Linear Algebra: Determine if V is a Vector space

[A--]

if u + v = 3n + r. Does that mean u = 3n and v = r?
To prove this is a vector space, you have to use the Axioms of a vector space.

[A1] Closure. The set is closed under addition
I am confused as how you would show this.
3n is an element of [MATH]N_0[/MATH] and r is an element of V. Therefore 3n + r is in V. Is this correct?

 

[Deleted member 4993]

View attachment 15069


if u + v = 3n + r. Does that mean u = 3n and v = r?
To prove this is a vector space, you have to use the Axioms of a vector space.

[A1] Closure. The set is closed under addition
I am confused as how you would show this.
3n is an element of [MATH]N_0[/MATH] and r is an element of V. Therefore 3n + r is in V. Is this correct?

if u + v = 3n + r. Does that mean u = 3n and v = r?
To prove this is a vector space, you have to use the Axioms of a vector space.

[A1] Closure. The set is closed under addition
I am confused as how you would show this.
3n is an element of N0N0 and r is an element of V. Therefore 3n + r is in V. Is this correct?

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment

 

[A--]

View attachment 15070
if u + v = 3n + r. Does that mean u = 3n and v = r?
To prove this is a vector space, you have to use the Axioms of a vector space.

[A1] Closure. The set is closed under addition
I am confused as how you would show this.
3n is an element of N0N0 and r is an element of V. Therefore 3n + r is in V. Is this correct?

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment

I did. Is there something wrong with the post?

 

[Deleted member 4993]

View attachment 15070
if u + v = 3n + r. Does that mean u = 3n and v = r?
To prove this is a vector space, you have to use the Axioms of a vector space.

[A1] Closure. The set is closed under addition
I am confused as how you would show this.
3n is an element of N0N0 and r is an element of V. Therefore 3n + r is in V. Is this correct?

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment

I did. Is there something wrong with the post?

The thread READ BEFORE POSTING tells you to share your work. This is because we need to know - what you know and we can start from there. Otherwise we have to start from very beginning - the definitions (e.g. vector space, closure, sets, elements, etc.). That could become very time consuming for you and us.

Where is work about this assignment?

 

[Dr.Peterson]

if u + v = 3n + r. Does that mean u = 3n and v = r?

No. It means only what it says, which is that the sum on the left equals the sum on the right.

The important thing is that there must exist some n that makes this true.

Have you ever seen modular arithmetic? That's what this really is: u ⊕ v = r means that u + v and r differ by a multiple of 3.

To prove this is a vector space, you have to use the Axioms of a vector space.

[A1] Closure. The set is closed under addition
I am confused as how you would show this.
3n is an element of [MATH]N_0[/MATH] and r is an element of V. Therefore 3n + r is in V. Is this correct?

Again, n is not a given number. It is not 3n + r that must be in V. What you need to show is that for any u and v, there is some n and r such that u + v = 3n + r.

 

[A--]

The thread READ BEFORE POSTING tells you to share your work. This is because we need to know - what you know and we can start from there. Otherwise we have to start from very beginning - the definitions (e.g. vector space, closure, sets, elements, etc.). That could become very time consuming for you and us.

Where is work about this assignment?

oh ok.

 

[A--]

No. It means only what it says, which is that the sum on the left equals the sum on the right.

The important thing is that there must exist some n that makes this true.

Have you ever seen modular arithmetic? That's what this really is: u ⊕ v = r means that u + v and r differ by a multiple of 3.

Again, n is not a given number. It is not 3n + r that must be in V. What you need to show is that for any u and v, there is some n and r such that u + v = 3n + r.

Yes i do see the modulo 3 now.
But how do I show it is a vector field for all u and v and there is some n and r that satisfies 3n + r?

By inspection, these axioms are satisfied:
[A1] Closure under addition and scalar multiplication
[A2] commutative w.r.t +
[A3] Associative w.r.t +
[A4] Zero vector such that u + 0 = u (is this satisfied by inspection? or Do I need to show there exists a zero vector and then add it to u?)

AXIOMS I need to show that are satisfied:
[A5] The exists an inverse element such that v + (-v) = 0 (not sure what the inverse element is here)
[A6] distributive a(b + u) = ab + au
[A7] there exist an identify element 1 such that 1v = v

 

[A--]

The thread READ BEFORE POSTING tells you to share your work. This is because we need to know - what you know and we can start from there. Otherwise we have to start from very beginning - the definitions (e.g. vector space, closure, sets, elements, etc.). That could become very time consuming for you and us.

Where is work about this assignment?

To be a vector space, a set must satisfy axioms that make a vector space;
This is what i have done so far

By inspection, these axioms are satisfied:
[A1] Closure under addition and scalar multiplication
[A2] commutative w.r.t +
[A3] Associative w.r.t +
[A4] Zero vector such that u + 0 = u (is this satisfied by inspection? or Do I need to show there exists a zero vector and then add it to u?)

