1,1>
____________________________________+
x_1+x_2=<2,0,2>+(a+c)<2,1,2>+(b+d)<1,1,1>

is that in S ???

Answer 16

I think so... could we say that since a+c and b+d are constants, and we know that the zero vector is in the set, we let a+c=-2 and b+d=2, then could we say that the conditions can be met and is therefore closed under addition? :o

Answer 17

let x_3=x_1+x_2
let p=a+c
let q=b+d

then we have from above
x_3=<2,0,2>+p<2,1,2>+q<1,1,1>
^^^^^^
this part doesn't fit the form x= <1,0,1>+t_1<2,1,2>+t_2<1,1,1>, so...?

Answer 18

to be closed under addition the set must be closed for the addition of any number of vectors, regardless of which values we choose for t_1 and t_2
but that first vector *has* to be <1,0,1> or it's not of the form
x= <1,0,1>+t_1<2,1,2>+t_2<1,1,1>
make sense? there is no t_1 and t_2 we could choose such that we can add two vectors and get <1,0,1> as the first vector

Answer 19

Ohh, is it necessary for it to fit the original form? because if we had p=-2, q=2, then the zero vector can be made, or does that not matter?

Answer 20

Ahh, I see.

Answer 21

I didn't know that, thank you :)

Answer 22

the zero vector is just one criteria for the set to be a subspace
closure under addition and scalar multiplication mean "do we get the same form, i.e. another vector belonging to the set, no matter which two vectors we add or what we multiply them by?"

hence to show closure you must show that the forms of the vectors do not change

Answer 23

what's that??

Answer 24

it'a also not closed under scalar multiplication since
2x=<2,0,2>+2t_1<2,1,2>+2t_2<1,1,1>

the second two are all right since we can call 2t_1=p and 2t_2=q and get

2x=<2,0,2>+p<2,1,2>+q<1,1,1>

but that still doesn't fit the form

x=<1,0,1>+p<2,1,2>+q<1,1,1>

so it's not closed under scalar multiplication either

do you disagree @experimentX or are you just asking me to clarify?

Answer 25

I think you or me or both of us are misunderstanding the problem. this goes into vector subspaces
r(t1, t1) = <2,0,2>+2t_1<2,1,2>+2t_2<1,1,1>
-----------------------------------------
I think this is a vector ... two vectors. two vectors never span R3. technically still it's a subspace it R3 ... but from your previous Question ... I should say No to what you mean.

Answer 26

I am doubtful on my interpretation of the problem, as I admitted at the beginning.

Answer 27

well ... I am also not well prepared on this topic. I'm merely beginnig it.

Answer 28

Oh, that might be error on my part. I stated the problem if the set was in R^3, but the question asks if it was a subspace in it's appropriate R^n, but I just assumed R^3 since in the vector x, there were three components in each of the vectors

Answer 29

if I interpreted the problem correctly I am quite sure it's not a subspace for the reasons I mentioned. If I misinterpreted it, who knows :P

Answer 30

hm... well it could be a subspace in R2 maybe
what do you think experiment?

Answer 31

hold on ... let me take care of another problem first. I'll be back!!

Answer 32

nah, it should be in R3 since there are 3 vectors with 3 components. I think my answer is right, but the wording is a bit strange...

Answer 33

technically it could be subspace of any space greater than 1. just need to check for these properties. I'll be back with details.

Answer 34

It's B2, e)

Answer 35

yes, that is in R3 and I think my answer is correct. I have been known to be wrong once or twice in my life, so I'd still like to hear experimentX's thoughts ;)

Answer 36

given that,
\[ x(t_1, t_2)=<1,0,1>+t_1<2,1,2>+t_2<1,1,1>\]