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Prove that linear operator is invertible

  1. Dec 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\mathcal{A}: \mathbb{R^3}\rightarrow \mathbb{R^3}[/itex] is a linear operator defined as [itex]\mathcal{A}(x_1,x_2,x_3)=(x_1+x_2-x_3, x_2+7x_3, -x_3)[/itex]
    Prove that [itex]\mathcal{A}[/itex] is invertible and find matrix of [itex]\mathcal{A},A^{-1}[/itex] in terms of canonical basis of [itex]\mathbb{R^3}[/itex].

    2. Relevant equations
    -Linear transformations
    -Jordan decomposition

    3. The attempt at a solution

    Linear mapping can be written as
    [tex]\mathcal{A}
    \begin{bmatrix}
    x_1 \\
    x_2 \\
    x_3 \\
    \end{bmatrix}
    =
    \begin{bmatrix}
    x_1+x_2-x_3 \\
    x_2+7x_3 \\
    -x_3 \\
    \end{bmatrix}
    [/tex]

    Matrix of linear operator [itex]\mathcal{A}[/itex] is [tex]T=

    \begin{bmatrix}
    1 & 1 & -1 \\
    0 & 1 & 7 \\
    0 & 0 & -1 \\
    \end{bmatrix}

    [/tex]

    Using Jordan decomposition method, it is possible to find [itex]T,T^{-1}[/itex] in terms of canonical basis.

    How to strictly prove that [itex]\mathcal{A}[/itex] is invertible (without matrix computation)?
     
  2. jcsd
  3. Dec 7, 2015 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    This belongs in the "Precalculus" math forum, not the Advanced Physics forum.

    Anyway, you do not need to use matrices in this problem: you just want to show that ##{\cal A}:\vec{x} \rightarrow \vec{y}## can be inverted, which means there is a map ##{\cal B}: \vec{y} \rightarrow \vec{x}## with the property that ##\vec{x} = {\cal B \, A} \vec{x}##. This amounts to simply solving the linear equations
    [tex] x_1+x_2-x_3 =y_1, \: x_2+7x_3 = y_2, \: -x_3 = y_3 [/tex]
    to determine the ##x_i## in terms of the ##y_j##. That is a simple high-school algebra exercise.
     
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