How to show for any symmetric matrices the quadratic mean of eigenvalues less than square of Frobenius norm?












1












$begingroup$


Let $A$ be a symmetric matrix which has $k$ non-zero eigenvalue. Show that the square of Frobenius norm is always bigger than the average of squared eigenvalues. That is:



$$|A|_F^2 geq frac{1}{k} (sum_{i=1}^k |lambda_i|)^2$$



My try: applying $|A|_F^2=sum_{i=1}^k lambda_i^2$.



Also, do we have the correspondence of this in a vector form?






linear-algebra eigenvalues-eigenvectors trace symmetric-matrices







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    1












    $begingroup$


    Let $A$ be a symmetric matrix which has $k$ non-zero eigenvalue. Show that the square of Frobenius norm is always bigger than the average of squared eigenvalues. That is:



    $$|A|_F^2 geq frac{1}{k} (sum_{i=1}^k |lambda_i|)^2$$



    My try: applying $|A|_F^2=sum_{i=1}^k lambda_i^2$.



    Also, do we have the correspondence of this in a vector form?






    linear-algebra eigenvalues-eigenvectors trace symmetric-matrices







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $A$ be a symmetric matrix which has $k$ non-zero eigenvalue. Show that the square of Frobenius norm is always bigger than the average of squared eigenvalues. That is:



      $$|A|_F^2 geq frac{1}{k} (sum_{i=1}^k |lambda_i|)^2$$



      My try: applying $|A|_F^2=sum_{i=1}^k lambda_i^2$.



      Also, do we have the correspondence of this in a vector form?






      linear-algebra eigenvalues-eigenvectors trace symmetric-matrices







      share|cite|improve this question











      $endgroup$




      Let $A$ be a symmetric matrix which has $k$ non-zero eigenvalue. Show that the square of Frobenius norm is always bigger than the average of squared eigenvalues. That is:



      $$|A|_F^2 geq frac{1}{k} (sum_{i=1}^k |lambda_i|)^2$$



      My try: applying $|A|_F^2=sum_{i=1}^k lambda_i^2$.



      Also, do we have the correspondence of this in a vector form?






      linear-algebra eigenvalues-eigenvectors trace symmetric-matrices




      linear-algebra eigenvalues-eigenvectors trace symmetric-matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 3:45







      Saeed

















      asked Jan 7 at 19:11









      SaeedSaeed

      763310




      763310






















          1 Answer
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          1












          $begingroup$

          We have
          begin{align*}
          k^{-1}|A|_F^2 = k^{-1}sum_{i=1}^{k}lambda_i^2 ge left(k^{-1}sum_{i=1}^{k}|lambda_i| right)^2
          end{align*}
          by the inequality between quadratic means and arithmetic means. Another proof would be, with $mathbf{lambda} = (|lambda_1|, cdots, |lambda_k|)$ and $mathbf{1} = (1, cdots, 1)$, we have
          begin{align*}
          |A|_F^2 = k^{-1}(mathbf{lambda}^intercalmathbf{lambda})(mathbf{1}^intercalmathbf{1}) ge k^{-1} (mathbf{lambda}^intercal mathbf{1})^2 = k^{-1}left(sum_{i=1}^{k}|lambda_i|right)^2
          end{align*}
          by Cauchy-Schwarz.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The second proof makes sense to me, but could you elaborate how we can use AM-GM to get the result because we have no multiplication hear?
            $endgroup$
            – Saeed
            Jan 7 at 19:56










          • $begingroup$
            The first proof uses AM-QM, not AM-GM. AM-QM is actually proved through exactly the same steps as seen with the CS proof seen in the second proof.
            $endgroup$
            – Tom Chen
            Jan 7 at 19:59










          • $begingroup$
            The second proof both $k^-1$ gets cancelled and does not give the claim? Can you elaborate that?
            $endgroup$
            – Saeed
            Jan 8 at 18:09










          • $begingroup$
            Oops, typo, the first $k^{-1}$ was not meant to be there. Fixed.
            $endgroup$
            – Tom Chen
            Jan 8 at 22:06











