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












0












$begingroup$


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











$endgroup$

















    0












    $begingroup$


    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











    $endgroup$















      0












      0








      0





      $begingroup$


      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











      $endgroup$




      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




      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

      41.2k1150102




      41.2k1150102










      asked Nov 29 '18 at 14:53









      Lucas TononLucas Tonon

      445




      445






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          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









          $endgroup$













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










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






          • 1




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










          • $begingroup$
            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
            $endgroup$
            – Lucas Tonon
            Nov 29 '18 at 15:11











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          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









          $endgroup$













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










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






          • 1




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










          • $begingroup$
            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
            $endgroup$
            – Lucas Tonon
            Nov 29 '18 at 15:11
















          0












          $begingroup$

          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









          $endgroup$













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










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






          • 1




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










          • $begingroup$
            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
            $endgroup$
            – Lucas Tonon
            Nov 29 '18 at 15:11














          0












          0








          0





          $begingroup$

          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









          $endgroup$



          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

          14.6k31130




          14.6k31130












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










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






          • 1




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










          • $begingroup$
            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
            $endgroup$
            – Lucas Tonon
            Nov 29 '18 at 15:11


















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










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






          • 1




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










          • $begingroup$
            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
            $endgroup$
            – Lucas Tonon
            Nov 29 '18 at 15:11
















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




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












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




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




          1




          1




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




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












          $begingroup$
          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
          $endgroup$
          – Lucas Tonon
          Nov 29 '18 at 15:11




          $begingroup$
          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
          $endgroup$
          – Lucas Tonon
          Nov 29 '18 at 15:11


















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