Do QR and RQ have the same eigenvalues [duplicate]

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  • Eigen Values Proof

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Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?



$det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$



so they have the same eigenvalue, is this ok?










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marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















    up vote
    2
    down vote

    favorite
    1













    This question already has an answer here:



    • Eigen Values Proof

      1 answer



    Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?



    $det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$



    so they have the same eigenvalue, is this ok?










    share|cite|improve this question















    marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
















      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1






      This question already has an answer here:



      • Eigen Values Proof

        1 answer



      Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?



      $det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$



      so they have the same eigenvalue, is this ok?










      share|cite|improve this question
















      This question already has an answer here:



      • Eigen Values Proof

        1 answer



      Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?



      $det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$



      so they have the same eigenvalue, is this ok?





      This question already has an answer here:



      • Eigen Values Proof

        1 answer







      linear-algebra eigenvalues-eigenvectors orthogonal-matrices






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      edited Sep 4 at 12:16









      Bernard

      112k635104




      112k635104










      asked Sep 4 at 11:15









      Marko Škorić

      3908




      3908




      marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















          1 Answer
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          Both matrices are similar:
          $$
          A = QR = Q(RQ)Q^-1,
          $$
          hence they have the same Jordan form.






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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote













            Both matrices are similar:
            $$
            A = QR = Q(RQ)Q^-1,
            $$
            hence they have the same Jordan form.






            share|cite|improve this answer
























              up vote
              1
              down vote













              Both matrices are similar:
              $$
              A = QR = Q(RQ)Q^-1,
              $$
              hence they have the same Jordan form.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                Both matrices are similar:
                $$
                A = QR = Q(RQ)Q^-1,
                $$
                hence they have the same Jordan form.






                share|cite|improve this answer












                Both matrices are similar:
                $$
                A = QR = Q(RQ)Q^-1,
                $$
                hence they have the same Jordan form.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 4 at 11:28









                daw

                22.2k1542




                22.2k1542












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