Vector multiplication with scalars

I'm a little confused as to the results of multiplying vectors and scalars. If I multiply a vector and a scalar, is it a vector? And if I multiply a vector and another vector, is the results also a vector? I looked at http://mathworld.wolfram.com/VectorMultiplication.html but still haven't figured it out.

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migrated from physics.stackexchange.comFeb 27 '11 at 16:46

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This is a basic question about linear algebra, but not about physics, please try math.stackexchange.com for this kind of question. – Tim van Beek Feb 27 '11 at 16:33

If $\alpha$ is a scalar and $\mathbf{v}$ is a vector that can be decomposed into some basis $\mathbf{e}_i$, then $$\alpha \mathbf{v} = \alpha v^i \mathbf{e}_i\, ,$$ thus remaining a vector.

Other vector multiplications where you get out a vector as well (e.g. the cross product) can be seen as $$\mathbf{v}\times \mathbf{w} = A^i_{kl}v^k w^l \mathbf{e}_i$$ but there are also other ways you can define a map from some $\mathbb{R}^n$ to a $\mathbb{R}^m$ like a scalar product etc.

It depends on what you want to calculate what you will get out. You may consider some Wikipedia articles such as linear algebra to specify your questions in the future.

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Ok, this is a math question really, but a 'physics answer' could go as follows:

Velocity is a vector: it has a size (speed) and a direction. If you multiply a velocity vector by - say - a factor of two, you go twice as fast in the same direction. So multiplying a (velocity) vector with a scalar gives again a (velocity) vector.

Now the multiplication of two vectors. There are several meaningful definitions of vector multiplications. Out of these, the inner product multiplication (or dot product) is most abundant in physics. We take force vectors and displacement vectors as examples. Dot multiplying the force vector exerted on a particle with its displacement vector yields the amount of work done. This amount of work is a scalar (a number without further attributes such as direction). So the dot product between two vectors gives a scalar.

As I mentioned, other definitions of products between vectors are possible. For instance, the direct product between two vectors yields a (rank two) tensor. No need to be bothered about this in basic physics though.

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I see, I wasn't sure if both would give me a vector as the result. Many thanks! – Teknophilia Feb 27 '11 at 16:44

Multiplication on scalar is an operation that must be defined on every vector space. Multiplicaton of two vectors is trickier, however, there is no single right way of doing it. Indeed, for any vector space $V$ over a field $k$ you can fix any $k$-bilinear mapping $f: V \times V \to V$ and call it multiplication. Then (V, f) is called a $k$-algebra. You can further require $f$ to satisfy some properties and get associative algebra, unital associative algebra, commutative associative algebra, division algebra, or maybe Lie algebra which is much different :)

Or you can fix a $k$-bilinear mapping $f: V \times V \to k$ and call it a scalar product. Depending on what additional properties you require, it may turn, for example, into (pseudo)-euclidean form or into symplectic form, which is much different. You get the picture :)

Some concrete examples with $\mathbb{R}^n$ as the vector space:
$(\mathbf{x}, \mathbf{y}) \mapsto \mathbf{x} \cdot \mathbf{y} = \sum_{i = 1}^n x^i y^i$ is a euclidean form,
$(\mathbf{x}, \mathbf{y}) \mapsto \sum_{i = 1}^n k_i x^i y^i$ $(k_i \in \lbrace 1, -1 \rbrace)$ is a pseudo-euclidean form,
$(\mathbf{x}, \mathbf{y}) \mapsto \mathbf{x} \times \mathbf{y}$ gives you a Lie algebra,
$((x^1, \ldots, x^n), (y^1, \ldots, y^n)) \mapsto (x^1 y^1, \ldots, x^n y^n)$ gives you a division algebra (not sure about that, though).

In particular, in $\mathbb{R}^2$ you can define a very well-behaved algebra :)

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God the answers above are mere showing off. Vector multiplied by a scalar is a scalar. vector multiplied by another vector can take two forms, known as scalar product (inner product) or cross product. The former (inner) is a scalar.

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I think the ambiguity stems from naming all those fundamentally different operators "multiplication". Let's go back to the definitions: a vector space over a field $K$ is a set $V$ together with additional structure provided by two binary operations $+:V \times V \rightarrow V$ and $\cdot: K \times V \rightarrow V$ that satisfy certain axioms I won't list here. One can equip a vector space with more structure still by equipping it with an inner product (of which the dot product is a single example/case) $<,>: V \times V \rightarrow K$ which, again, satisfies certain axioms. The binary operator often called vector cross product is a handy mathematical tool for certain geometrical calculations in 3-dimensional Euclidean Space (which is a very specific example of a vector space over the field of real numbers $\mathbb{R}$).

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