- #1

- 7

- 0

(Scalar)·(Vector) = Scalar

(Vector)·(Vector) = Scalar

(Scalar)x(Scalar) = Not valid

(Scalar)x(Vector) = Vector

(Vector)x(Vector) = Vector

Did I get them right, if not why?

Thanks

- #1

- 7

- 0

(Scalar)·(Vector) = Scalar

(Vector)·(Vector) = Scalar

(Scalar)x(Scalar) = Not valid

(Scalar)x(Vector) = Vector

(Vector)x(Vector) = Vector

Did I get them right, if not why?

Thanks

- #2

- 8,499

- 2,634

- Scalar times scalar to produce a scalar (ordinary multiplication)
- Scalar times vector to produce a vector (scaling a vector)
- Vector times vector to produce a scalar (scalar or "dot" product)
- Vector times vector to produce a vector ("cross" product)

- #3

fresh_42

Mentor

- 14,413

- 11,726

6. Scalar cross vector also denotes a pair and no vector. If it is supposed to be a vector, the cross has to be explicitly defined, e.g as scaling.

- #4

jbriggs444

Science Advisor

Homework Helper

- 9,612

- 4,260

You are interpreting AxB for scalars A and B in the sense of the Cartesian product of a pair of sets? [the set of ordered pairs (a,b) where a is an element of A and b is an element of B]. That is rather a stretch since A and B are not sets and their cross product would not be an ordered pair but would, rather, be a [possibly singleton] set of ordered pairs.5. Scalar cross scalar to denote the pair (scalar,scalar). "Not valid" is misleading here.

- #5

fresh_42

Mentor

- 14,413

- 11,726

Agreed. I simply found "not valid" a bit harsh, esp. because the word scalar has been put into brackets. Sometimes I've also read lines like ##2 \times 2 = 4##.You are interpreting AxB for scalars A and B in the sense of the Cartesian product of a pair of sets? [the set of ordered pairs (a,b) where a is an element of A and b is an element of B]. That is rather a stretch since A and B are not sets and their cross product would not be an ordered pair but would, rather, be a [possibly singleton] set of ordered pairs.

- #6

jbriggs444

Science Advisor

Homework Helper

- 9,612

- 4,260

If one insists on intepreting ##\times## in the sense of the Cartesian product and interpreting 2 in the sense of set theory and the Von Neumann construction then it would be more correct to say that ##|2 \times 2| = 4## and that ##2 \times 2## = { (0,0), (0,1), (1,0), (1,1) }Agreed. I simply found "not valid" a bit harsh, esp. because the word scalar has been put into brackets. Sometimes I've also read lines like ##2 \times 2 = 4##.

This follows since, in the Von Neumann construction, 2 = {0,1}.

Bringing us back on topic for this thread... One should interpret the ##\times## notation according to context. In the context of ##2 \times 2## and a discussion of scalars and vectors, it cannot reasonably denote either a cross product of vectors or a Cartesian product of sets. The most reasonable interpretation would as an ordinary product of integers.

Last edited:

- #7

- 7

- 0

Thank you all for your feedback.

The comments certainly help a lot.

The comments certainly help a lot.

- #8

- 631

- 132

It works for 7D vectors as well.

- Scalar times scalar to produce a scalar (ordinary multiplication)
- Scalar times vector to produce a vector (scaling a vector)
- Vector times vector to produce a scalar (scalar or "dot" product)
- Vector times vector to produce a vector ("cross" product)

- #9

- 22,089

- 3,296

It works in any dimension with geometric algebra as well. Also, the 7D cross product is unsatisfactory: no Jacobi identify and it's not unique.It works for 7D vectors as well.

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 978

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 7K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 0

- Views
- 497

- Replies
- 2

- Views
- 2K

- Replies
- 1

- Views
- 1K