A question about the symmetric positive definite matrix A and $D^{-1/2}AD^{-1/2}$












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Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.



What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?






linear-algebra matrices positive-definite







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    1












    $begingroup$


    Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.



    What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?






    linear-algebra matrices positive-definite







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.



      What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?






      linear-algebra matrices positive-definite







      share|cite|improve this question











      $endgroup$




      Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.



      What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?






      linear-algebra matrices positive-definite




      linear-algebra matrices positive-definite






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 15:16









      gt6989b

      34.3k22455




      34.3k22455










      asked Dec 26 '18 at 15:15









      like_mathlike_math

      29917




      29917






















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          $begingroup$

          There is not much to say.



          $D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.






          share|cite|improve this answer









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          • $begingroup$
            I think you are right. Thank you
            $endgroup$
            – like_math
            Dec 26 '18 at 22:00











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          1 Answer
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          active

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          oldest

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          active

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          $begingroup$

          There is not much to say.



          $D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think you are right. Thank you
            $endgroup$
            – like_math
            Dec 26 '18 at 22:00
















          2












          $begingroup$

          There is not much to say.



          $D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think you are right. Thank you
            $endgroup$
            – like_math
            Dec 26 '18 at 22:00














          2












          2








          2





          $begingroup$

          There is not much to say.



          $D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.






          share|cite|improve this answer









          $endgroup$



          There is not much to say.



          $D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 15:26









          Robert IsraelRobert Israel

          324k23214468




          324k23214468












          • $begingroup$
            I think you are right. Thank you
            $endgroup$
            – like_math
            Dec 26 '18 at 22:00


















          • $begingroup$
            I think you are right. Thank you
            $endgroup$
            – like_math
            Dec 26 '18 at 22:00
















          $begingroup$
          I think you are right. Thank you
          $endgroup$
          – like_math
          Dec 26 '18 at 22:00




          $begingroup$
          I think you are right. Thank you
          $endgroup$
          – like_math
          Dec 26 '18 at 22:00


















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