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Homework Help: Linear algebra and matrix operations: correct me if im wrong

  1. Oct 17, 2005 #1
    as far as intro linear algebra is concerned, there's no such thing as matrix division and its not valid to say that for a linear transformation T(x) and its standard matrix A,

    [tex]A = \frac{T(\vec{x})}{\vec{x}} [/tex]

    just because [itex]T(\vec{x}) = A\vec{x}[/itex]

    Am I right?
  2. jcsd
  3. Oct 17, 2005 #2


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    It's been awhile since I've done linear algebra, but this seems like more of a semantic question than anything else. Division is simply multiplication by the inverse. (My LaTEX skills also stink, so I'll have to do this in words.) It may be more correct to say

    A = T(x)*inv(x) than A = T(x)/x,

    but I think you'd be in the right to say that's division. It assumes, of course, that the inverse of x exists - but if it doesn't, you couldn't divide by it anyway.

    Your professor may have other opinions than mine - I'd suggest you ask her.
  4. Oct 17, 2005 #3


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    Actually, there IS such a thing as "matrix division" if the matrix has an inverse- multiply by the inverse matrix.

    There is, however, no such thing as "vector division" which is what your second question is about. It certainly would make no sense to write what you have above.
  5. Oct 18, 2005 #4

    Thanks. I didn't think so.
  6. Oct 18, 2005 #5


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    If Ax = T(x), and x has a right inverse y (a matrix such that xy is the identity matrix with the same number of rows as x), then:

    Axy = T(x)y
    A = T(x)y

    If you want to denote right multiplication by the right inverse of x as division by x, you can do so, but this only makes sense when x has a right inverse, and when the number of columns of whatever you're dividing has the same number of rows as the right inverse of x. If x is a column vector, it won't have a right inverse unless it is a 1x1 non-zero matrix.
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