trace of symmetric matrix problems

Multi tool use
Multi tool use












3












$begingroup$


I have the two problems below from a practice exam. I can prove them on their own but am not exactly sure if/how to show that they only hold for symmetric matrices and for '3)' showing that it only holds for a matrix with only positive eigenvalues. I know that if the eigenvalues are all positive the determinant will be positive and the trace but cant see how that affects whether '3)' is true or not.




  1. Show that $operatorname{Tr}(A^2) leq operatorname{Tr}(A)^2$ holds for any symmetric matrix $A$ whose eigenvalues are all non-negative.

  2. Show that $operatorname{Tr}(AB)^2 le operatorname{Tr}(A^2)operatorname{Tr}(B^2)$ holds for any symmetric matrices $A$ and $B$.






linear-algebra matrices trace







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $lambda_1, ldots, lambda_n$ are the eigenvalues of $A$, what are the eigenvalues of $A^2$? And how the trace of a matrix is related to the eigenvalues?
    $endgroup$
    – thanasissdr
    Oct 28 '15 at 9:08










  • $begingroup$
    for both questions i just worked through an example which proved the questions right. I understand the eigenvalues for $A^2$ are the same as for $A$ but squared and when you add the trace you should get the same value as the eigenvalues added. I just wasn't sure in the second question as it says about all non-negative eigenvalues if just working through an example that proves it right really proves it for all cases and therefore had to go about it a different way
    $endgroup$
    – dmnte
    Oct 28 '15 at 12:25










  • $begingroup$
    for the fourth question I cant really prove it as much as I just show that it is true through a worked example, im guessing this is wrong but im not sure how else to do it
    $endgroup$
    – dmnte
    Oct 28 '15 at 14:49
















3












$begingroup$


I have the two problems below from a practice exam. I can prove them on their own but am not exactly sure if/how to show that they only hold for symmetric matrices and for '3)' showing that it only holds for a matrix with only positive eigenvalues. I know that if the eigenvalues are all positive the determinant will be positive and the trace but cant see how that affects whether '3)' is true or not.




  1. Show that $operatorname{Tr}(A^2) leq operatorname{Tr}(A)^2$ holds for any symmetric matrix $A$ whose eigenvalues are all non-negative.

  2. Show that $operatorname{Tr}(AB)^2 le operatorname{Tr}(A^2)operatorname{Tr}(B^2)$ holds for any symmetric matrices $A$ and $B$.






linear-algebra matrices trace







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $lambda_1, ldots, lambda_n$ are the eigenvalues of $A$, what are the eigenvalues of $A^2$? And how the trace of a matrix is related to the eigenvalues?
    $endgroup$
    – thanasissdr
    Oct 28 '15 at 9:08










  • $begingroup$
    for both questions i just worked through an example which proved the questions right. I understand the eigenvalues for $A^2$ are the same as for $A$ but squared and when you add the trace you should get the same value as the eigenvalues added. I just wasn't sure in the second question as it says about all non-negative eigenvalues if just working through an example that proves it right really proves it for all cases and therefore had to go about it a different way
    $endgroup$
    – dmnte
    Oct 28 '15 at 12:25










  • $begingroup$
    for the fourth question I cant really prove it as much as I just show that it is true through a worked example, im guessing this is wrong but im not sure how else to do it
    $endgroup$
    – dmnte
    Oct 28 '15 at 14:49














3












3








3


1



$begingroup$


I have the two problems below from a practice exam. I can prove them on their own but am not exactly sure if/how to show that they only hold for symmetric matrices and for '3)' showing that it only holds for a matrix with only positive eigenvalues. I know that if the eigenvalues are all positive the determinant will be positive and the trace but cant see how that affects whether '3)' is true or not.




  1. Show that $operatorname{Tr}(A^2) leq operatorname{Tr}(A)^2$ holds for any symmetric matrix $A$ whose eigenvalues are all non-negative.

  2. Show that $operatorname{Tr}(AB)^2 le operatorname{Tr}(A^2)operatorname{Tr}(B^2)$ holds for any symmetric matrices $A$ and $B$.






linear-algebra matrices trace







share|cite|improve this question











$endgroup$




I have the two problems below from a practice exam. I can prove them on their own but am not exactly sure if/how to show that they only hold for symmetric matrices and for '3)' showing that it only holds for a matrix with only positive eigenvalues. I know that if the eigenvalues are all positive the determinant will be positive and the trace but cant see how that affects whether '3)' is true or not.




