If $T$ is an invertible linear transformation and $vec{v}$ is an eigenvector of $T$, then $vec{v}$ is an...












1












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I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?






linear-algebra eigenvalues-eigenvectors transformation







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  • $begingroup$
    Can't you just fix bases and consider your linear transformation as a matrix?
    $endgroup$
    – dcolazin
    yesterday










  • $begingroup$
    yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
    $endgroup$
    – DGR
    yesterday










  • $begingroup$
    Wouldn’t the proof be exactly the same as the one for matrices?
    $endgroup$
    – amd
    yesterday
















1












$begingroup$


I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?






linear-algebra eigenvalues-eigenvectors transformation







share|cite|improve this question









New contributor




DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Can't you just fix bases and consider your linear transformation as a matrix?
    $endgroup$
    – dcolazin
    yesterday










  • $begingroup$
    yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
    $endgroup$
    – DGR
    yesterday










  • $begingroup$
    Wouldn’t the proof be exactly the same as the one for matrices?
    $endgroup$
    – amd
    yesterday














1












1








1





$begingroup$


I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?






linear-algebra eigenvalues-eigenvectors transformation







share|cite|improve this question









New contributor




DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?






linear-algebra eigenvalues-eigenvectors transformation




linear-algebra eigenvalues-eigenvectors transformation






share|cite|improve this question









New contributor




DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




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share|cite|improve this question




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edited yesterday









YuiTo Cheng

2,41341037




2,41341037






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asked yesterday









DGRDGR

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163




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  • $begingroup$
    Can't you just fix bases and consider your linear transformation as a matrix?
    $endgroup$
    – dcolazin
    yesterday










  • $begingroup$
    yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
    $endgroup$
    – DGR
    yesterday










  • $begingroup$
    Wouldn’t the proof be exactly the same as the one for matrices?
    $endgroup$
    – amd
    yesterday


















  • $begingroup$
    Can't you just fix bases and consider your linear transformation as a matrix?
    $endgroup$
    – dcolazin
    yesterday










  • $begingroup$
    yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
    $endgroup$
    – DGR
    yesterday










  • $begingroup$
    Wouldn’t the proof be exactly the same as the one for matrices?
    $endgroup$
    – amd
    yesterday
















$begingroup$
Can't you just fix bases and consider your linear transformation as a matrix?
$endgroup$
– dcolazin
yesterday




$begingroup$
Can't you just fix bases and consider your linear transformation as a matrix?
$endgroup$
– dcolazin
yesterday












$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday




$begingroup$
yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
$endgroup$
– DGR
yesterday












$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday




$begingroup$
Wouldn’t the proof be exactly the same as the one for matrices?
$endgroup$
– amd
yesterday










2 Answers
2






active

oldest

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4












$begingroup$

Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
begin{align*}
mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
end{align*}
Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$






share|cite|improve this answer









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    1












    $begingroup$

    OK, so Iv'e found a solution!



    begin{align*}\
    Tv &= lambda v vert *T^{-1} \
    T^{-1}Tv &= T^{-1}lambda v \
    v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
    lambda^{-1}v &= T^{-1}v
    end{align*}



    For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).



    Hope this helps!






    share|cite|improve this answer










    New contributor




    DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






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      2 Answers
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      2 Answers
      2






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      4












      $begingroup$

      Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
      begin{align*}
      mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
      end{align*}
      Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
        begin{align*}
        mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
        end{align*}
        Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
          begin{align*}
          mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
          end{align*}
          Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$






          share|cite|improve this answer









          $endgroup$



          Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
          begin{align*}
          mathbf v &= I mathbf v\&=A^{-1}A mathbf v\&=lambda A^{-1} mathbf v
          end{align*}
          Hence $A^{-1} mathbf v = lambda^{-1}mathbf v$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          P. QuintonP. Quinton

          2,0001214




          2,0001214























              1












              $begingroup$

              OK, so Iv'e found a solution!



              begin{align*}\
              Tv &= lambda v vert *T^{-1} \
              T^{-1}Tv &= T^{-1}lambda v \
              v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
              lambda^{-1}v &= T^{-1}v
              end{align*}



              For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).



              Hope this helps!






              share|cite|improve this answer










              New contributor




              DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$


















                1












                $begingroup$

                OK, so Iv'e found a solution!



                begin{align*}\
                Tv &= lambda v vert *T^{-1} \
                T^{-1}Tv &= T^{-1}lambda v \
                v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
                lambda^{-1}v &= T^{-1}v
                end{align*}



                For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).



                Hope this helps!






                share|cite|improve this answer










                New contributor




                DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  OK, so Iv'e found a solution!



                  begin{align*}\
                  Tv &= lambda v vert *T^{-1} \
                  T^{-1}Tv &= T^{-1}lambda v \
                  v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
                  lambda^{-1}v &= T^{-1}v
                  end{align*}



                  For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).



                  Hope this helps!






                  share|cite|improve this answer










                  New contributor




                  DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$



                  OK, so Iv'e found a solution!



                  begin{align*}\
                  Tv &= lambda v vert *T^{-1} \
                  T^{-1}Tv &= T^{-1}lambda v \
                  v &= lambda T^{-1}v vert *lambda^{-1} & text{(Invertible transformation, $lambdaneq 0$)} \
                  lambda^{-1}v &= T^{-1}v
                  end{align*}



                  For $T^{-1}$, eigenvalue $lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).



                  Hope this helps!







                  share|cite|improve this answer










                  New contributor




                  DGR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited yesterday









                  Widawensen

                  4,80531447




                  4,80531447






                  New contributor




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                  Check out our Code of Conduct.









                  answered yesterday









                  DGRDGR

                  163




                  163




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