What Defines the Zero Vector in Modified Vector Space Operations?

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Homework Help Overview

The discussion revolves around defining the zero vector in a modified vector space with non-standard operations for vector addition and scalar multiplication. Participants are exploring the implications of these operations on the identification of the zero vector and the additive inverse.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants attempt to determine the zero vector by setting up equations based on the modified addition operation. There is confusion regarding the definition of the zero vector and how it differs from standard definitions in vector spaces.

Discussion Status

Some participants have provided guidance on how to approach finding the zero vector by suggesting to solve the equation derived from the modified addition operation. There is ongoing exploration of the additive identity and its properties, with participants questioning their own reasoning and seeking confirmation of their findings.

Contextual Notes

Participants express confusion about the non-standard operations and their implications for traditional vector space concepts. There is a recognition that the zero vector in this context may not be (0,0) as typically expected, leading to further investigation of the definitions involved.

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Homework Statement



If we give X = R2 the non-standard operations

(x, y,) ⊕ (x' , y') = (x + x' - 1, y + y' +2) (vector addition)
and
k~(x, y) = (kx - k + 1, ky + 2k -2) (multiplication by scalars)

then X is a real vector space.

- What is the zero vector of X?
- If v = (x, y) is in X then what is -v?

Homework Equations


The Attempt at a Solution



So attempted to answer both these questions. For the first one if you make x and y = 0 then you are left with (-1, 2) so that is what I thought the zero vector of X was. For the second answer I did (-1)x(x,y) so I got -v = (-x+2, -y-4). Just wondering if someone could help me out here and confirm if these are correct.

Thanks
 
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v1ru5 said:

Homework Statement



If we give X = R2 the non-standard operations

(x, y,) ⊕ (x' , y') = (x + x' - 1, y + y' +2) (vector addition)
and
k~(x, y) = (kx - k + 1, ky + 2k -2) (multiplication by scalars)

then X is a real vector space.

- What is the zero vector of X?
- If v = (x, y) is in X then what is -v?

Homework Equations





The Attempt at a Solution



So attempted to answer both these questions. For the first one if you make x and y = 0 then you are left with (-1, 2) so that is what I thought the zero vector of X was. For the second answer I did (-1)x(x,y) so I got -v = (-x+2, -y-4). Just wondering if someone could help me out here and confirm if these are correct.

Thanks

To get the zero vector (a,b) you need to have (x,y) ⊕ (a,b) = (x,y). Solve that for a and b.
 
LCKurtz said:
To get the zero vector (a,b) you need to have (x,y) ⊕ (a,b) = (x,y). Solve that for a and b.

So therefore (a,b) = (0,0)? Also, is my answer for the second question correct?
 
v1ru5 said:
So therefore (a,b) = (0,0)? Also, is my answer for the second question correct?

No. Show your work how you got (a,b). No point in worrying about your second question until you get the first one correct because you will need -v ⊕ v = 0 and you don't have 0 figured out yet.
 
LCKurtz said:
No. Show your work how you got (a,b). No point in worrying about your second question until you get the first one correct because you will need -v ⊕ v = 0 and you don't have 0 figured out yet.

so then would it be (x,y,) ⊕ (a,b) = (x+a-1, y+b+2) then if (a,b) = (0,0) then = (x-1, y+2). Is that how it should be? Thanks for your help!
 
LCKurtz said:
To get the zero vector (a,b) you need to have (x,y) ⊕ (a,b) = (x,y). Solve that for a and b.

v1ru5 said:
so then would it be (x,y,) ⊕ (a,b) = (x+a-1, y+b+2) then if (a,b) = (0,0) then = (x-1, y+2). Is that how it should be? Thanks for your help!

I don't know where you are getting (0,0) from. You are trying to find the additive identity -- when you ⊕ it to a vector you get that vector back. The red tells you exactly what to do.
 
LCKurtz said:
I don't know where you are getting (0,0) from. You are trying to find the additive identity -- when you ⊕ it to a vector you get that vector back. The red tells you exactly what to do.

Sorry lol I am really confused. If (x,y) ⊕ (a,b) = (x,y) then (x+a, y+b) = (x,y).Is that how it should be?
 
v1ru5 said:
Sorry lol I am really confused. If (x,y) ⊕ (a,b) = (x,y) then (x+a, y+b) = (x,y). Is that how it should be?

