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Divide Two Vectors

  1. Mar 31, 2013 #1
    Can we divide two vector ? If we can't why
  2. jcsd
  3. Mar 31, 2013 #2


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    It is not as much that we can't, but that there is little point in doing so.
    To understand that, remember that "division" is supposed to be the "opposite" of multiplication, so that if a/b=c, that simply means that it follows that a=b*c.
    Furthermore, we want "c" to be a UNIQUE quantity, rather than that there exists lots of quantities that might be said to be the result of the division a/b.

    So, to answer your question on vectors, we need to see what sort of vector multiplications we've got.
    We have for example, the so-called cross product of vectors, but in a=b(cross)c, there are lots of c's we can use, that cross multiplied with "b" yields "a".

    Thus, an "opposite" operation of cross multiplication does not provide us with a unique answer, and we choose therefore not to define such an operations.

    There are other troubles with trying to define vectorial division as well, so it is typically not defined as an operation at all, at least to my knowledge.
  4. Mar 31, 2013 #3
    Obviously we can define something similar, as we can define multiplication pointwise of vectors.

    [tex](a,b,c)(d,e,f)= (ad,be,cf)[/tex]

    divison when [tex]d,e,f\neq 0[/tex] :

    Obviously, you will see this popping up in Functional Analysis, which is a generalization of linear algebra.
  5. Mar 31, 2013 #4
    How do you prove it ?
  6. Mar 31, 2013 #5


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    The best way to deal with vectors is to work in a vector space. A vector space has no canonical division operation. So it does not make sense to divide two vectors. I suppose you can make sense of it in specific situations, but nobody will call it division.
  7. Mar 31, 2013 #6
    Prove what? He's giving you a definition.
  8. Mar 31, 2013 #7


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    Really? The product you defined is not coordinate independent and since nearly all of the spaces arising in functional analysis have no natural choice of basis, it would surprise me to learn that component-wise products play an important role anywhere in functional analysis.
  9. Mar 31, 2013 #8
    Well, you can check the subject of Banach Algebras, I first encoutered this subject in the second course in Functional analysis which was given at my school.
  10. Mar 31, 2013 #9


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    I am familiar with Banach algebras, and while I get where you are coming from now, the claim that component-wise products are honestly studied in functional analysis still seems silly to me. As you certainly know, almost all of the Banach algebras that occur "in nature" have coordinate invariant products (i.e. they do not depend on a particular choice of basis) and as a result, are not given by component-wise multiplication. The examples you gave are certainly subsumed into the theory of Banach algebras, but as a matter of taste I would not really say they come up in functional analysis.

    N.B. Since I do not intend to come across as argumentative, the one thing I want to make clear is that I do concede your point on this, just that I probably misinterpreted what you meant by "popping up in functional analysis"
  11. Mar 31, 2013 #10
    Can I say something.In matrices (2x1)/(2x1)=(2x2) we can say this because If we want to control that we must multiply (2x2)x(2x1) and we get (2x1) so I understand that a/b question's answer is two vector system.Can we say this ?
    (HERE a and b vector) and (2x1) or (2x2) is matrices)
    ( (2x1) matrice symbolize vector)
  12. Mar 31, 2013 #11
    Can I say something.In matrices (2x1)/(2x1)=(2x2) we can say this because If we want to control that we must multiply (2x2)x(2x1) and we get (2x1) so I understand that a/b question's answer is two vector system.Can we say this ?
    (HERE a and b vector) and (2x1) or (2x2) is matrices)
    ( (2x1) matrice symbolize vector)
  13. Mar 31, 2013 #12
    I don't quite understand what you mean. Can you rephrase?

    The point others are making that you don't seem to be getting, however, is that there is no reason to have multiplication or division of vectors.

    There are lots of operations on vectors which can seem like a form of multiplication (dot products, cross products etc...) but none of them are "one to one", meaning for a given output there is more than one possible input.

    Now if you know that the input does not depend on the output, how are you going to define an inverse operation (akin to divison) that gives you the input based on the output?
  14. Apr 1, 2013 #13
    Sure,we symbolize vector in matrix (2x1) so If we try divide two vectors in matrix system, (2x1)/(2x1) we get (2x2) so if we want control this,we will multiply (2x2)x(2x1) and we get (2x1)
    (2x1) is one vector (2x2) is two vector system

    so we can say

    If we want write this in vector system
    pointwise of vectors
    or [tex]a/b=((c,d))[/tex]
    (a,b,c,d) vectors
  15. Apr 1, 2013 #14


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    In words: the product of a 2 x 2 matrix and a 2 x 1 matrix is a 2 x 1 matrix.
    I don't see how this makes sense. Matrix multiplication is defined if the multiplication is conformable. IOW, AB makes sense if the number of columns of A is the same as the number of rows of B.

    However, there is no concept of dividing one matrix by another. The property you are showing about the equivalence of multiplication and division is one that applies to real numbers, not matrices.
  16. Apr 2, 2013 #15
    I am just trying to explain it.I wrote it because its a way to how can I found it.And my idea can be true ?
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