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Vector product VS dot product in matrix

  1. Apr 25, 2010 #1
    hi, i don't really understand whats the difference between vector product and dot product in matrix form.

    for example

    (1 2) X (1 2)
    (3 4) (3 4) = ?

    so when i take rows multiply by columns, to get a 2x2 matrix, i am doing vector product?

    so what then is dot producT?

    lastly, my notes says |detT| = final area of basic box/ initial area of basic box

    where detT = (Ti) x (Tj) . (Tk)

    so, whats the difference between how i should work out x VS . ?

    also, |detT| = magnitude of T right? so is there a formula i should use to find magnitude?

    so why is |k . k| = 1?
  2. jcsd
  3. Apr 25, 2010 #2


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    No, you are doing a "matrix product". There are no vectors here.

    With matrices? It isn't anything. The matrix product is the only multiplication defined for matrices. The dot product is defined for vectors, not matrices.

    Well, we don't have your notes so we have no idea what "T", "Ti", "Tj", "Tk" are nor do we know what a "basic box" is.

    I do know that if you have a "parallelpiped" with adjacent sides given by the vectors [itex]\vec{u}[/itex], [itex]\vec{v}[/itex], and [itex]\vec{w}[/itex], then the volume (not area) of the parallelpiped is given by the "triple product", [itex](\vec{u}\times\vec{v})\cdot\vec{w}[/itex] which can be represented by determinant having the components of the vectors as rows. That has nothing at all to do with matrices.

    No, "det" applies only to square arrays for which "magnitude" is not defined.

    I guess you mean "k" to be the unit vector in the z direction in a three dimensional coordinate system. If so, then |k.k| is, by definition, the length of k which is, again by definition of "unit vector", 1.

    You seem to be confusing a number of very different concepts. Go back and review.
    Last edited by a moderator: Apr 27, 2010
  4. Apr 27, 2010 #3
    oh.. em..

    ok lets say we have

    (1 2) x (4 5)
    (3 4) (6 7) = so this is just rows multiply by column to get a 2x2 matrix right? so what is the difference if i replace the x sign with the dot sign now. do i still get the same?
    i presume one is cross (x) product , one is dot (.) product? or is it for matrix there is no such things as cross or dot product? thats weird. my tutor tells us to know the difference between cross and dot matrix product

    so for the case of the parallelpiped, whats the significance of the triple product (u x v) .w? why do we use x for u&v but . for w?

    is it just to tell us that we have to use sin and cos respectively? but if u v and w were square matrix, then there won't be any sin and cos to use? so we just multiply as usual rows by columns?

    oh by definition . so that means |k.k| = (k)(k)cos(0) = (1)(1)cos(0) = 1
    so |i.k| = (1)(1)cos(90) = 0 ?
    so if i x k gives us -j by the right hand rule, then does it mean the magnitude, which is |i.k| = 0 is 0? in the direction of the -j?? or are they 2 totally different aspects?

    btw, sry for another question,
    why is e(w)(A),
    where A =
    (0 -1)
    (1 0)

    can be expressed as
    ( cosw -sinw)
    ( sinw cosw)
    which is the rotational matrix anti-clockwise about the x-axis right?

  5. Apr 27, 2010 #4


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    You can replace it by whatever symbol you like. As long as your multiplication is "matrix multiplication" you will get the same result.

    No, just changing the symbol doesn't make it one or the other.

    I suspect your tutor was talking about vectors not matrices.

    Because you are talking about vectors not matrices!

    They are NOT matrices, they are vectors!!

    You can think of vectors as "row matrices" (n by 1) or "column matrices" (1 by n) but they still have properties that matrices in general do not have.

    Yes, that is correct.

    No, the length of i x k is NOT |i.k|, it is [itex]|i||j|= 1[/itex].

    In general, the length of [itex]\vec{u}\times\vec{v}[/itex] is [itex]|u||v| sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex].

    For objects other than numbers, where we have a notion of addition and multiplication, we define higher functions by using their "Taylor series", power series that are equal to the functions. In particular, [itex]e^x= 1+ x+ (1/2)x^2+ \cdot\cdot\cdot+ (1/n!)x^n+ \cdot\cdot\cdot[/itex].

    It should be easy to calculate that
    [tex]A^2= \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}[/tex]
    [tex]A^3= \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}[/tex]
    and, since that is the identity matrix, it all repeats:
    [tex]A^4= \begin{pmatrix}0 & -1 \\ 1 & 0}\end{pmatrix}= A[/tex]

    That gives
    [tex]e^{Aw}= \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}+ \begin{pmatrix}0 & -w \\ w & 0\end{pmatrix}+ \frac{1}{2}\begin{pmatrix}-w^2 & 0 \\ 0 & -w^2\end{pmatrix}[/tex][tex]+ \frac{1}{3!}\begin{pmatrix}0 & w^3 \\ -w^3 & 0\end{pmatrix}+ \frac{1}{4!}\begin{pmatrix}w^4 & 0 \\ 0 & w^4\end{pmatrix}+ \cdot\cdot\cdot[/tex]

    [tex]= \begin{pmatrix}1- \frac{1}{2}w^2+ \frac{1}{4}w^4+ \cdot\cdot\cdot & -w+ \frac{1}{3!} w^3+ \cdot\cdot\cdot \\ w- \frac{1}{3!}w^3+ \cdot\cdot\cdot & -1+ \frac{1}{2}w^2- \frac{1}{4}w^4+ \cdot\cdot\cdot\end{pmatrix}[/tex]

    and you should be able to recognise those as the Taylor's series about 0 for cos(w) and sin(w).
    Last edited by a moderator: Apr 27, 2010
  6. Apr 27, 2010 #5

    ok, i went to check again what my tutor said and it was
    "scalar and vector products in terms of matrices". so what does he mean by this?

    the scalar product is (A B C) x (D E F)T, (so we can take the transpose of DEF because it is symmetric matrix? or is it for some other reason? )
    so rows multiply by columns again?
    but what about vector product?

    for the parallelpiped, (u x v).w
    so lets say u = (1,1) , v = (2,2), w = (3,3)
    so u x v = (1x2, 1x2)sin(angle between vectors)
    so .w = (2x3,2x3) cos(angle) ?
    so if it yields 0, that vector w lies in the plane define by u and v, but if its otherwise, then w doesn't lie in the plane of u v ?

    for i x k, why is the length |i||j|? why is j introduced here? shouldn't it be |i||k|sin(90) = 1?
    oh i see.. so the right hand rule gives the direection but the magnitude for i x k = |i||k|sin(90) = 1?

    thanks a ton!
  7. Apr 27, 2010 #6


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    Okay, you think of one vector as a row matrix and the other as a column matrix then the "dot product" is the matrix product
    [tex]\begin{bmatrix}A & B & C\end{bmatrix}\begin{bmatrix}D \\ E \\ F\end{bmatrix}= AD+ BE+ CF[/tex]
    But the dot product is commutative isn't it? Does it really make sense to treat the two vectors as different kinds of matrices? It is really better here to think of this not as the product of two vectors but a vector in a vector space and functional in the dual space.

    Yes, that was a typo. I meant |i||k|.


  8. Apr 27, 2010 #7
    i see. thank you very much
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