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          We have
          begin{align*}
          k^{-1}|A|_F^2 = k^{-1}sum_{i=1}^{k}lambda_i^2 ge left(k^{-1}sum_{i=1}^{k}|lambda_i| right)^2
          end{align*}
          by the inequality between quadratic means and arithmetic means. Another proof would be, with $mathbf{lambda} = (|lambda_1|, cdots, |lambda_k|)$ and $mathbf{1} = (1, cdots, 1)$, we have
          begin{align*}
          |A|_F^2 = k^{-1}(mathbf{lambda}^intercalmathbf{lambda})(mathbf{1}^intercalmathbf{1}) ge k^{-1} (mathbf{lambda}^intercal mathbf{1})^2 = k^{-1}left(sum_{i=1}^{k}|lambda_i|right)^2
          end{align*}
          by Cauchy-Schwarz.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The second proof makes sense to me, but could you elaborate how we can use AM-GM to get the result because we have no multiplication hear?
            $endgroup$
            – Saeed
            Jan 7 at 19:56










          • $begingroup$
            The first proof uses AM-QM, not AM-GM. AM-QM is actually proved through exactly the same steps as seen with the CS proof seen in the second proof.
            $endgroup$
            – Tom Chen
            Jan 7 at 19:59










          • $begingroup$
            The second proof both $k^-1$ gets cancelled and does not give the claim? Can you elaborate that?
            $endgroup$
            – Saeed
            Jan 8 at 18:09










          • $begingroup$
            Oops, typo, the first $k^{-1}$ was not meant to be there. Fixed.
            $endgroup$
            – Tom Chen
            Jan 8 at 22:06
















          1












          $begingroup$

          We have
          begin{align*}
          k^{-1}|A|_F^2 = k^{-1}sum_{i=1}^{k}lambda_i^2 ge left(k^{-1}sum_{i=1}^{k}|lambda_i| right)^2
          end{align*}
          by the inequality between quadratic means and arithmetic means. Another proof would be, with $mathbf{lambda} = (|lambda_1|, cdots, |lambda_k|)$ and $mathbf{1} = (1, cdots, 1)$, we have
          begin{align*}
          |A|_F^2 = k^{-1}(mathbf{lambda}^intercalmathbf{lambda})(mathbf{1}^intercalmathbf{1}) ge k^{-1} (mathbf{lambda}^intercal mathbf{1})^2 = k^{-1}left(sum_{i=1}^{k}|lambda_i|right)^2
          end{align*}
          by Cauchy-Schwarz.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The second proof makes sense to me, but could you elaborate how we can use AM-GM to get the result because we have no multiplication hear?
            $endgroup$
            – Saeed
            Jan 7 at 19:56










          • $begingroup$
            The first proof uses AM-QM, not AM-GM. AM-QM is actually proved through exactly the same steps as seen with the CS proof seen in the second proof.
            $endgroup$
            – Tom Chen
            Jan 7 at 19:59










          • $begingroup$
            The second proof both $k^-1$ gets cancelled and does not give the claim? Can you elaborate that?
            $endgroup$
            – Saeed
            Jan 8 at 18:09










          • $begingroup$
            Oops, typo, the first $k^{-1}$ was not meant to be there. Fixed.
            $endgroup$
            – Tom Chen
            Jan 8 at 22:06














          1












          1








          1





          $begingroup$

          We have
          begin{align*}
          k^{-1}|A|_F^2 = k^{-1}sum_{i=1}^{k}lambda_i^2 ge left(k^{-1}sum_{i=1}^{k}|lambda_i| right)^2
          end{align*}
          by the inequality between quadratic means and arithmetic means. Another proof would be, with $mathbf{lambda} = (|lambda_1|, cdots, |lambda_k|)$ and $mathbf{1} = (1, cdots, 1)$, we have
          begin{align*}
          |A|_F^2 = k^{-1}(mathbf{lambda}^intercalmathbf{lambda})(mathbf{1}^intercalmathbf{1}) ge k^{-1} (mathbf{lambda}^intercal mathbf{1})^2 = k^{-1}left(sum_{i=1}^{k}|lambda_i|right)^2
          end{align*}
          by Cauchy-Schwarz.