No. That isn't how you do ⊕. Do it correctly. Then figure out the a and b that make it work.

On another note, one thing that may be confusing you is the fact the problem as calling what you seek the "zero vector", which you are thinking of as (0,0). But this is a different vector space, and ⊕ is not + and the "zero vector" would better be called the "additive identity". And it isn't (0,0). You can see that by taking (x,y)⊕(0,0) and seeing you don't get (x,y).
 
Sorry I am really trying to understand this but I can't :(

The ⊕ symbol is basically implying vector addition. So if you want me to do (x,y) ⊕ (a,b) = (x,y) it would basically be (x+a, y+b) = (x,y). Is that right?

Sorry for not understanding but I am new to all this linear algebra stuff. Everything else is pretty easy.
 
  • #10
v1ru5 said:

Homework Statement



If we give X = R2 the non-standard operations

(x, y,) ⊕ (x' , y') = (x + x' - 1, y + y' +2) (vector addition)
and
k~(x, y) = (kx - k + 1, ky + 2k -2) (multiplication by scalars)

Thanks

v1ru5 said:
Sorry I am really trying to understand this but I can't :(

The ⊕ symbol is basically implying vector addition. So if you want me to do (x,y) ⊕ (a,b) = (x,y) it would basically be (x+a, y+b) = (x,y). Is that right?

NO! (x,y) ⊕ (a,b) is not (x+a, y+b). Your first post tells how to do ⊕. Look at your formula above. It is not standard vector addition. Use that formula.
 
  • #11
LCKurtz said:
NO! (x,y) ⊕ (a,b) is not (x+a, y+b). Your first post tells how to do ⊕. Look at your formula above. It is not standard vector addition. Use that formula.

Oh okay, so like this:

(x, y,) ⊕ (a , b) = (x + a - 1, y + b +2)

then a = 1 - x and b = -2 - y?
 
  • #12
LCKurtz said:
To get the zero vector (a,b) you need to have (x,y) ⊕ (a,b) = (x,y). Solve that for a and b.

v1ru5 said:
Oh okay, so like this:

(x, y,) ⊕ (a , b) = (x + a - 1, y + b +2)

then a = 1 - x and b = -2 - y?

No. Do exactly what I suggested in post #2, which I have quoted above.
 
  • #13
Okay so this is what you quoted:

(x, y,) ⊕ (x' , y') = (x + x' - 1, y + y' +2)

To find the zero vector (a, b), I need to do this (x, y) ⊕ (a, b). So basically I am replacing (x', y') with (a, b) which would give me

(x, y) ⊕ (a, b) = (x + a - 1, y + b - + 2)

Like that?
 
  • #14
Man, you are making this way too difficult. What part of (x,y) ⊕ (a,b) = (x,y) don't you understand? It is an equation with two sides and an = sign. And you know how to calculate the left side but you keep ignoring the right side.
 
  • #15
LCKurtz said:
Man, you are making this way too difficult. What part of (x,y) ⊕ (a,b) = (x,y) don't you understand? It is an equation with two sides and an = sign. And you know how to calculate the left side but you keep ignoring the right side.

So then (x + a - 1, y + b +2) = (x, y)?
 
  • #16
v1ru5 said:
So then (x + a - 1, y + b +2) = (x, y)?

and so...?
 
  • #17
LCKurtz said:
and so...?

(x + a - 1, y + b +2) = (x, y)

x + a - 1 = x
a - 1 = 0

y + b + 2 = y
b + 2 = 0

so (a-1, b+2) is the zero vector?
 
  • #18
Read my post #2 again. What does it say to do with the equation?
 
  • #19
LCKurtz said:
Read my post #2 again. What does it say to do with the equation?

Oh so a = 1 and b = -2 so (1, -2) is the zero vector?
 
  • #20
Now, you don't have to ask me that. I want you to check it yourself. If it is the additive identity then if you ⊕ it with any vector (c,d) you should get (c,d) back. That's what an additive identity does. Does (1,-2) do that?
 
  • #21
LCKurtz said:
Now, you don't have to ask me that. I want you to check it yourself. If it is the additive identity then if you ⊕ it with any vector (c,d) you should get (c,d) back. That's what an additive identity does. Does (1,-2) do that?

Ya so (c, d) ⊕ (1, -2) = (c + 1 - 1, d + -2 + 2) = (c, d). I get it now :) thanks. Now for the second part. So 0 + v = v or v+(-v) = 0. I know that the zero vector (is 1, -2) so:

(x, y) + -(x, y) = (1, -2)

like that?
 
