What Defines the Zero Vector in Modified Vector Space Operations?

[v1ru5]

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

 

[LCKurtz]

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.

 

[v1ru5]

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?

 

[LCKurtz]

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.

 

[v1ru5]

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]

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

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.

 

[v1ru5]

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?

 

[LCKurtz]

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).

 

[v1ru5]

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.

 

[LCKurtz]

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

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.

 

[v1ru5]

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?

 

[LCKurtz]

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

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.

 

[v1ru5]

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?

 

[LCKurtz]

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.

 

[v1ru5]

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)?

 

[LCKurtz]

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

and so...?

 

[v1ru5]

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?

 

[LCKurtz]

Read my post #2 again. What does it say to do with the equation?

 

[v1ru5]

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?

 

[LCKurtz]

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?

 

[v1ru5]

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?

 

[LCKurtz]

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.

 

[v1ru5]

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)

 

[v1ru5]

so is that correct?

 

[LCKurtz]

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.

 

[LCKurtz]

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.

 

[v1ru5]

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?

 

[LCKurtz]

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?

 

[v1ru5]

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.

 

[LCKurtz]

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.

 

[v1ru5]

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

Oh okay so

k~(x, y) = (kx - k + 1, ky + 2k -2)
-1~(x,y) = -x + 1 + 1, -y -2 - 2
= -x + 2, -y - 4

So -(x, y) = (-x + 2, -y - 4)

and that is correct since I got that as my answer when I did it the other way. So I can conclude that if I want to get the additive inverse of any vector, I can multiply that vector by scalar $-1$ using the rules for scalar multiplication given.

 

[LCKurtz]

Yes that is correct. I don't think we need to beat this horse any more. But using these techniques you could verify all the axioms of a vector space work even though its arithmetic rules may seem bizarre. To paraphrase, "This ain't your daddy's vector space".

 

[v1ru5]

Yes that is correct. I don't think we need to beat this horse any more. But using these techniques you could verify all the axioms of a vector space work even though its arithmetic rules may seem bizarre. To paraphrase, "This ain't your daddy's vector space".

Thanks for your all your help Professor LCKurtz.

 

[s_nirmit]

how do you find the -v vector then ?

 

[s_nirmit]

what is the correct awnser for -v

 

[LCKurtz]

The additive inverse of (x,y) is the vector (a,b) you can ⊕ to (x,y) and get the additive identity.

how do you find the -v vector then ?

The answer is in the thread or you can do it yourself starting with the above statement.

 

[s_nirmit]

is it -v= (1-x), (-2-y) ?

 

[LCKurtz]

is it -v= (1-x), (-2-y) ?

No. Read the thread and/or show your work.