Visualizing SPD cone for $3times3$ matrices












4












$begingroup$


Can anyone see a good way to visualize the SPD cone for 3x3 symmetric matrices?
I'm interested in something that would highlight it's special structure, like non-smoothness.



Here's one attempt, looks pretty smooth to me








convex-optimization visualization







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Can anyone see a good way to visualize the SPD cone for 3x3 symmetric matrices?
    I'm interested in something that would highlight it's special structure, like non-smoothness.



    Here's one attempt, looks pretty smooth to me








    convex-optimization visualization







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Can anyone see a good way to visualize the SPD cone for 3x3 symmetric matrices?
      I'm interested in something that would highlight it's special structure, like non-smoothness.



      Here's one attempt, looks pretty smooth to me








      convex-optimization visualization







      share|cite|improve this question











      $endgroup$




      Can anyone see a good way to visualize the SPD cone for 3x3 symmetric matrices?
      I'm interested in something that would highlight it's special structure, like non-smoothness.



      Here's one attempt, looks pretty smooth to me








      convex-optimization visualization




      convex-optimization visualization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 8:23









      Glorfindel

      3,41981830




      3,41981830










      asked Jan 23 '12 at 23:11









      Yaroslav BulatovYaroslav Bulatov

      1,87411526




      1,87411526






















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          $begingroup$

          I think the "symmetric" projections to three dimensions do a pretty good job. If you plot against the diagonal entries, $begin{bmatrix}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & zend{bmatrix}$, you get three nonsmooth edges along the coordinate axes. With the off-diagonal entries, $begin{bmatrix}1 & x & y \ x & 1 & z \ y & z & 1end{bmatrix}$, you get four nonsmooth vertices in a tetrahedral shape at coordinates $(pm1,pm1,pm1)$.



          enter image description here $quad$ enter image description here



          Any point at which the boundary is nonsmooth corresponds to a matrix with more than one vanishing eigenvalue. This is easy to check for the nonsmooth points in both these particular cases.



          (P.S. For the benefit of future searchers, I think the phrase "cone of positive semidefinite matrices" should appear somewhere on this page :) ...)






          share|cite|improve this answer









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            $begingroup$

            I think the "symmetric" projections to three dimensions do a pretty good job. If you plot against the diagonal entries, $begin{bmatrix}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & zend{bmatrix}$, you get three nonsmooth edges along the coordinate axes. With the off-diagonal entries, $begin{bmatrix}1 & x & y \ x & 1 & z \ y & z & 1end{bmatrix}$, you get four nonsmooth vertices in a tetrahedral shape at coordinates $(pm1,pm1,pm1)$.



            enter image description here $quad$ enter image description here



            Any point at which the boundary is nonsmooth corresponds to a matrix with more than one vanishing eigenvalue. This is easy to check for the nonsmooth points in both these particular cases.



            (P.S. For the benefit of future searchers, I think the phrase "cone of positive semidefinite matrices" should appear somewhere on this page :) ...)






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              I think the "symmetric" projections to three dimensions do a pretty good job. If you plot against the diagonal entries, $begin{bmatrix}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & zend{bmatrix}$, you get three nonsmooth edges along the coordinate axes. With the off-diagonal entries, $begin{bmatrix}1 & x & y \ x & 1 & z \ y & z & 1end{bmatrix}$, you get four nonsmooth vertices in a tetrahedral shape at coordinates $(pm1,pm1,pm1)$.



              enter image description here $quad$ enter image description here



              Any point at which the boundary is nonsmooth corresponds to a matrix with more than one vanishing eigenvalue. This is easy to check for the nonsmooth points in both these particular cases.



              (P.S. For the benefit of future searchers, I think the phrase "cone of positive semidefinite matrices" should appear somewhere on this page :) ...)






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                I think the "symmetric" projections to three dimensions do a pretty good job. If you plot against the diagonal entries, $begin{bmatrix}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & zend{bmatrix}$, you get three nonsmooth edges along the coordinate axes. With the off-diagonal entries, $begin{bmatrix}1 & x & y \ x & 1 & z \ y & z & 1end{bmatrix}$, you get four nonsmooth vertices in a tetrahedral shape at coordinates $(pm1,pm1,pm1)$.



                enter image description here $quad$ enter image description here



                Any point at which the boundary is nonsmooth corresponds to a matrix with more than one vanishing eigenvalue. This is easy to check for the nonsmooth points in both these particular cases.



                (P.S. For the benefit of future searchers, I think the phrase "cone of positive semidefinite matrices" should appear somewhere on this page :) ...)






                share|cite|improve this answer









                $endgroup$



                I think the "symmetric" projections to three dimensions do a pretty good job. If you plot against the diagonal entries, $begin{bmatrix}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & zend{bmatrix}$, you get three nonsmooth edges along the coordinate axes. With the off-diagonal entries, $begin{bmatrix}1 & x & y \ x & 1 & z \ y & z & 1end{bmatrix}$, you get four nonsmooth vertices in a tetrahedral shape at coordinates $(pm1,pm1,pm1)$.



                enter image description here $quad$ enter image description here



                Any point at which the boundary is nonsmooth corresponds to a matrix with more than one vanishing eigenvalue. This is easy to check for the nonsmooth points in both these particular cases.



                (P.S. For the benefit of future searchers, I think the phrase "cone of positive semidefinite matrices" should appear somewhere on this page :) ...)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '12 at 7:35









                RahulRahul

                33.2k568173




                33.2k568173






























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