'vector spaces' (1 Viewer)

[shannonm]

'vector spaces' II

show that the system S with the usual rules for addition and multiplication by a scalar in R^3, and where:
S = { x E R³ : 2x₁ +3x₂³ -4x₃² =0 },
is not a vector space


let (x1, x2, x3) and (y1, y2, y3) be in S
then
2x₁ +3x₂³ -4x₃² =0
and
2y₁ +3y₂³ -4y₃² =0

now
(x1, x2, x3) + (y1, y2, y3) = (x1 + y1, x2+y2, x3+y3)

2(x1+y1) +3(x2+y2)³ -4(x3+y3)² = 0
binomial expansion this
2x₁ + 2y₁ +3(x³₂ + 3x²₂y₂ +3x₂y₂²+y³₂) -4(x²₃+2x₃y₃+y²₃)
2x1 + 2y1 + 3x³₂ + 9x²₂y₂ +9x₂y₂²+3y³₂ -4x²₃-8x₃y³-4y²₃

(2x₁+3x₂³-4x₃²) + (2y₁+3y₂³-4y₃²) + 9x²₂y₂ +9x₂y₂² -8x₃y₃
(0)+(0)+9x²₂y₂ +9x₂y₂² -8x₃y₃

now i can get that far..
would i say: "since the above is not equal to 0, then this set is 'open' under addition, and is therefor not a vector space?
1) is the above even correct?
2) is the wording of that explanation of the answer correct?

























george's response below was to my earlier question:


hi this question is #1 in the unsw yellow book
the concept hasnt sunk in yet so im just wondering if this would be correct

Show that the set
S = {x E R³ : x₁ < / 0, x₂ >/ 0}
with the usual rules for addition and multiplication by a scaler in R³ is not a vector space.
in the question < / means less than or equal to, and
>/ means greater than, or equal to

now can i let
X = (-1,1,0)
and
LX = L(-1,1,0)
and let L be a scalar quantity (-1)
so LX = -1(-1,1,0)
=(1,-1,0)

and since x₁ is positive, therefor
x₁ is not part of set S
therefor S is not a vector space?

 

[gman03]

hi this question is #1 in the unsw yellow book
the concept hasnt sunk in yet so im just wondering if this would be correct

Show that the set
S = {x E R³ : x₁ < / 0, x₂ >/ 0}
with the usual rules for addition and multiplication by a scaler in R³ is not a vector space.
in the question < / means less than or equal to, and
>/ means greater than, or equal to

now can i let
X = (-1,1,0)
and
LX = L(-1,1,0)
and let L be a scalar quantity (-1)
so LX = -1(-1,1,0)
=(1,-1,0)

and since x₁ is positive, therefor
x₁ is not part of set S
therefor S is not a vector space?

yes, and i think posting it here is not quite appropriate...

To any admins reading this, a new session "University" under Maths would appreciated... or anything simiar

 

[shannonm]

yes, and i think posting it here is not quite appropriate...

To any admins reading this, a new session "University" under Maths would appreciated... or anything simiar

sweet thanks george

and if this isnt appropriate where do i post it?
i agree with the uni maths forum thing thou

 

[gman03]

show that the system S with the usual rules for addition and multiplication by a scalar in R^3, and where:
S = { x E R³ : 2x₁ +3x₂³ -4x₃² =0 },
is not a vector space


let (x1, x2, x3) and (y1, y2, y3) be in S
then
2x₁ +3x₂³ -4x₃² =0
and
2y₁ +3y₂³ -4y₃² =0

now
(x1, x2, x3) + (y1, y2, y3) = (x1 + y1, x2+y2, x3+y3)

2(x1+y1) +3(x2+y2)³ -4(x3+y3)² = 0
binomial expansion this
2x₁ + 2y₁ +3(x³₂ + 3x²₂y₂ +3x₂y₂²+y³₂) -4(x²₃+2x₃y₃+y²₃)
2x1 + 2y1 + 3x³₂ + 9x²₂y₂ +9x₂y₂²+3y³₂ -4x²₃-8x₃y³-4y²₃

(2x₁+3x₂³-4x₃²) + (2y₁+3y₂³-4y₃²) + 9x²₂y₂ +9x₂y₂² -8x₃y₃
(0)+(0)+9x²₂y₂ +9x₂y₂² -8x₃y₃

now i can get that far..
would i say: "since the above is not equal to 0, then this set is 'open' under addition, and is therefor not a vector space?
1) is the above even correct?
2) is the wording of that explanation of the answer correct?

Is this Q9 or Q10? if it is Q9 then it should be 2x₁ + 3x₂ - 4x₃ = <b>6</b>
If it is Q10 then I believe it is a vector space

sweet thanks george

and if this isnt appropriate where do i post it?
i agree with the uni maths forum thing thou

No worrys. Atm this place is alrite, but this stuff really isn't 4U in my opinion.. perhaps just in the UNSW thread and make a topic like "math1231 tut problems "

 

[gman03]

I was in hunt's lecture the other day and Q10 is a vector space because it passes thru the origin and also addition of any two vectors in S will also be in S (i.e. the resultant vector lies on the plane described by S)

 

[shannonm]

Is this Q9 or Q10? if it is Q9 then it should be 2x₁ + 3x₂ - 4x₃ = <b>6</b>
If it is Q10 then I believe it is a vector space

its question 2

 

[gman03]

My Bad, as i was on the other page...

I believe you are on the right track

let (x1, x2, x3) and (y1, y2, y3) be in S
then
...
now
(x1, x2, x3) + (y1, y2, y3) = (x1 + y1, x2+y2, x3+y3)

At this point, you kinda assume axiom 1 to be true, but we have not proved axiom 1 yet I think this is the reason why your answer gets to nowhere (or somewhere )

Anyway, what I did was I prove axiom 6 to be wrong and hence S is not a vector space...

To start you off: choose (x1,x2,x3) from S and let lamda as -1
What you do: show lamda * (x1, x2, x3) may not be in S

From here, your original steps will lead you to somewhere

 

[gman03]

I should have read your post carefully, in fact ingnore my previous post, your statement is right... (i.e. not close under addition)... Guilty of spamming

 

[turtle_2468]

yes, and i think posting it here is not quite appropriate...

To any admins reading this, a new session "University" under Maths would appreciated... or anything simiar

I'll msg laz about it... admins can only do threads, not forums

 

[shannonm]

thanks alot for the help george!