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







share|cite|improve this question











$endgroup$

















    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
          1






          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065379%2fhow-to-show-for-any-symmetric-matrices-the-quadratic-mean-of-eigenvalues-less-th%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065379%2fhow-to-show-for-any-symmetric-matrices-the-quadratic-mean-of-eigenvalues-less-th%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
          Read more