Proof on unitary diagonalisable matrices












1












$begingroup$


My task is the following:



$Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



My approach was:



$M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.






linear-algebra proof-writing diagonalization







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$endgroup$

















    1












    $begingroup$


    My task is the following:



    $Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



    My approach was:



    $M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



    But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.






    linear-algebra proof-writing diagonalization







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      My task is the following:



      $Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



      My approach was:



      $M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



      But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.






      linear-algebra proof-writing diagonalization







      share|cite|improve this question









      $endgroup$




      My task is the following:



      $Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



      My approach was:



      $M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



      But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.






      linear-algebra proof-writing diagonalization




      linear-algebra proof-writing diagonalization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 16:57









      Fo Young Areal LoFo Young Areal Lo

      345




      345






















          1 Answer
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          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}
          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01











          Your Answer





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          1 Answer
          1






          active

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          active

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          active

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          2












          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}
          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01
















          2












          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}
          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01














          2












          2








          2





          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}
          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$



          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}
          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 17:14

























          answered Jan 8 at 17:12









          jobejobe

          994615




          994615












          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01


















          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01
















          $begingroup$
          Why is A* = U M* U* shouldn't it be A* = U* M* U
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:46






          $begingroup$
          Why is A* = U M* U* shouldn't it be A* = U* M* U
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:46














          $begingroup$
          If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:51




          $begingroup$
          If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:51












          $begingroup$
          $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:53




          $begingroup$
          $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:53












          $begingroup$
          Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:55






          $begingroup$
          Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:55














          $begingroup$
          But I looked it up, seems like I got something wrong. Thank you
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 18:01




          $begingroup$
          But I looked it up, seems like I got something wrong. Thank you
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 18:01


















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