# Vectors have no signs?

1. Jun 2, 2009

### 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?

2. Jun 2, 2009

### 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 $$x$$.

3. Jun 2, 2009

### D H

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

4. Jun 2, 2009

### Klockan3

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.

5. Jun 3, 2009

### dE_logics

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

6. Jun 3, 2009

### D H

Staff Emeritus
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.

7. Jun 3, 2009

### 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.

8. Jun 3, 2009

### D H

Staff Emeritus
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.

9. Jun 3, 2009

### tiny-tim

Hi D H!
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

10. Jun 3, 2009

### D H

Staff Emeritus
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.

11. Jun 3, 2009

### 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.

12. Jun 4, 2009

### dE_logics

So doing this is not right -

13. Jun 4, 2009

### dE_logics

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.

14. Jun 4, 2009

### tiny-tim

??

All numbers (positve zero or negative) are scalars.

15. Jun 4, 2009

### D H

Staff Emeritus
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.

16. Jun 4, 2009

### HallsofIvy

???? 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!

17. Jun 4, 2009

### Staff: Mentor

Is scaler a scalar used for scaling?

18. Jun 4, 2009

### 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 -

19. Jun 4, 2009

### dE_logics

20. Jun 4, 2009

### D H

Staff Emeritus
Borek was being mean regarding your misspelling of scalar.