AXIOMS I need to show that are satisfied:
[A5] The exists an inverse element such that v + (-v) = 0 (not sure what the inverse element is here)
[A6] distributive a(b + u) = ab + au
[A7] there exist an identify element 1 such that 1v = v

 

[Dr.Peterson]

Yes i do see the modulo 3 now.
But how do I show it is a vector field for all u and v and there is some n and r that satisfies 3n + r?

You have a number u+v. You want to find a quotient n and a remainder r. Do you know a theorem that says you can always do that?

By inspection, these axioms are satisfied:
[A1] Closure under addition and scalar multiplication
[A2] commutative w.r.t +
[A3] Associative w.r.t +
[A4] Zero vector such that u + 0 = u (is this satisfied by inspection? or Do I need to show there exists a zero vector and then add it to u?)

AXIOMS I need to show that are satisfied:
[A5] The exists an inverse element such that v + (-v) = 0 (not sure what the inverse element is here)
[A6] distributive a(b + u) = ab + au
[A7] there exist an identify element 1 such that 1v = v

[A5] Give it a try. The additive inverse of 0 is obvious; given v = 1, try adding 0, 1, and 2 to it, and see which produces 0. Same with v = 2. Then you'll know the inverses. (Note: it's not "the inverse element", but "the inverse of any particular element v or V".)

[A6] This is typically easier to show than associativity. Show us how you did the latter, and we may be able to use that to suggest how to prove distributivity.

[A7] Have you tried to see if the scalar 1 is the identity? What do you find?

 

[A--]

You have a number u+v. You want to find a quotient n and a remainder r. Do you know a theorem that says you can always do that?

I can't think of one. But is it rearranging:

((u+v)-r)/3 ? This doesn't seem right, I'm not sure what I am supposed to do.

 

[Dr.Peterson]

Would the division algorithm do the job?

 

[Steven G]

To be a vector space, a set must satisfy axioms that make a vector space;
This is what i have done so far

By inspection, these axioms are satisfied:
[A1] Closure under addition and scalar multiplication
[A2] commutative w.r.t +
[A3] Associative w.r.t +
[A4] Zero vector such that u + 0 = u (is this satisfied by inspection? or Do I need to show there exists a zero vector and then add it to u?)

AXIOMS I need to show that are satisfied:
[A5] The exists an inverse element such that v + (-v) = 0 (not sure what the inverse element is here)
[A6] distributive a(b + u) = ab + au
[A7] there exist an identify element 1 such that 1v = v

[A4] I just get the feeling that you think that the zero vector is always 0. That is why I prefer to call it the identity element instead of the 0 vector.
You need to find a single element, 0 or otherwise, such that 0 plus this element is 0, 1 plus this element is 1 and 2 plus this element is 2 and the other way around as well.

Can you please find the zero vector for *. Let the set be real number such that a*b= a+b -1? Is it 0?


[A5]Since 0 is the zero vector, the inverse of v will be some element of the set, -v, such that v + (-v) = 0. Just look at your table and see if there is an element for 0, 1 and 2 so 0+x=0 and 1+y=0 and 2+z=0.

 

[A--]

The identity element is 0 because from the table: 0 +1 = 1, 0 + 2 = 2 and 0 + 0 = 0 and their commutative

 

[Dr.Peterson]

Correct. Now, what is the inverse of 1? of 2? of 0?

But have you completed the part about closure?

 

[A--]

is the inverse n = 3/-r ?

 

[Dr.Peterson]

is the inverse n = 3/-r ?

The inverse of what? Are you saying that if r is in V, then its inverse is 3/-r? That won't even be an integer.

Look at the table, just as you did to recognize the identity. What do you add to 1 to get 0? That is the inverse of 1.

I think you are not understanding the concepts here. We may need to back up and have you tell us in words what you understand the inverse of an element to be, so we can correct any wrong ideas.

 

[A--]

The additive inverse -v in the element of V, such that v+(-v) = 0 but the additive inverse of 1 is -1 which its not an element of V

 

[A--]

Oh wait it is any two numbers that sum up to a multiple of 3 so is it v + u such that the sum = multiple of 3 ? because 1 + 2 = 3 so 3 is a multiple of 3.

 

[Dr.Peterson]

Oh wait it is any two numbers that sum up to a multiple of 3 so is it v + u such that the sum = multiple of 3 ? because 1 + 2 = 3 so 3 is a multiple of 3.

Great! Now you're thinking.

So the additive inverse of 1 is 2. What is the additive inverse of 2?

This is what you have to do. Don't think about inverses in Z; think about inverses in V, which sometimes have to be worked out one by one. When you have done this, you may see a formula you could write, and then justify it; but that isn't necessary.

 

[A--]

Great! Now you're thinking.

So the additive inverse of 1 is 2. What is the additive inverse of 2?

This is what you have to do. Don't think about inverses in Z; think about inverses in V, which sometimes have to be worked out one by one. When you have done this, you may see a formula you could write, and then justify it; but that isn't necessary.

The inverse is 3n - 1 for all n in the set N_0?