          share|cite|improve this answer











          $endgroup$



          We have
          begin{align*}
          k^{-1}|A|_F^2 = k^{-1}sum_{i=1}^{k}lambda_i^2 ge left(k^{-1}sum_{i=1}^{k}|lambda_i| right)^2
          end{align*}
          by the inequality between quadratic means and arithmetic means. Another proof would be, with $mathbf{lambda} = (|lambda_1|, cdots, |lambda_k|)$ and $mathbf{1} = (1, cdots, 1)$, we have
          begin{align*}
          |A|_F^2 = k^{-1}(mathbf{lambda}^intercalmathbf{lambda})(mathbf{1}^intercalmathbf{1}) ge k^{-1} (mathbf{lambda}^intercal mathbf{1})^2 = k^{-1}left(sum_{i=1}^{k}|lambda_i|right)^2
          end{align*}
          by Cauchy-Schwarz.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 22:04









          Saeed

          763310




          763310










          answered Jan 7 at 19:25









          Tom ChenTom Chen

          548212




          548212












          • $begingroup$
            The second proof makes sense to me, but could you elaborate how we can use AM-GM to get the result because we have no multiplication hear?
            $endgroup$
            – Saeed
            Jan 7 at 19:56










          • $begingroup$
            The first proof uses AM-QM, not AM-GM. AM-QM is actually proved through exactly the same steps as seen with the CS proof seen in the second proof.
            $endgroup$
            – Tom Chen
            Jan 7 at 19:59










          • $begingroup$
            The second proof both $k^-1$ gets cancelled and does not give the claim? Can you elaborate that?
            $endgroup$
            – Saeed
            Jan 8 at 18:09










          • $begingroup$
            Oops, typo, the first $k^{-1}$ was not meant to be there. Fixed.
            $endgroup$
            – Tom Chen
            Jan 8 at 22:06


















          • $begingroup$
            The second proof makes sense to me, but could you elaborate how we can use AM-GM to get the result because we have no multiplication hear?
            $endgroup$
            – Saeed
            Jan 7 at 19:56










          • $begingroup$
            The first proof uses AM-QM, not AM-GM. AM-QM is actually proved through exactly the same steps as seen with the CS proof seen in the second proof.
            $endgroup$
            – Tom Chen
            Jan 7 at 19:59










          • $begingroup$
            The second proof both $k^-1$ gets cancelled and does not give the claim? Can you elaborate that?
            $endgroup$
            – Saeed
            Jan 8 at 18:09










          • $begingroup$
            Oops, typo, the first $k^{-1}$ was not meant to be there. Fixed.
            $endgroup$
            – Tom Chen
            Jan 8 at 22:06
















          $begingroup$
          The second proof makes sense to me, but could you elaborate how we can use AM-GM to get the result because we have no multiplication hear?
          $endgroup$
          – Saeed
          Jan 7 at 19:56




          $begingroup$
          The second proof makes sense to me, but could you elaborate how we can use AM-GM to get the result because we have no multiplication hear?
          $endgroup$
          – Saeed
          Jan 7 at 19:56












          $begingroup$
          The first proof uses AM-QM, not AM-GM. AM-QM is actually proved through exactly the same steps as seen with the CS proof seen in the second proof.
          $endgroup$
          – Tom Chen
          Jan 7 at 19:59




          $begingroup$
          The first proof uses AM-QM, not AM-GM. AM-QM is actually proved through exactly the same steps as seen with the CS proof seen in the second proof.
          $endgroup$
          – Tom Chen
          Jan 7 at 19:59












          $begingroup$
          The second proof both $k^-1$ gets cancelled and does not give the claim? Can you elaborate that?
          $endgroup$
          – Saeed
          Jan 8 at 18:09




          $begingroup$
          The second proof both $k^-1$ gets cancelled and does not give the claim? Can you elaborate that?
          $endgroup$
          – Saeed
          Jan 8 at 18:09












          $begingroup$
          Oops, typo, the first $k^{-1}$ was not meant to be there. Fixed.
          $endgroup$
          – Tom Chen
          Jan 8 at 22:06




          $begingroup$
          Oops, typo, the first $k^{-1}$ was not meant to be there. Fixed.
          $endgroup$
          – Tom Chen
          Jan 8 at 22:06


















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