  1. Show that $operatorname{Tr}(A^2) leq operatorname{Tr}(A)^2$ holds for any symmetric matrix $A$ whose eigenvalues are all non-negative.

  2. Show that $operatorname{Tr}(AB)^2 le operatorname{Tr}(A^2)operatorname{Tr}(B^2)$ holds for any symmetric matrices $A$ and $B$.






linear-algebra matrices trace




linear-algebra matrices trace






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 29 '15 at 7:14









thanasissdr

5,55111325




5,55111325










asked Oct 28 '15 at 9:03









dmntedmnte

1129




1129








  • 1




    $begingroup$
    If $lambda_1, ldots, lambda_n$ are the eigenvalues of $A$, what are the eigenvalues of $A^2$? And how the trace of a matrix is related to the eigenvalues?
    $endgroup$
    – thanasissdr
    Oct 28 '15 at 9:08










  • $begingroup$
    for both questions i just worked through an example which proved the questions right. I understand the eigenvalues for $A^2$ are the same as for $A$ but squared and when you add the trace you should get the same value as the eigenvalues added. I just wasn't sure in the second question as it says about all non-negative eigenvalues if just working through an example that proves it right really proves it for all cases and therefore had to go about it a different way
    $endgroup$
    – dmnte
    Oct 28 '15 at 12:25










  • $begingroup$
    for the fourth question I cant really prove it as much as I just show that it is true through a worked example, im guessing this is wrong but im not sure how else to do it
    $endgroup$
    – dmnte
    Oct 28 '15 at 14:49














  • 1




    $begingroup$
    If $lambda_1, ldots, lambda_n$ are the eigenvalues of $A$, what are the eigenvalues of $A^2$? And how the trace of a matrix is related to the eigenvalues?
    $endgroup$
    – thanasissdr
    Oct 28 '15 at 9:08










  • $begingroup$
    for both questions i just worked through an example which proved the questions right. I understand the eigenvalues for $A^2$ are the same as for $A$ but squared and when you add the trace you should get the same value as the eigenvalues added. I just wasn't sure in the second question as it says about all non-negative eigenvalues if just working through an example that proves it right really proves it for all cases and therefore had to go about it a different way
    $endgroup$
    – dmnte
    Oct 28 '15 at 12:25










  • $begingroup$
    for the fourth question I cant really prove it as much as I just show that it is true through a worked example, im guessing this is wrong but im not sure how else to do it
    $endgroup$
    – dmnte
    Oct 28 '15 at 14:49








1




1




$begingroup$
If $lambda_1, ldots, lambda_n$ are the eigenvalues of $A$, what are the eigenvalues of $A^2$? And how the trace of a matrix is related to the eigenvalues?
$endgroup$
– thanasissdr
Oct 28 '15 at 9:08




$begingroup$
If $lambda_1, ldots, lambda_n$ are the eigenvalues of $A$, what are the eigenvalues of $A^2$? And how the trace of a matrix is related to the eigenvalues?
$endgroup$
– thanasissdr
Oct 28 '15 at 9:08












$begingroup$
for both questions i just worked through an example which proved the questions right. I understand the eigenvalues for $A^2$ are the same as for $A$ but squared and when you add the trace you should get the same value as the eigenvalues added. I just wasn't sure in the second question as it says about all non-negative eigenvalues if just working through an example that proves it right really proves it for all cases and therefore had to go about it a different way
$endgroup$
– dmnte
Oct 28 '15 at 12:25




$begingroup$
for both questions i just worked through an example which proved the questions right. I understand the eigenvalues for $A^2$ are the same as for $A$ but squared and when you add the trace you should get the same value as the eigenvalues added. I just wasn't sure in the second question as it says about all non-negative eigenvalues if just working through an example that proves it right really proves it for all cases and therefore had to go about it a different way
$endgroup$
– dmnte
Oct 28 '15 at 12:25












$begingroup$
for the fourth question I cant really prove it as much as I just show that it is true through a worked example, im guessing this is wrong but im not sure how else to do it
$endgroup$
– dmnte
Oct 28 '15 at 14:49




$begingroup$
for the fourth question I cant really prove it as much as I just show that it is true through a worked example, im guessing this is wrong but im not sure how else to do it
$endgroup$
– dmnte
Oct 28 '15 at 14:49










2 Answers
2






active

oldest

votes


















1












$begingroup$

About question $#4$:



Notation: Let $C= AB$ and $K^2_{ii}$ denote the element at the $(i,i)$ position of the matrix $K^2$.



