Are Vectors Always Positive in a 1-Dimensional System?

[dE_logics]

I know they don't...there is no reason.

If they did have a sign, values would have been positive only on the first quadrant (2-d).

But suppose we have a vector B...if another vector is radically opposite in direction but same in magnitude, we call it -B, then what's this sign?

 

[pbandjay]

I don't really see why not.

Let's use x = (x_1,x_2). If we want a vector that points in the opposite direction with the same magnitude we can just use scalar multiplication of vectors:

(-x_1,-x_2) = (-1)(x_1,x_2) = -(x_1,x_2) = -x

Then -x is, in a sense, the negation of vector x but only of x and vectors equal to x.

 

[D H]

What is the sign of the vector (1,-1)?

 

[Klockan3]

What is the sign of the vector (1,-1)?

You do not assign a sign to it, instead if you say that B=(1,-1) then -B=-(1,-1)=(-1,1). If then A=(-1,1), then A=-B and B=-A. Minus isn't an object, it is an operation which mirrors the vector in its origin. Like, in reality we just have the numbers 0 to infinity, the negative numbers are just mirrored using the minus operator. But you can still say that A=-5, then -A=5, it is exactly the same thing as you are doing with vectors.

 

[dE_logics]

What do you mean by (a,b)...those are coordinates right?

 

[D H]

Back to the original post.

But suppose we have a vector B...if another vector is radically opposite in direction but same in magnitude, we call it -B, then what's this sign?

-B is the additive inverse of the vector B. That minus sign does not "belong" to B, it is an operation applied to B. Compare to 1 versus -1. Like B and -B, 1 and -1 are additive inverses of each other. However, the minus sign attached to -1 is an integral part of the number "-1". Real numbers have a sign. Vectors don't.

 

[tiny-tim]

Vectors are elements of a vector space,

and one of the conditions for a vector space is that anything may be multiplied by a scalar,

and "scalar" means any real number (ok, technically we can also have complex vector spaces, but don't bother with them) …

so for any vector B, there's a vector kB, for any real number k.

It isn't just k = ±1 …

k isn't a sign.

 

[D H]

Vectors are elements of a vector space,

and one of the conditions for a vector space is that anything may be multiplied by a scalar,

and "scalar" means any real number (ok, technically we can also have complex vector spaces, but don't bother with them) …

so for any vector B, there's a vector kB, for any real number k.

It isn't just k = ±1 …

k isn't a sign.

We write -B rather than -1*B because the existence of an additive inverse is also a condition for a space to be a vector space. The field of scalars might even not have an element called "-1". Think of a vector of Booleans.

 

[tiny-tim]

Hi D H!

We write -B rather than -1*B because the existence of an additive inverse is also a condition for a space to be a vector space. The field of scalars might even not have an element called "-1". Think of a vector of Booleans.

ah, but we're talking about a space with four quadrants, and I'm pretty sure the Booleans all live in a circular wormhole, where they survive by pure logic …

… values would have been positive only on the first quadrant (2-d). …

 

[D H]

That elements of a set can be multiplied by a scalar is one prerequisite for calling the elements of that set "vectors". However, we write the additive inverse of a vector B as -B rather than -1*B. Note that -2B is obviously shorthand for -2*B, and it looks perfectly natural to write -2B. We don't do that with -1*B. -1B looks incredibly clunky! We write -B instead, and this is not just shorthand for -1*B.

There is an even more basic prerequisite than multiplication by a scalar that must be satisfied in order to call the elements of that set "vectors": The set must form an abelian group with addition as the operator. The standard symbol for the inverse operator for addition is to prefix the element with a minus sign. In other words, -B.

 

[HallsofIvy]

One proves that, for vector spaces, "-1 times v" is the same as "the additive inverse of B", of course. They are not defined as the same thing.

As for the "sign" of an object, that normally implies sets of "positive" and "negative" objects which in turn implies an order. A vector space of dimension greater than 1 cannot be ordered.

 

[dE_logics]

Back to the original post.


-B is the additive inverse of the vector B. That minus sign does not "belong" to B, it is an operation applied to B. Compare to 1 versus -1. Like B and -B, 1 and -1 are additive inverses of each other. However, the minus sign attached to -1 is an integral part of the number "-1". Real numbers have a sign. Vectors don't.

