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Does this work. Matrices

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data

    (a) Show that for any invertible matrix A,

    [itex](A^{-1})^TA^T=I[/itex] and [itex]A^T(A^{-1})^T=I[/itex]

    (b) Deduce that AT is invertible and that its inverse is the transpose of [itex]A^{-1}[/itex]

    (c) Deduce also that if A is symmetric then A-1 is also symmetric.


    2. Relevant equations (AB)T=BTAT



    3. The attempt at a solution

    (a) If A is invertible,

    [itex]AA^{-1}=A^{-1}A=I[/itex]
    [itex]\Rightarrow (AA^{-1})^T=I^T[/itex]
    [itex]\Rightarrow (AA^{-1})^T=I[/itex]
    [itex]\Rightarrow (A^{-1})^TA^T=I[/itex]


    Now for part (b) and (c)
     
  2. jcsd
  3. Jul 8, 2008 #2

    Dick

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    Yes, that works GREAT! Doesn't that make b) pretty easy?
     
  4. Jul 8, 2008 #3
    Oh yeah. I guess it kind of takes care of it right? Since this last line [itex](A^{-1})^TA^T=I[/itex] is the definition of an Inverse? That, is: if AB=BA=I then B=A^{-1}

    So I also have to show, somehow, that [itex]A^T(A^{-1})^T=I[/itex] as well?
     
  5. Jul 8, 2008 #4

    Dick

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    The definition does say 'if AB=BA=I'. So you'd better show both. There's nothing hard about it.
     
  6. Jul 9, 2008 #5


    If A is invertible, [itex]AA^{-1}=A^{-1}A=I[/itex]


    [itex]\Rightarrow A^{-1}A=I[/itex]
    [itex]\Rightarrow (A^{-1}A)^T=I^T[/itex]
    [itex]\Rightarrow (A^{-1}A)^T=I[/itex]
    [itex]\Rightarrow A^T(A^{-1})^T=I[/itex]

    Alright-then :smile:

    Now how about part (c). . . Deduce also that if A is symmetric then A-1 is also symmetric.

    If A is symmetric, A=AT and if A-1 is symmetric, A-1=(A-1)T

    Let me think for a minute here...
     
  7. Jul 9, 2008 #6
    Remember:

    [tex](\mathbf{A}^\mathrm{T})^{-1} = (\mathbf{A}^{-1})^\mathrm{T}[/tex]
     
  8. Jul 9, 2008 #7
    Is this a property of matrices, or just of exponents in general? Is this just saying that exponents can 'commute'? Sorry if that seems like a stupid question :redface:

    :smile:
     
  9. Jul 9, 2008 #8
    It's a property of the transpose of matrices I think you need it here.
     
  10. Jul 9, 2008 #9

    Dick

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    No, you can't 'commute' everything that's written as a superscript. But you have already shown that [tex]A^T(A^{-1})^T=I[/tex]. That means that the second matrix is the inverse of the first. The inverse of the first is [tex](A^T)^{-1}[/tex]. It's pretty easy to show for powers as well.
     
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