Firstly, due to symmetry of the $ntimes n$ matrices $A,B$, it is easy to prove that:
$$c_{ii} le left(A^2_{ii}right)^{1/2} cdot left(B^2_{ii}right)^{1/2},quad i = 1,ldots, n.tag 1$$




Proof of $(1)$



$c_{ii} =^color{red}{starstar} a_{i1} cdot b_{i1} + cdots +a_{in}cdot b_{in}= sumlimits_{j=1}^{n}a_{ij}cdot b_{ij}color{blue}{le^star} left(sumlimits_{j=1}^{n}a^2_{ij}right)^{1/2}cdot left(sumlimits_{j=1}^nb_{ij}^2right)^{1/2}=left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}quad text{QED}$



Thus, we have:
$$big[operatorname{trace} (AB)big]^2=left(sum_{i=1}^n c_{ii}right)^2leleft[ sum_{i=1}^n left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}right]^2color{blue}{le^star}sum_{i=1}^n A^2_{ii} cdot sum_{i=1}^n B^2_{ii}=operatorname{trace} (A^2) cdot operatorname{trace} (B^2)$$




$^color{blue}{star}$ We have applied the Cauchy-Schwarz inequality.



$^color{red}{starstar}$ Normally, it is:
$c_{ii} = a_{i1}b_{1i} + a_{i2}b_{2i} +cdots + a_{in}b_{ni},$
but notice that $b_{kell} = b_{ell k}$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Question No.3 is more related to the fact that for given numbers $x_1,dots,x_n$, the following inequality $$x_1^2+dots+x_n^2leq ,(x_1+dots+x_n)^2$$ holds only if $x_i$ are non-negative. Let $A$ be any diagonalizable matrix so that $A=TLambda T^{-1}$ and $A^2=TLambda^2 T^{-1}$. Thus, if $x_1,dots,x_n$ are the eigenvalues, then $mathrm{trace}(A^2)=x_1^2+dots+x_n^2 $ and $mathrm{trace}(A)^2=(x_1+dots+x_n)^2 $. Note that symmetric matrices are readily diagonalizable since they are normal.



    Question No.4 is more related to the fact that trace is an inner product in the space of symmetric matrices. In fact, that inequality you have given is Cauchy-Schwartz indeed.






    share|cite|improve this answer











    $endgroup$













      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%2f1501499%2ftrace-of-symmetric-matrix-problems%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      About question $#4$:



      Notation: Let $C= AB$ and $K^2_{ii}$ denote the element at the $(i,i)$ position of the matrix $K^2$.



      Firstly, due to symmetry of the $ntimes n$ matrices $A,B$, it is easy to prove that:
      $$c_{ii} le left(A^2_{ii}right)^{1/2} cdot left(B^2_{ii}right)^{1/2},quad i = 1,ldots, n.tag 1$$




      Proof of $(1)$



      $c_{ii} =^color{red}{starstar} a_{i1} cdot b_{i1} + cdots +a_{in}cdot b_{in}= sumlimits_{j=1}^{n}a_{ij}cdot b_{ij}color{blue}{le^star} left(sumlimits_{j=1}^{n}a^2_{ij}right)^{1/2}cdot left(sumlimits_{j=1}^nb_{ij}^2right)^{1/2}=left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}quad text{QED}$



      Thus, we have:
      $$big[operatorname{trace} (AB)big]^2=left(sum_{i=1}^n c_{ii}right)^2leleft[ sum_{i=1}^n left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}right]^2color{blue}{le^star}sum_{i=1}^n A^2_{ii} cdot sum_{i=1}^n B^2_{ii}=operatorname{trace} (A^2) cdot operatorname{trace} (B^2)$$




      $^color{blue}{star}$ We have applied the Cauchy-Schwarz inequality.



      $^color{red}{starstar}$ Normally, it is:
      $c_{ii} = a_{i1}b_{1i} + a_{i2}b_{2i} +cdots + a_{in}b_{ni},$
      but notice that $b_{kell} = b_{ell k}$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        About question $#4$:



        Notation: Let $C= AB$ and $K^2_{ii}$ denote the element at the $(i,i)$ position of the matrix $K^2$.



