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Row of a matrix is a vector along the same degree of freedom

  1. Oct 29, 2009 #1
    A particular introduction to matrices involved viewing them as an array/list of vectors (column vectors) in Rn. The problem I see in this is that it is sort of like saying that a row of a matrix is a vector along the same degree of freedom (elements of the same row are elements of different vectors all in the same dimension). So from this, technically, the scalar product of a column vector v and row1 of a matrix A should only exist as a product between the elements of row1 of the matrix A and row1 of the column vector v...which doesn't seem right (since matrix-vector multiplication Av is defined as a column vector of dot products between the vector v and rows of A). How would one geometrically interpret a matrix?
     
    Last edited: Oct 29, 2009
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  3. Oct 30, 2009 #2

    tiny-tim

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    Hi Gear300! :smile:

    Technically, row vectors are transpose vectors,

    so the first row of A is is not the vector a1 (say), but the transpose vector a1T.

    Then your scalar product in matrix form is (a1T)v,

    but in ordinary form is a.v :wink:
    Dunno :redface:, except that I always think of a matrix as being a rule that converts one vector into another vector. :smile:
     
  4. Oct 30, 2009 #3
    Re: Matrices

    I understand matrices as operators, like you may have basis vectors (x,y,z all normal), you can have basis functions (such as sin(nx) for n=1,2,3...) which any periodic function can be decomposed into using Fourier series. Then the function can be represented completely by a column vector containing the amplitude of each frequency, and the matrix multiplication will output a new function. So in that sense you might represent a matrix geometrically by a series of "before and after" functions!

    Then again this is just fun speculation... the most fundamental way of expressing a matrix geometrically is probably by a discrete 2-D plot, f(m,n) = the (m,n)th element of the matrix.
     
  5. Oct 30, 2009 #4
    Re: Matrices

    I see...good stuff so far...Thanks for all the replies.
     
  6. Nov 1, 2009 #5

    tiny-tim

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    Hi Gear300! :smile:

    I've found http://farside.ph.utexas.edu/teaching/336k/lectures/node78.html#mom" on john baez's excellent website which shows the relation between the ordinary and the matrix representation of the same equation, and where the transpose fits in …

    if ω is the (instantaneous) angular velocity, then the rotational kinetic energy is both:

    1/2 ω.L
    and (without the "dot")
    1/2 ωTL = 1/2 ωTÎω

    where L is the (instantaneous) angular momentum, and Î is the moment of inertia tensor, both measured relative to the centre of mass.
     
    Last edited by a moderator: Apr 24, 2017
  7. Nov 1, 2009 #6
    Re: Matrices

    Thanks a lot for the link tiny-tim.

    So, its sort of like saying that for an m x p matrix with n being a dimensional space, you could label the rows (going down) from n = 1 to n = m and you could also label the columns (going across) from n = 1 to n = p, right?
     
    Last edited: Nov 1, 2009
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