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Matrix multiplication of an nxn matrix is the scaling

  1. Jul 15, 2008 #1
    Geometrically, matrix multiplication of an nxn matrix is the scaling, and rotation of a vector in n dimensions true? So when you find the inverse of a matrix, what you're actually doing is finding a transformation such that in the 'transformed space' the vector is a unit vector.

    If the inverse matrix ([tex]A^{-1}[/tex])is plotted in the original space, then does it have any relation to the original matrix([tex]A[/tex])?

    What I mean by that is, if you have a function [tex]y=f(x)[/tex] in 2 D space, and you find the inverse function [tex]x=f^{-1}(y)[/tex] the inverse function is a reflection of the function [tex]y=f(x)[/tex] about the line [tex]y=x[/tex]. Does the inverse matrix ([tex]A^{-1}[/tex]) have any such relation to the original matrix([tex]A[/tex])?
  2. jcsd
  3. Jul 15, 2008 #2
    Re: Matrices

    The relation between [tex]A^{-1}[/tex] & [tex]A^[/tex]... the inverse of [tex]A^[/tex], where it exists, is denoted by [tex]A^{-1}[/tex] and [tex]AA^{-1}[/tex] is equal to I ie. identity matrix = [tex]A^{-1}A[/tex]
  4. Jul 16, 2008 #3


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    Re: Matrices

    True, but I don't believe that was the question he was asking.

    What do you mean by "plotted in the original space"? An n by n matrix operates on n dimensional vectors and itself exist in an n2 vector space. In order to "graph" A and A-1 you would need an n4 dimensional graph.
  5. Jul 16, 2008 #4
    Re: Matrices

    Why would you need an n4 dimensional graph? Say for a 3x3 matrix, you would need a 9 D space then. Its kind of hard for me to imagine that, but say if we took each row or column to be a plane in 3D (where I suppose each plane corresponds to a projection of 9D space on 3D), would there be a relation between the planes of [tex]A[/tex] and [tex]A^{-1}[/tex] ?
  6. Jul 16, 2008 #5
    Re: Matrices

    A plane? Wouldn't each row or column of a three by three matrix simply be a vector in [tex]F^3[/tex]?
  7. Jul 17, 2008 #6
    Re: Matrices

    Yeah, each row or column of a 3x3 matrix would be a vector, but if you assume that vector to be the normal vector passing through a given point then you get three planes depending on whatever matrix equation you use. I dont think a standalone 3x3 matrix could represent planes by itself.

    Even so, I'm guessing here, the projection would give you three vectors instead of three planes. So you would have a set of three vectors corresponding to [tex]A[/tex] and three vectors corresponding to [tex]A^{-1}[/tex], would the vectors of A and A-1 be co-related in any way?
  8. Jul 23, 2008 #7
    Re: Matrices

    Anyone? Please :p ?
  9. Jul 23, 2008 #8
    Re: Matrices

    In general, any function (including e.g. matrices) [tex]f:R^n\rightarrow R^n[/tex] has a graph in [tex]R^{2n}[/tex] which is the set [tex]\{(\textbf{x},f(\textbf{x})):\textbf{x}\in R^n\}[/tex]. If f is invertible then the graph of [tex]f^{-1}[/tex] is [tex]\{(f(\textbf{x}),\textbf{x}):\textbf{x}\in R^n\}[/tex]. You get one from the other by reflection through a hyperplane and you can easily write out the matrix for this.

    For a function on R this is very usefull as a visual aid. In general, not so. I think what you want to know is if we can read off the value that A^{-1} assigns to x using the graph of A? Because we can't visualise the graph we can't.
    Last edited: Jul 23, 2008
  10. Jul 23, 2008 #9
    Re: Matrices

    Thats exactly it. How would you go about writing a matrix for this? Perhaps we cant use it as a visual aid, but it would simplify a lot of other calculations.
  11. Aug 2, 2008 #10
    Re: Matrices


    [tex]B:=\left[ \begin{array}{cc} 0 & I \\
    I & 0 \end{array} \right] [/tex]

    where I is the identity on [tex]R^n[/tex]. The coordinates of a general point on the graph are [tex](Ix,Ax)[/tex], so the coords of a general point on the graph of the inverse are [tex]B(Ix,Ax)=(Ax,Ix)[/tex]. One usually writes out the matrix [tex](A,I):R^n\rightarrow R^{2n}[/tex] and uses gaussian elimination to convert it to [tex](I,A^{-1})[/tex], which again gives points on the graph of the inverse, but in a more useful fashion.
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