        Firstly, due to symmetry of the $ntimes n$ matrices $A,B$, it is easy to prove that:
        $$c_{ii} le left(A^2_{ii}right)^{1/2} cdot left(B^2_{ii}right)^{1/2},quad i = 1,ldots, n.tag 1$$




        Proof of $(1)$



        $c_{ii} =^color{red}{starstar} a_{i1} cdot b_{i1} + cdots +a_{in}cdot b_{in}= sumlimits_{j=1}^{n}a_{ij}cdot b_{ij}color{blue}{le^star} left(sumlimits_{j=1}^{n}a^2_{ij}right)^{1/2}cdot left(sumlimits_{j=1}^nb_{ij}^2right)^{1/2}=left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}quad text{QED}$



        Thus, we have:
        $$big[operatorname{trace} (AB)big]^2=left(sum_{i=1}^n c_{ii}right)^2leleft[ sum_{i=1}^n left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}right]^2color{blue}{le^star}sum_{i=1}^n A^2_{ii} cdot sum_{i=1}^n B^2_{ii}=operatorname{trace} (A^2) cdot operatorname{trace} (B^2)$$




        $^color{blue}{star}$ We have applied the Cauchy-Schwarz inequality.



        $^color{red}{starstar}$ Normally, it is:
        $c_{ii} = a_{i1}b_{1i} + a_{i2}b_{2i} +cdots + a_{in}b_{ni},$
        but notice that $b_{kell} = b_{ell k}$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          About question $#4$:



          Notation: Let $C= AB$ and $K^2_{ii}$ denote the element at the $(i,i)$ position of the matrix $K^2$.



          Firstly, due to symmetry of the $ntimes n$ matrices $A,B$, it is easy to prove that:
          $$c_{ii} le left(A^2_{ii}right)^{1/2} cdot left(B^2_{ii}right)^{1/2},quad i = 1,ldots, n.tag 1$$




          Proof of $(1)$



          $c_{ii} =^color{red}{starstar} a_{i1} cdot b_{i1} + cdots +a_{in}cdot b_{in}= sumlimits_{j=1}^{n}a_{ij}cdot b_{ij}color{blue}{le^star} left(sumlimits_{j=1}^{n}a^2_{ij}right)^{1/2}cdot left(sumlimits_{j=1}^nb_{ij}^2right)^{1/2}=left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}quad text{QED}$



          Thus, we have:
          $$big[operatorname{trace} (AB)big]^2=left(sum_{i=1}^n c_{ii}right)^2leleft[ sum_{i=1}^n left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}right]^2color{blue}{le^star}sum_{i=1}^n A^2_{ii} cdot sum_{i=1}^n B^2_{ii}=operatorname{trace} (A^2) cdot operatorname{trace} (B^2)$$




          $^color{blue}{star}$ We have applied the Cauchy-Schwarz inequality.



          $^color{red}{starstar}$ Normally, it is:
          $c_{ii} = a_{i1}b_{1i} + a_{i2}b_{2i} +cdots + a_{in}b_{ni},$
          but notice that $b_{kell} = b_{ell k}$.






          share|cite|improve this answer











          $endgroup$



          About question $#4$:



          Notation: Let $C= AB$ and $K^2_{ii}$ denote the element at the $(i,i)$ position of the matrix $K^2$.



          Firstly, due to symmetry of the $ntimes n$ matrices $A,B$, it is easy to prove that:
          $$c_{ii} le left(A^2_{ii}right)^{1/2} cdot left(B^2_{ii}right)^{1/2},quad i = 1,ldots, n.tag 1$$




          Proof of $(1)$



          $c_{ii} =^color{red}{starstar} a_{i1} cdot b_{i1} + cdots +a_{in}cdot b_{in}= sumlimits_{j=1}^{n}a_{ij}cdot b_{ij}color{blue}{le^star} left(sumlimits_{j=1}^{n}a^2_{ij}right)^{1/2}cdot left(sumlimits_{j=1}^nb_{ij}^2right)^{1/2}=left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}quad text{QED}$



          Thus, we have:
          $$big[operatorname{trace} (AB)big]^2=left(sum_{i=1}^n c_{ii}right)^2leleft[ sum_{i=1}^n left(A^2_{ii}right)^{1/2}cdot left(B^2_{ii}right)^{1/2}right]^2color{blue}{le^star}sum_{i=1}^n A^2_{ii} cdot sum_{i=1}^n B^2_{ii}=operatorname{trace} (A^2) cdot operatorname{trace} (B^2)$$




          $^color{blue}{star}$ We have applied the Cauchy-Schwarz inequality.



