How to show $sqrt{text{Tr}(A^2)} leq text{Tr}(A)$?












2












$begingroup$


Let $A$ be a positive semi-definite matrix. How to show that Frobenius norm is less than trace of the matrix? Formally,
$$sqrt{text{Tr}(A^2)} leq text{Tr}(A)$$
Also, show when $A$ is an $n times m$ the following is true
$$sqrt{text{Tr}(A^TA)} leq |A|_*$$
where $|cdot|_*$ is nuclear norm which is the summation of the singular values.






matrices inequality trace







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    2












    $begingroup$


    Let $A$ be a positive semi-definite matrix. How to show that Frobenius norm is less than trace of the matrix? Formally,
    $$sqrt{text{Tr}(A^2)} leq text{Tr}(A)$$
    Also, show when $A$ is an $n times m$ the following is true
    $$sqrt{text{Tr}(A^TA)} leq |A|_*$$
    where $|cdot|_*$ is nuclear norm which is the summation of the singular values.






    matrices inequality trace







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $A$ be a positive semi-definite matrix. How to show that Frobenius norm is less than trace of the matrix? Formally,
      $$sqrt{text{Tr}(A^2)} leq text{Tr}(A)$$
      Also, show when $A$ is an $n times m$ the following is true
      $$sqrt{text{Tr}(A^TA)} leq |A|_*$$
      where $|cdot|_*$ is nuclear norm which is the summation of the singular values.






      matrices inequality trace







      share|cite|improve this question









      $endgroup$




      Let $A$ be a positive semi-definite matrix. How to show that Frobenius norm is less than trace of the matrix? Formally,
      $$sqrt{text{Tr}(A^2)} leq text{Tr}(A)$$
      Also, show when $A$ is an $n times m$ the following is true
      $$sqrt{text{Tr}(A^TA)} leq |A|_*$$
      where $|cdot|_*$ is nuclear norm which is the summation of the singular values.






      matrices inequality trace




      matrices inequality trace






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 16 at 3:12









      SaeedSaeed

      1,036310




      1,036310






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          For the first you can assume that $A$ is diagonal with diagonal entries
          $a_1,ldots,a_n$, all $ge0$. Then your inequality becomes
          $$sum_{i=1}^n a_i^2leleft(sum_{i=1}^n a_iright)^2$$
          which is clearly true on expanding the right side, recalling all variables
          are non-negative.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How about the other?
            $endgroup$
            – Saeed
            Jan 16 at 3:20










          • $begingroup$
            How about applying the first to the psd square root of $A^TA$? @Saeed
            $endgroup$
            – Lord Shark the Unknown
            Jan 16 at 3:25





















          2












          $begingroup$

          Let $sigma_1, ldots, sigma_r$ be the singular values of $A$.
          Then
          $$sqrt{text{Tr}(A^top A)} = sqrt{sum_i sigma^2_r} le sum_i |sigma_r| = |A|_*.$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            For the first you can assume that $A$ is diagonal with diagonal entries
            $a_1,ldots,a_n$, all $ge0$. Then your inequality becomes
            $$sum_{i=1}^n a_i^2leleft(sum_{i=1}^n a_iright)^2$$
            which is clearly true on expanding the right side, recalling all variables
            are non-negative.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How about the other?
              $endgroup$
              – Saeed
              Jan 16 at 3:20










            • $begingroup$
              How about applying the first to the psd square root of $A^TA$? @Saeed
              $endgroup$
              – Lord Shark the Unknown
              Jan 16 at 3:25


















            1












            $begingroup$

            For the first you can assume that $A$ is diagonal with diagonal entries
            $a_1,ldots,a_n$, all $ge0$. Then your inequality becomes
            $$sum_{i=1}^n a_i^2leleft(sum_{i=1}^n a_iright)^2$$
            which is clearly true on expanding the right side, recalling all variables
            are non-negative.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How about the other?
              $endgroup$
              – Saeed
              Jan 16 at 3:20










            • $begingroup$
              How about applying the first to the psd square root of $A^TA$? @Saeed
              $endgroup$
              – Lord Shark the Unknown
              Jan 16 at 3:25
















            1












            1








            1





            $begingroup$

            For the first you can assume that $A$ is diagonal with diagonal entries
            $a_1,ldots,a_n$, all $ge0$. Then your inequality becomes
            $$sum_{i=1}^n a_i^2leleft(sum_{i=1}^n a_iright)^2$$
            which is clearly true on expanding the right side, recalling all variables
            are non-negative.






            share|cite|improve this answer









            $endgroup$



            For the first you can assume that $A$ is diagonal with diagonal entries
            $a_1,ldots,a_n$, all $ge0$. Then your inequality becomes
            $$sum_{i=1}^n a_i^2leleft(sum_{i=1}^n a_iright)^2$$
            which is clearly true on expanding the right side, recalling all variables
            are non-negative.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 16 at 3:18









            Lord Shark the UnknownLord Shark the Unknown

            105k1160132




            105k1160132












            • $begingroup$
              How about the other?
              $endgroup$
              – Saeed
              Jan 16 at 3:20










            • $begingroup$
              How about applying the first to the psd square root of $A^TA$? @Saeed
              $endgroup$
              – Lord Shark the Unknown
              Jan 16 at 3:25




















            • $begingroup$
              How about the other?
              $endgroup$
              – Saeed
              Jan 16 at 3:20










            • $begingroup$
              How about applying the first to the psd square root of $A^TA$? @Saeed
              $endgroup$
              – Lord Shark the Unknown
              Jan 16 at 3:25


















            $begingroup$
            How about the other?
            $endgroup$
            – Saeed
            Jan 16 at 3:20




            $begingroup$
            How about the other?
            $endgroup$
            – Saeed
            Jan 16 at 3:20












            $begingroup$
            How about applying the first to the psd square root of $A^TA$? @Saeed
            $endgroup$
            – Lord Shark the Unknown
            Jan 16 at 3:25






            $begingroup$
            How about applying the first to the psd square root of $A^TA$? @Saeed
            $endgroup$
            – Lord Shark the Unknown
            Jan 16 at 3:25













            2












            $begingroup$

            Let $sigma_1, ldots, sigma_r$ be the singular values of $A$.
            Then
            $$sqrt{text{Tr}(A^top A)} = sqrt{sum_i sigma^2_r} le sum_i |sigma_r| = |A|_*.$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Let $sigma_1, ldots, sigma_r$ be the singular values of $A$.
              Then
              $$sqrt{text{Tr}(A^top A)} = sqrt{sum_i sigma^2_r} le sum_i |sigma_r| = |A|_*.$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Let $sigma_1, ldots, sigma_r$ be the singular values of $A$.
                Then
                $$sqrt{text{Tr}(A^top A)} = sqrt{sum_i sigma^2_r} le sum_i |sigma_r| = |A|_*.$$






                share|cite|improve this answer









                $endgroup$



                Let $sigma_1, ldots, sigma_r$ be the singular values of $A$.
                Then
                $$sqrt{text{Tr}(A^top A)} = sqrt{sum_i sigma^2_r} le sum_i |sigma_r| = |A|_*.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 3:26









                angryavianangryavian

                41.5k23381




                41.5k23381






























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