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A question on invertible matrixes

  1. Nov 4, 2012 #1
    I'm making notes for linear algebra, and I'm using the weaker definition for the invertible matrix:

    "A matrix A is called invertible it there exists a matrix B such that BA = I"

    How do we prove that AB = I using only matrix theory?
     
  2. jcsd
  3. Nov 4, 2012 #2
    You mean how one would prove that [itex]AB = I[/itex] provided that [itex]BA = I[/itex]?

    If you assume that [itex]BA = I[/itex] and thus that [itex]A^{-1} = B[/itex] and [itex]A = B^{-1}[/itex] you could try multiplying from the left with [itex]B^{-1}[/itex] and from the right with [itex]A^{-1}[/itex], which gives you

    [itex]B^{-1}BAA^{-1} = B^{-1}IA^{-1} \ .[/itex]​
     
  4. Nov 4, 2012 #3

    Erland

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    Square, you are assuming what should be proved: You assume that there is a matrix ##A^{-1}## such that ##AA^{-1}=A^{-1}A=I##, but this is what should be proved.

    To prove that ##BA=I## implies ##AB=I## is not entirely easy. First of all, we must assume that ##A## and/or ##B## are square matrices. Otherwise, this is false. For example, if

    ##B=[1\quad 1]## and ##A=[2\quad -1]^{T}##, then ##BA=[1]=I_1## and ##AB\neq I_2##.

    So, assume that ##A## and ##B## are ##n\times n##-matrices, for some ##n\ge 1##, such that ##BA=I##.

    Now, consider the homogeneous system

    ##A\bf x= 0## (1).

    If ##\bf x## is a solution of (1), then

    ##{\bf x}=I{\bf x}=(BA){\bf x}=B(A{\bf x})=B{\bf 0}={\bf 0}##.

    This means that the homogeneous system (1) has only the trivial solution ##{\bf x} ={\bf 0}##. Thus, if we solve (1) by elimination, transforming ##A## to reduced echelon form by a sequence of elementary row operations, the resulting reduced echelon form must be ##I## (otherwise, there would be non-pivot columns which correspond to free variables).

    Now, consider the inhomogeneous system

    ##A{\bf x} ={\bf b}##, (2),

    where ##\bf b## is an arbitrary vector in ##R^n##.

    If we perform the same sequence of row operations as in the solution of (1), the resulting coefficient matrix is again ##I##, since we started with the same coefficient matrix ##A##.
    This means that (2) has a unique solution.

    This holds for all vectors ##{\bf b}\in R^n##. In particular, it holds for the standard basis vectors ##{\bf e_1}, {\bf e_2}\dots, {\bf e_n}##. Let us denote the solutions for these vectors by ##{\bf x_1}, {\bf x_2},\dots,{\bf x_n}##, respectively.

    Let ##X## be the ##n\times n##-matrix with the vectors ##{\bf x_1}, {\bf x_2},\dots,{\bf x_n}## as columms, that is ##X=[{\bf x_1}\, {\bf x_2}\dots{\bf x_n}]##. Also, notice that ##[{\bf e_1}\,{\bf e_2}\dots{\bf e_n}]=I##.

    It follows that ##AX=I##.

    Next, ##B=BI=B(AX)=(BA)X=IX=X##, that is, ##B=X##.
    Hence, ##AB=I##.
     
    Last edited: Nov 4, 2012
  5. Nov 5, 2012 #4

    mathwonk

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    it is not true unless the matrices are square.
     
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