So doing this is not right -

Suppose we have a vector B

Assume C = -B (this step should be wrong)

Then if B = 3

Then C = -3

 

[dE_logics]

Vectors are elements of a vector space,

and one of the conditions for a vector space is that anything may be multiplied by a scalar,

and "scalar" means any real number (ok, technically we can also have complex vector spaces, but don't bother with them) …

so for any vector B, there's a vector kB, for any real number k.

It isn't just k = ±1 …

k isn't a sign.

BTW that's only a scaler value, so it obviously doens't have any sign.

So -1 * B should not exist...since, obviously -1 is a scaler and should not have a sign.

 

[tiny-tim]

BTW that's only a scaler value, so it obviously doens't have any sign.

So -1 * B should not exist...since, obviously -1 is a scaler and should not have a sign.

??

All numbers (positve zero or negative) are scalars.

 

[D H]

Scalars, at least in the context of vector spaces, most certainly can have a sign. The scalars must be a field: Something with addition and multiplication, identity elements (for addition and multiplication (e.g., zero and one), and additive and multiplicative inverses.

 

[HallsofIvy]

BTW that's only a scaler value, so it obviously doens't have any sign.

So -1 * B should not exist...since, obviously -1 is a scaler and should not have a sign.

? A "scalar" is a member of the field the vector space is defined over. In many applications, the field is the field of rational numbers or field of real numbers. And those certainly do have signs!

 

[Borek]

Is scaler a scalar used for scaling?

 

[dE_logics]

Ok.

But when we take suppose, motion in one dimension, the displacement/velocity/acceleration are vectors, yet they have a sign...most probably cause they are taken WRT another inverse vector.

But problem is we actually use those sign conventions in the numerical analysis...shouldn't that give the wrong answer?

Cause as I said before...doing this should be wrong -

Suppose we have a vector B

Assume C = -B (this step should be wrong)

Then if B = 3

Then C = -3

So vector C has a negative value.

 

[dE_logics]

Is scaler a scalar used for scaling?

 

[D H]

Is scaler a scalar used for scaling?

Borek was being mean regarding your misspelling of scalar.

 

[Borek]

Borek was being mean regarding your misspelling of scalar.

Hey, English is my second language, I have right to have doubts, don't I?

 

[Klockan3]

Ok.

But when we take suppose, motion in one dimension, the displacement/velocity/acceleration are vectors, yet they have a sign...most probably cause they are taken WRT another inverse vector.

But problem is we actually use those sign conventions in the numerical analysis...shouldn't that give the wrong answer?

Cause as I said before...doing this should be wrong -

No, that is 100% correct, vectors can have signs like that. A vector just have a length and a direction, however the sign tells us the direction.

In that way (-1,1) is not equal to (1,-1) since they point in opposite directions. Another way to describe the vector would be to say that they got a length sqrt2 and the first got an angle of 3pi/4 and the second got an angle of -pi/4. But saying both of those have equal meanings, and the signs are necessary since they tell you the direction.

 

[D H]

No, that is 100% correct, vectors can have signs like that.

That is 100% incorrect.The additive inverse operator is an operation that can applied to a vector. It is not an intrinsic part of the vector. Vectors have a magnitude and a direction. Those are intrinsic parts of the vector. Rather than repeating what HallsofIvy said, I'll simply quote him:

As for the "sign" of an object, that normally implies sets of "positive" and "negative" objects which in turn implies an order. A vector space of dimension greater than 1 cannot be ordered.

 

[dE_logics]

No doubt...best humour.

No, that is 100% correct, vectors can have signs like that. A vector just has a length and a direction, however the sign tells us the direction.

What exactly is happening here?

Let me guess in this 1-d case, we assume one vector as a coordinate system and in 2-d or more, this assumed vector is gone...

I'm sorry but I'm sort of not getting it...

The angle for all vectors should be taken in either clockwise or anticlockwise direction a vector cannot have 2 angles considering one reference frame of either clockwise or anticlockwise.

That is 100% incorrect.

Do we have a reference book? I'm downloading though...that 4.7 GB collection.

A vector space of dimension greater than 1 cannot be ordered.

Actually I did not get what it means...'ordered'?...you mean sign convention?

 

[D H]

A set is orderable means you can establish a total ordering on the set. In lay terms, you can put the set in some kind of ordered arrangement. The set {2 0 1 -1 -2} can be sorted via ≤ to form the ordered list {-2 -1 0 1 2}. There is no way to come up with a way to order vectors of dimension 2 or higher.

 

[belliott4488]

EDIT: oops. I posted this before I realized there was a second page to this thread ... I'll keep it here in case it's of any value.