  • #22
v1ru5 said:
Ya so (c, d) ⊕ (1, -2) = (c + 1 - 1, d + -2 + 2) = (c, d). I get it now :) thanks. Now for the second part. So 0 + v = v or v+(-v) = 0. I know that the zero vector (is 1, -2) so:

(x, y) + -(x, y) = (1, -2)

like that?

Start with -(x,y) = (a,b) unknown just like before. And use ⊕, not + in that equation.
 
  • #23
LCKurtz said:
Start with -(x,y) = (a,b) unknown just like before. And use ⊕, not + in that equation.

SO it would be:
-(x, y) ⊕ (a, b) = 0
(-x, -y) ⊕ (a, b) = 0
(-x + a - 1, -y + b +2) = (1, -2)
So
-x + a - 1 = 1
a = x + 2

-y + b + 2 = -2
b = y + 4

So -v = (x + 2, y + 4)
 
  • #24
so is that correct?
 
  • #25
v1ru5 said:
so is that correct?

Again, you don't have to ask me that. Check that it does what it is supposed to like you did before for the additive identity.
 
  • #26
v1ru5 said:
SO it would be:
-(x, y) ⊕ (a, b) = 0

No. I was in a hurry this morning but I presume you may have discovered by now your answer is wrong. The additive inverse of (x,y) is the vector (a,b) you can ⊕ to (x,y) and get the additive identity. That isn't what you wrote above to start.
 
  • #27
LCKurtz said:
No. I was in a hurry this morning but I presume you may have discovered by now your answer is wrong. The additive inverse of (x,y) is the vector (a,b) you can ⊕ to (x,y) and get the additive identity. That isn't what you wrote above to start.

so then since (a, b) is the inverse of (x, y) then

(a, b) ⊕ (x, y) = 0
(a + x - 1, b + y +2) = (1, -2)

a + x - 1 = 1
a = -x + 2

b + y + 2 = -2
b = -y - 4

so -v = (-x + 2, -y - 4)

I can check this by doing v⊕(-v) = 0 so
(x, y)⊕(a, b) = 0

(x + a - 1, y + b +2) = (1, -2)

x + a - 1 = 1
x = -a + 2
x = x-2+2
x-x=0

y + b + 2 = -2
y = -2 - b - 2
y = -2 + y + 4 - 2
y-y=0

So = (0, 0) so I guess I cam correct?
 
  • #28
Yes. You are catching on. So, if you like, here's an "extra credit" problem to see if you understand it all. For a vector space like ##R^2## with the usual operations, where the additive identity is (0,0) and the additive inverse of (a,b) is (-a,-b), if you multiply (a,b) by the scalar ##-1##, you get (-a,-b), which is the additive inverse of (a,b).

So I wonder, and hope that you wonder, if you multiply (a,b) in your vector space by the scalar ##-1## using your rules for scalar multiplication, do you get the additive inverse of (a,b) in your space?
 
  • #29
LCKurtz said:
Yes. You are catching on. So, if you like, here's an "extra credit" problem to see if you understand it all. For a vector space like ##R^2## with the usual operations, where the additive identity is (0,0) and the additive inverse of (a,b) is (-a,-b), if you multiply (a,b) by the scalar ##-1##, you get (-a,-b), which is the additive inverse of (a,b).

So I wonder, and hope that you wonder, if you multiply (a,b) in your vector space by the scalar ##-1## using your rules for scalar multiplication, do you get the additive inverse of (a,b) in your space?

So if (x, y) is the vector in ##R^2## then to get its additive inverse I would multiply it by ##-1## which would give me (-x, -y). So if I do (x, y) ##+## (-x, -y) I should get the zero vector (1, -2). So

(x, y) ##+## (-x, -y)
(x - x - 1, y - y +2) = (1, -2) So

x - x - 1 = 1
-1 ≠ 1

y -y + 2 = - 2
2 ≠ -2

Therefore it is not possible to multiply this vector in ##R^2## by ##-1## to get its additive inverse because we get a contradiction.
 
  • #30
v1ru5 said:
So if (x, y) is the vector in ##R^2## then to get its additive inverse I would multiply it by ##-1## which would give me (-x, -y).

Yes in ##R^2## but not in your problem. That isn't your rule for scalar multiplication.
 

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