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Homework Help: Diagonalizable matrix A^-1=a

  1. Apr 24, 2010 #1
    Let A be a diagonalizable matrix whose eigenvalues are all either 1 or -1. Show that [tex]A^{-1}=A[/tex].

    [tex]A=X\begin{bmatrix}
    \pm 1 & \cdots & 0 \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & \pm 1
    \end{bmatrix}X^{-1}
    [/tex] and [tex]A^{-1}=X\begin{bmatrix}
    \pm 1 & \cdots & 0 \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & \pm 1
    \end{bmatrix}^{-1}X^{-1}
    [/tex]

    How do I show they are equal?
     
  2. jcsd
  3. Apr 24, 2010 #2

    rock.freak667

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    Re: A^-1=a

    How would you compute the diagonal matrix raised to any power n?
     
  4. Apr 24, 2010 #3
    Re: A^-1=a

    [tex]
    A^n=X\begin{bmatrix}
    \pm 1 & \cdots & 0 \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & \pm 1
    \end{bmatrix}^nX^{-1}

    [/tex]
     
  5. Apr 24, 2010 #4

    gabbagabbahey

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    Re: A^-1=a

    He was asking how you compute/simplify

    [tex]\begin{bmatrix}
    \lambda_1 & \cdots & 0 \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & \lambda_m
    \end{bmatrix}^n[/tex]
     
  6. Apr 24, 2010 #5
    Re: A^-1=a

    Raise the diagonal terms by n since this just a diagonal matrix.
     
  7. Apr 24, 2010 #6

    gabbagabbahey

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    Re: A^-1=a

    Right, so do that with [itex]n=-1[/itex]....what is [itex](\pm 1)^{-1}[/itex]?
     
  8. Apr 24, 2010 #7
    Re: A^-1=a

    I don't understand what you mean with your latex code
     
  9. Apr 24, 2010 #8

    gabbagabbahey

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    Re: A^-1=a

    It should be displaying properly now, try refreshing your page.
     
  10. Apr 24, 2010 #9
    Re: A^-1=a

    [tex](\pm 1)^{-1}=\pm 1[/tex]
     
  11. Apr 24, 2010 #10

    gabbagabbahey

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    Re: A^-1=a

    Right, so

    [tex]\begin{bmatrix}
    \pm 1 & \cdots & 0 \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & \pm 1
    \end{bmatrix}^{-1}=\begin{bmatrix}
    \pm 1 & \cdots & 0 \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & \pm 1
    \end{bmatrix}[/tex]

    ...Plug that into your equation for [itex]A^{-1}[/itex]
     
  12. Apr 24, 2010 #11
    Re: A^-1=a

    Ok I understand thanks.
     
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