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Dot Product

  1. Dec 11, 2008 #1
    If you look up dot product in Wikipedia, under 'properties' it states the following:

    "The dot product is not associative, however with the help of the matrix-multiplication one can derive:
    [tex]
    \left(\vec{a} \cdot \vec{b}\right) \vec{c} = \left(\vec{c}\vec{b}^{T}\right)\vec{a}
    [/tex]"​

    I simply dont see how this can be true for any vector [tex]\vec{c}[/tex]. Is it?

    Thanks in advance,
     
  2. jcsd
  3. Dec 11, 2008 #2

    tiny-tim

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    Hi Apteronotus! :smile:

    cTb is a scalar, c.b, but cbT is a matrix.

    (cbT)a = ∑∑(cbT)ijajei
    = ∑∑cibjajei
    = (∑bjaj)∑ciei = (b.a)c :smile:
     
  4. Dec 11, 2008 #3

    HallsofIvy

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    cTb, the dot product, is often called the "inner product" and is a scalar while bcT is called the "outer product" and is a matrix.
     
  5. Dec 22, 2008 #4
    Thank you both for your replies. But I still think there may be a problem.
    Consider three vectors [tex]\vec{a}, \vec{b}[/tex] and [tex]\vec{c}[/tex], where

    [tex]\vec{a}=\left[a_{1}, a_{1}, a_{3}\right][/tex]
    [tex]\vec{b}=\left[b_{1}, b_{1}, b_{3}\right] [/tex]

    and
    [tex]\vec{c}=\left[c_{1}, c_{1}, c_{3}\right][/tex]

    then
    [tex]
    \left(\vec{a}\cdot\vec{b}\right)\vec{c}=\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)\left[c_{1}, c_{1}, c_{3}\right]
    [/tex]

    and

    [tex]
    \left(\vec{c}\vec{b}^{T}\right)\vec{a}=\left(c_{1}b_{1}+c_{2}b_{2}+c_{3}b_{3}\right)\left[a_{1}, a_{2}, a_{3}\right]
    [/tex]

    Now these vectors are not necessarily equal. Consider the first element of each.
    It is clear that in general,

    [tex]
    \left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)c_{1}\neq\left(c_{1}b_{1}+c_{2}b_{2}+c_{3}b_{3}\right)a_{1}
    [/tex]
     
  6. Dec 22, 2008 #5

    tiny-tim

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    But tcbT is not (c1b2+c2b2+c3b3) …

    it's a matrix.
     
  7. Dec 23, 2008 #6
    Specifically, it is the matrix
    [tex]
    cb^T = \begin{bmatrix}
    b_1 c_1 & b_2 c_1 & b_3 c_1 \\
    b_1 c_2 & b_2 c_2 & b_3 c_2 \\
    b_1 c_3 & b_2 c_3 & b_3 c_3
    \end{bmatrix}.
    [/tex]
     
    Last edited: Dec 23, 2008
  8. Dec 25, 2008 #7
    Yes! Thank you both very much.
    My error was in taking the vectors as row vectors.
    Thanks again,
     
  9. Dec 27, 2008 #8

    HallsofIvy

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    This is more abstract and more advanced than the "inner product" but if you are using the "outer product", you may want to think about this.

    The "dual" of a finite dimensional vector space, V, (the space of linear functionals from V to the base field) is isomorphic to v with a "natural" isomorphism: given a basis [itex]{e_1, e_2, \cdot\cdot\cdot, e_n}[/itex], map each basis vector [itex]e_i[/itex] to the functional, [itex]f_{e_i}(v)[/itex] that maps [itex]e_i[/itex] to 1, all other [itex]e_j[/itex] to 0. Then extend it to the entire space by "linearity": if [itex]v= a_1e_1+ a_2e_2+ \cdot\cdot\cdot a_ie_i+ \cdot\cdot\cdot+ a_ne_n[/itex], [itex]f(v)= a_1 f(e_1)+ a_2f(e_2)+\cdot\cdot\cdot+ a_if(e_i)+ \cdot\cdot\cdot+ a_nf(e_n)[/itex][itex]= a_1(0)+ a_2(0)+ \cdot\cdot\cdot+ a_i(1)+ \cdot\cdot\cdot+ a_n(0)= a_i[/itex].

    Since that is an isomorphism, given any vector u, that isomorphism maps it to f_u(x). Given any two vectors, u, and v, the functional f_u(v) takes v to the real number [itex]u\cdot v[/itex], their dot product as defined in that particular basis. On the other hand "[itex]v f_v(x)[/itex] can be interpreted as a linear transformation that maps each vector , w, into the vector [itex](f_v(w))u[/itex] an numeric multiple of u. If we agree to write vectors as column matrices, say
    [tex] v= \left[\begin{array}{c}a_1 \\ a_2\\ \cdot \\ \cdot \\ \cdot \\ a_n\end{array}\right][/tex]
    and functionals in the dual space as row matrices, say
    [tex]f_u= \left[\begin{array}{ccccc}b_1 & b_2 & \cdot\cdot\cdot & b_n\end{array}\right][/tex]
    Then the operation of the functional, [itex]f_u[/itex] on v is the matrix product
    [tex]\left[\begin{array}{ccccc}b_1 & b_2 & \cdot\cdot\cdot & b_n\end{array}\right]\left[\begin{array}{c}a_1 \\ a_2\\ \cdot \\ \cdot \\ \cdot \\ a_n\end{array}\right][/tex]
    while the linear transformation corresponding to [itex]v f_u[/itex] is give by the matrix product

    [tex]\left[\begin{array}{c}a_1 \\ a_2\\ \cdot \\ \cdot \\ \cdot \\ a_n\end{array}\right]\left[\begin{array}{ccccc}b_1 & b_2 & \cdot\cdot\cdot & b_n\end{array}\right][/tex]
     
    Last edited: Dec 27, 2008
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