Is a Polar Decomposition of Normal Operators Unique?












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Here I am concerned with finite dimensional complex inner product spaces.



We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)










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    0












    $begingroup$


    Here I am concerned with finite dimensional complex inner product spaces.



    We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)










    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Here I am concerned with finite dimensional complex inner product spaces.



      We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)










      share|cite|improve this question









      $endgroup$




      Here I am concerned with finite dimensional complex inner product spaces.



      We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)







      linear-algebra functional-analysis






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      asked Dec 15 '18 at 4:19









      CF8432CF8432

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          $begingroup$

          The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.



          For a concrete example: take
          $$
          A = pmatrix{1&0\0&0\&&ddots\&&&0}
          $$

          Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.






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            1 Answer
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            1 Answer
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            $begingroup$

            The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.



            For a concrete example: take
            $$
            A = pmatrix{1&0\0&0\&&ddots\&&&0}
            $$

            Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.



              For a concrete example: take
              $$
              A = pmatrix{1&0\0&0\&&ddots\&&&0}
              $$

              Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.



                For a concrete example: take
                $$
                A = pmatrix{1&0\0&0\&&ddots\&&&0}
                $$

                Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.






                share|cite|improve this answer









                $endgroup$



                The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.



                For a concrete example: take
                $$
                A = pmatrix{1&0\0&0\&&ddots\&&&0}
                $$

                Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 15 '18 at 4:26









                OmnomnomnomOmnomnomnom

                127k790178




                127k790178






























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