# Difference between scalars and one-dimensional vectors?

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1. Sep 20, 2015

### Mr Davis 97

What is the difference between scalars and one-dimensional vectors? I know that we represent the set of all two-dimensional vectors as $\mathbb{R}^2$, so doesn't this mean that we would represent the set of all one-dimensional vectors as as just $\mathbb{R}$? However, doesn't this also refer to scalars that are real numbers? Does this show that scalars and one-dimensional vectors are equivalent?

2. Sep 20, 2015

### Staff: Mentor

A scalar value can have a positive, zero or negative value whereas a one dimensional vector will always have a positive or zero magnitude and a direction.

While its true that you can represent the vector in component form using any real number that component value isn't the vector itself ie for the real value of 5 and the one dimensional vector <5>

5 =/= <5>

3. Sep 20, 2015

### tommyxu3

Yes, you can say that the space composed of one-dimensional vectors is $\mathbb{R},$ which you can realize with the concept of the number line. Different from scalars, the signals of vectors tell us their directions, while the signals of scalars just mean more and less(the latter is how I get them).

4. Sep 20, 2015

### Mr Davis 97

Okay, that makes sense. But, for example, what if I were trying to define a function that mapped one-dimensional vectors to one-dimensional vectors. What notation could I use to specify this without the reader getting confused with whether I mean real numbers? $f: \mathbb{R} \to \mathbb{R}$ seems to be ambiguous...

5. Sep 20, 2015

### Staff: Mentor

6. Sep 20, 2015

### mathman

One dimensional vectors and real scalars are in one to one correspondence, as long as you identify one vector direction with + and the other vector direction with -.

7. Sep 20, 2015

### Mr Davis 97

Is this why in 1D kinematics we solve problems as if the equations were a scalar equations rather than a vector ones, when we really mean vectors? They are in one-to-one correspondence, so are we just being lazy by not wring something like $\vec{v_0} = v_0\hat{j}$?

8. Sep 20, 2015

### lavinia

A field can always be considered to be a one dimensional vector space over itself. The field is both the field of scalars for this vector space and it is the vector space itself.

So the real numbers are a one dimensional vector space over itself and the complex numbers are a one dimensional vector space over itself. In the first case, the field of scalars is the real numbers, in the second, the field of complex numbers.

Last edited: Sep 20, 2015
9. Sep 21, 2015

### Ssnow

The $1$-dimensional vectors in $\mathbb{R}^{1}$ are in bijection with the scalars $\mathbb{R}$.

10. Sep 25, 2015

### mathwonk

To me the key point is that in a one dimensional vector space, one is not necessarily given a basis vector. So without a given basis vector, there is no way to naturally identify vectors with scalars. I.e. R^1 is more than a one dimensional vector space, it is a one dimensional vector space plus a given basis vector, namely the scalar 1. A better example as a one dimensional vector space, would be a choice of a line through the origin of 3 space. Then where is the vector corresponding to 1? In particular one can only add vectors in a one dimensional vector space, but not multiply them, as one can do for scalars. Also length (as a number) is not necessarily defined for vectors in a more general vector space.