As for the "sign" of an object, that normally implies sets of "positive" and "negative" objects which in turn implies an order. A vector space of dimension greater than 1 cannot be ordered.

This is the key point, isn't it? Asking whether a 2-d vector in a given quadrant is positive or negative is a meaningless question.

Yes, for a vector B we can define -B, but that doesn't mean one is positive and the other negative. B might point in the direction of anyone of the four quadrants, so you could have B=(-1,-1) and -B=(1,1), or B=(1,-1) and -B=(-1,1), which is even more ambiguous.

As for this:

So doing this is not right -

Suppose we have a vector B

Assume C = -B (this step should be wrong)

Then if B = 3

Then C = -3

Saying B = 3 doesn't make a lot of sense if B is a vector. B must have a direction (unless it's a 1-d vector), so it must have as many components as there are dimensions in the space in which it lives - so you could say B = (3,0), or B = 3i. Otherwise, there's nothing wrong with the statements above.

 

[belliott4488]

What exactly is happening here?

Let me guess in this 1-d case, we assume one vector as a coordinate system and in 2-d or more, this assumed vector is gone...

I'm sorry but I'm sort of not getting it...

The angle for all vectors should be taken in either clockwise or anticlockwise direction a vector cannot have 2 angles considering one reference frame of either clockwise or anticlockwise.

Whoa ... what?

"in this 1-d case, we assume one vector as a coordinate system" -- a vector and a coordinate system are two different things. You can express a vector in terms of its coordinates in some chosen coordinate system.

"in 2-d or more, this assumed vector is gone" -- Given a Cartesian coordinate system, you can express vectors in terms of their Cartesian coordinates, which people have been implicitly doing here when they've written expressions like "B = (1,-1)", where 1 and -1 are the x- and y-components, respectively. You could also write B = i - j for the same vector, where i and j are unit vectors in the x and y directions.

You can also define a vector in terms of its length (magnitude) and direction, where the direction requires N-1 angles to be defined (N = number of dimensions). So, in 2-d you need length and one angle (typically anticlockwise from x-axis by convention), or in 3-d you need length and two angles, e.g. polar and azimuth angles, or latitude and longitude angles. Our vector above could then be defined as having length SQRT(2) and direction -45 degrees (using the convention I just mentioned).

 

[tiny-tim]

Hi dE_logics!

The angle for all vectors should be taken in either clockwise or anticlockwise direction a vector cannot have 2 angles considering one reference frame of either clockwise or anticlockwise.

?? looks like this is where you're misunderstanding vectors …

let's try to sort this out …

are you saying that you think a vector v is defined by a magnitude (|v|) and an angle?

that's certainly not the standard way of doing it: the standard way is to specify a basis (eg i j or -i -j), and to write it in the form v = xi + yj … or v = (-x)(-i) + (-y)(-j)

(and btw, angles in 2D are always anti-clockwise)

 

[belliott4488]

I kind of looks like tiny-tim and I are disagreeing, but I don't think we really are.

are you saying that you think a vector v is defined by a magnitude (|v|) and an angle?

that's certainly not the standard way of doing it: the standard way is to specify a basis (eg i j or -i -j), and to write it in the form v = xi + yj … or v = (-x)(-i) + (-y)(-j)

That's might be the way it's first taught, but I expect tiny-tim will agree that it's certainly possible to define vectors in terms of their magnitudes and directions, and there are plenty of situations where's that's a desirable way to do it. It's also taught fairly early-on in most intro. to vector courses, I believe.

(and btw, angles in 2D are always anti-clockwise)

Again, that's probably the most typical convention, especially when talking about a 2-d Cartesian space where we've labeled the axes by x and y. It's not universal, however - for example heading angles for velocity vectors are often given as degrees clockwise from North.

These are just little nitpicks, however - on the whole tiny-tim and I are saying the same things. I just wanted to clear up any potential confusion in case it looked like weren't.

 

[dE_logics]

There is no way to come up with a way to order vectors of dimension 2 or higher.

Since in 2-d or 3-d the magnitude of a vector is different along different directions, since in 1-d the directions are only 2 (just like a scalar) the magnitudes can be ordered...right?[/quote]

Saying B = 3 doesn't make a lot of sense if B is a vector. B must have a direction

Another example, in polar coordinate system 3 cos \theta

are you saying that you think a vector v is defined by a magnitude (|v|) and an angle?

In polar coordinate system i.e

I've seemed to forget vectors completely...