How do I find eigenvalues of an hessian matrix that depends on (x,y,z)?












0














For example:



$f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$



Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$



Well, how do I calculate the eigen values?
Is it going to be in function of y and z? So they won't be fixed?



If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?



I'm a little bit confused:
my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.



Thank you all!!










share|cite|improve this question





























    0














    For example:



    $f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$



    Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$



    Well, how do I calculate the eigen values?
    Is it going to be in function of y and z? So they won't be fixed?



    If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?



    I'm a little bit confused:
    my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.



    Thank you all!!










    share|cite|improve this question



























      0












      0








      0







      For example:



      $f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$



      Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$



      Well, how do I calculate the eigen values?
      Is it going to be in function of y and z? So they won't be fixed?



      If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?



      I'm a little bit confused:
      my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.



      Thank you all!!










      share|cite|improve this question















      For example:



      $f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$



      Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$



      Well, how do I calculate the eigen values?
      Is it going to be in function of y and z? So they won't be fixed?



      If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?



      I'm a little bit confused:
      my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.



      Thank you all!!







      linear-algebra analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 15:13









      greedoid

      38.6k114797




      38.6k114797










      asked Nov 29 '18 at 14:53









      Lucas TononLucas Tonon

      445




      445






















          1 Answer
          1






          active

          oldest

          votes


















          0














          A quick comment first,



          $$
          frac{partial |y|^3}{partial y} not= 2y
          $$



          as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.






          share|cite|improve this answer





















          • actualy it is 6y, sorry my bad, but anyways the problem stands still
            – Lucas Tonon
            Nov 29 '18 at 15:05










          • Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
            – Lucas Tonon
            Nov 29 '18 at 15:06






          • 1




            @LucasTonon Yes, absolutely. There's not a problem with that
            – caverac
            Nov 29 '18 at 15:06










          • it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
            – Lucas Tonon
            Nov 29 '18 at 15:11











          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%2f3018724%2fhow-do-i-find-eigenvalues-of-an-hessian-matrix-that-depends-on-x-y-z%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









          0














          A quick comment first,



          $$
          frac{partial |y|^3}{partial y} not= 2y
          $$



          as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.






          share|cite|improve this answer





















          • actualy it is 6y, sorry my bad, but anyways the problem stands still
            – Lucas Tonon
            Nov 29 '18 at 15:05










          • Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
            – Lucas Tonon
            Nov 29 '18 at 15:06






          • 1




            @LucasTonon Yes, absolutely. There's not a problem with that
            – caverac
            Nov 29 '18 at 15:06










          • it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
            – Lucas Tonon
            Nov 29 '18 at 15:11
















          0














          A quick comment first,



          $$
          frac{partial |y|^3}{partial y} not= 2y
          $$



          as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.






          share|cite|improve this answer





















          • actualy it is 6y, sorry my bad, but anyways the problem stands still
            – Lucas Tonon
            Nov 29 '18 at 15:05










          • Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
            – Lucas Tonon
            Nov 29 '18 at 15:06






          • 1




            @LucasTonon Yes, absolutely. There's not a problem with that
            – caverac
            Nov 29 '18 at 15:06










          • it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
            – Lucas Tonon
            Nov 29 '18 at 15:11














          0












          0








          0






          A quick comment first,



          $$
          frac{partial |y|^3}{partial y} not= 2y
          $$



          as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.






          share|cite|improve this answer












          A quick comment first,



          $$
          frac{partial |y|^3}{partial y} not= 2y
          $$



          as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 15:02









          caveraccaverac

          14k21130




          14k21130












          • actualy it is 6y, sorry my bad, but anyways the problem stands still
            – Lucas Tonon
            Nov 29 '18 at 15:05










          • Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
            – Lucas Tonon
            Nov 29 '18 at 15:06






          • 1




            @LucasTonon Yes, absolutely. There's not a problem with that
            – caverac
            Nov 29 '18 at 15:06










          • it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
            – Lucas Tonon
            Nov 29 '18 at 15:11


















          • actualy it is 6y, sorry my bad, but anyways the problem stands still
            – Lucas Tonon
            Nov 29 '18 at 15:05










          • Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
            – Lucas Tonon
            Nov 29 '18 at 15:06






          • 1




            @LucasTonon Yes, absolutely. There's not a problem with that
            – caverac
            Nov 29 '18 at 15:06










          • it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
            – Lucas Tonon
            Nov 29 '18 at 15:11
















          actualy it is 6y, sorry my bad, but anyways the problem stands still
          – Lucas Tonon
          Nov 29 '18 at 15:05




          actualy it is 6y, sorry my bad, but anyways the problem stands still
          – Lucas Tonon
          Nov 29 '18 at 15:05












          Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
          – Lucas Tonon
          Nov 29 '18 at 15:06




          Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
          – Lucas Tonon
          Nov 29 '18 at 15:06




          1




          1




          @LucasTonon Yes, absolutely. There's not a problem with that
          – caverac
          Nov 29 '18 at 15:06




          @LucasTonon Yes, absolutely. There's not a problem with that
          – caverac
          Nov 29 '18 at 15:06












          it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
          – Lucas Tonon
          Nov 29 '18 at 15:11




          it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
          – Lucas Tonon
          Nov 29 '18 at 15:11


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3018724%2fhow-do-i-find-eigenvalues-of-an-hessian-matrix-that-depends-on-x-y-z%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

          Ellipse (mathématiques)

          Quarter-circle Tiles

          Mont Emei