Show the set S is not a vector space with the usual rules of (x+)

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Bronn

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H i, so I'm working my way through course notes over the uni break and its a new topic and hard for me to clearly understand the notes as I don't have a lecturer going over them yet. So forgive my butchering of this as I'm trying to understand it.

Problem:
Show that the set S={x element of R^3: x1 <=0 , x2 >=0} is not a vector space.

unfinished solution:

So to start I wanted to test to see if it was closed under addition.

So I say let u, v be an element of S where u1 , v1 <=0 and u2 , v2 >=0

then


u + v = [ (u1+v1) , (u2 + v2) , (u3 + v3) ] = w

now if it is closed under addition then w is also an element of S. So to show w is an element of S you need to show that

(u1+v1) <= 0 and (u2 + v2) >= 0


now is it sufficient to say because u1<= 0 and v1 <= 0 then (u1+v1) <= 0 ?

or how do you prove this otherwise?

am I doing things correctly so far?

note, I haven't finished this question and proved S is not a vector space, I was just a bit stuck on the first axiom so far.

Thanks
 
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Bronn said:
H i, so I'm working my way through course notes over the uni break and its a new topic and hard for me to clearly understand the notes as I don't have a lecturer going over them yet. So forgive my butchering of this as I'm trying to understand it. problem: Show that the set S={x element of R^3: x1 <=0 , x2 >=0} is not a vector space. unfinished solution: So to start I wanted to test to see if it was closed under addition. So I say let u, v be an element of S where u1 , v1 <=0 and u2 , v2 >=0 then u + v = [ (u1+v1) , (u2 + v2) , (u3 + v3) ] = w now if it is closed under addition then w is also an element of S. So to show w is an element of S you need to show that (u1+v1) <= 0 and (u2 + v2) >= 0 now is it sufficient to say because u1<= 0 and v1 <= 0 then (u1+v1) <= 0 ? or how do you prove this otherwise? am I doing things correctly so far? note, I haven't finished this question and proved S is not a vector space, I was just a bit stuck on the first axiom so far.
Click to expand...
Do you know that in order to show that \(\displaystyle \mathscr{S}\) is a subsace of \(\displaystyle \mathscr{R}^3\) you must show that:
1) \(\displaystyle (0,0,0)\in\mathscr{S}\)
2) \(\displaystyle (\forall \alpha\in\mathcal{R}~\&~\forall \{B,C\}\subset\mathscr{S})[\alpha B+ C\in \mathscr{S}]\)

What if \(\displaystyle \alpha=-5,~B=(-2,0,0),~\&~C=(-1,2,3)\)
 
Bronn said:
H i, so I'm working my way through course notes over the uni break and its a new topic and hard for me to clearly understand the notes as I don't have a lecturer going over them yet. So forgive my butchering of this as I'm trying to understand it.

Problem:
Show that the set S={x element of R^3: x1 <=0 , x2 >=0} is not a vector space.

unfinished solution:

So to start I wanted to test to see if it was closed under addition.

So I say let u, v be an element of S where u1 , v1 <=0 and u2 , v2 >=0

then


u + v = [ (u1+v1) , (u2 + v2) , (u3 + v3) ] = w

now if it is closed under addition then w is also an element of S. So to show w is an element of S you need to show that

(u1+v1) <= 0 and (u2 + v2) >= 0


now is it sufficient to say because u1<= 0 and v1 <= 0 then (u1+v1) <= 0 ?

or how do you prove this otherwise?

am I doing things correctly so far?

note, I haven't finished this question and proved S is not a vector space, I was just a bit stuck on the first axiom so far.

Thanks
Click to expand...
The components of elements in R3 are real number. So YES, when you add two non-negative numbers (non-positive numbers) you get non-negative numbers (non-positive numbers)!!!!

But what happens when you multiply U by a real scalar. Will that be closed??!!
 
pka said:
Do you know that in order to show that \(\displaystyle \mathscr{S}\) is a subsace of \(\displaystyle \mathscr{R}^3\) you must show that:
1) \(\displaystyle (0,0,0)\in\mathscr{S}\)
2) \(\displaystyle (\forall \alpha\in\mathcal{R}~\&~\forall \{B,C\}\subset\mathscr{S})[\alpha B+ C\in \mathscr{S}]\)

What if \(\displaystyle \alpha=-5,~B=(-2,0,0),~\&~C=(-1,2,3)\)
Click to expand...

sorry i never explained it in the OP, I'm just methodically going through each individual axiom which the textbook says defines a vector space

1. closure under addition
2. associative law of addition
3. commutative law of addition
4. existence of zero
5. existence of a negative
6. closure under multiplication by scalar
7. associative law of multiplication by scalar
8. 1v=v
9. scalar distributive law
10. vector distributive law


I wanted to go through each one as an exercise even though it may be redundant to answer the question, just to lay some foundations
 
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Jomo said:
The components of elements in R3 are real number. So YES, when you add two non-negative numbers (non-positive numbers) you get non-negative numbers (non-positive numbers)!!!!

But what happens when you multiply U by a real scalar. Will that be closed??!!
Click to expand...
thanks

I haven't worked up to that one yet but by inspection no.


for 5. the existence of a negative, which I posted above.

if v is an element of S then there must be a vector -v such that v+(-v) = 0

where -v= [-v1 , -v2 , -v3 ]

but -v2 is not >= 0 therfore -v is not an element of S.


is this the correct approach?
 
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Bronn said:
thanks

I haven't worked up to that one yet but by inspection no.


for 5. the existence of a negative, which I posted above.

if v is an element of S then there must be a vector -v such that v+(-v) = 0

where -v= [-v1 , -v2 , -v3 ]

but -v2 is not >= 0 therfore -v is not an element of S.


is this the correct approach?
Click to expand...
Looks good to be. I have only one concern. You keep using bold letters for vectors (a good habit!), yet you wrote v+(-v) = 0. When you add two vectors YOU GET a vector. So, v+(-v) = (0,0,0) = 0, not 0
 
Jomo said:
Looks good to be. I have only one concern. You keep using bold letters for vectors (a good habit!), yet you wrote v+(-v) = 0. When you add two vectors YOU GET a vector. So, v+(-v) = (0,0,0) = 0, not 0
Click to expand...
ah yes, of course, thanks.

is there plugins or something to type with mathematical symbols or do you have to know the code or something?
Because I'm probably gonna have lots of posts here in the coming months lol
 
mmm4444bot said:
Check out this notice, from the News board. :cool:
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oh thanks!











one more follow up question on this problem.

It says give a geometric interpretation of your results.
I'm not really sure about this.
if x1 is less than or eqaul to 0
x2 is greater than or equal to 0
and x3 is all R

then its like a big cube covering

x axis <= 0

y axis >= 0

and z axis R ? I don't really know what it means otherwise
 
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Bronn said:
oh thanks!

one more follow up question on this problem.

It says give a geometric interpretation of your results.
I'm not really sure about this.
if x1 is less than or eqaul to 0
x2 is greater than or equal to 0
and x3 is all R

then its like a big cube covering

x axis <= 0

y axis >= 0

and z axis R ? I don't really know what it means otherwise
Click to expand...
What exactly do you mean by z axis R???? Think projection onto the x-y plane.
 
Jomo said:
What exactly do you mean by z axis R???? Think projection onto the x-y plane.
Click to expand...
well in R^3 if the axis are x,y, and z and z is the entire z axis for the set S.
Im still not sure with your clue sorry
 
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