          $^color{red}{starstar}$ Normally, it is:
          $c_{ii} = a_{i1}b_{1i} + a_{i2}b_{2i} +cdots + a_{in}b_{ni},$
          but notice that $b_{kell} = b_{ell k}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 28 '15 at 21:59

























          answered Oct 28 '15 at 19:16









          thanasissdrthanasissdr

          5,55111325




          5,55111325























              1












              $begingroup$

              Question No.3 is more related to the fact that for given numbers $x_1,dots,x_n$, the following inequality $$x_1^2+dots+x_n^2leq ,(x_1+dots+x_n)^2$$ holds only if $x_i$ are non-negative. Let $A$ be any diagonalizable matrix so that $A=TLambda T^{-1}$ and $A^2=TLambda^2 T^{-1}$. Thus, if $x_1,dots,x_n$ are the eigenvalues, then $mathrm{trace}(A^2)=x_1^2+dots+x_n^2 $ and $mathrm{trace}(A)^2=(x_1+dots+x_n)^2 $. Note that symmetric matrices are readily diagonalizable since they are normal.



              Question No.4 is more related to the fact that trace is an inner product in the space of symmetric matrices. In fact, that inequality you have given is Cauchy-Schwartz indeed.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Question No.3 is more related to the fact that for given numbers $x_1,dots,x_n$, the following inequality $$x_1^2+dots+x_n^2leq ,(x_1+dots+x_n)^2$$ holds only if $x_i$ are non-negative. Let $A$ be any diagonalizable matrix so that $A=TLambda T^{-1}$ and $A^2=TLambda^2 T^{-1}$. Thus, if $x_1,dots,x_n$ are the eigenvalues, then $mathrm{trace}(A^2)=x_1^2+dots+x_n^2 $ and $mathrm{trace}(A)^2=(x_1+dots+x_n)^2 $. Note that symmetric matrices are readily diagonalizable since they are normal.



                Question No.4 is more related to the fact that trace is an inner product in the space of symmetric matrices. In fact, that inequality you have given is Cauchy-Schwartz indeed.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Question No.3 is more related to the fact that for given numbers $x_1,dots,x_n$, the following inequality $$x_1^2+dots+x_n^2leq ,(x_1+dots+x_n)^2$$ holds only if $x_i$ are non-negative. Let $A$ be any diagonalizable matrix so that $A=TLambda T^{-1}$ and $A^2=TLambda^2 T^{-1}$. Thus, if $x_1,dots,x_n$ are the eigenvalues, then $mathrm{trace}(A^2)=x_1^2+dots+x_n^2 $ and $mathrm{trace}(A)^2=(x_1+dots+x_n)^2 $. Note that symmetric matrices are readily diagonalizable since they are normal.



                  Question No.4 is more related to the fact that trace is an inner product in the space of symmetric matrices. In fact, that inequality you have given is Cauchy-Schwartz indeed.






                  share|cite|improve this answer











                  $endgroup$



                  Question No.3 is more related to the fact that for given numbers $x_1,dots,x_n$, the following inequality $$x_1^2+dots+x_n^2leq ,(x_1+dots+x_n)^2$$ holds only if $x_i$ are non-negative. Let $A$ be any diagonalizable matrix so that $A=TLambda T^{-1}$ and $A^2=TLambda^2 T^{-1}$. Thus, if $x_1,dots,x_n$ are the eigenvalues, then $mathrm{trace}(A^2)=x_1^2+dots+x_n^2 $ and $mathrm{trace}(A)^2=(x_1+dots+x_n)^2 $. Note that symmetric matrices are readily diagonalizable since they are normal.



                  Question No.4 is more related to the fact that trace is an inner product in the space of symmetric matrices. In fact, that inequality you have given is Cauchy-Schwartz indeed.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Oct 29 '15 at 9:18

























                  answered Oct 29 '15 at 7:33









                  dineshdileepdineshdileep

                  5,96611735




                  5,96611735






























                      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%2f1501499%2ftrace-of-symmetric-matrix-problems%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







                      aC8yFAHZkhzUypJUq,CZACzvBAP2NT1UoUEClrpAOU38PAx V7dq6ZzMA50AYpSYr
                      -
                      KzVBrR7vPCu0bfyAFjvDonh6tBHURICm 81 2ZIfvOy

                      Popular posts from this blog

                      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

                      menu

                      Image
                      menu    Wanted List (rus) Main Projects    FAMI Dumping project    Pirate cart's ID numbers list    My carts collection    Unpublished NES cheats    >All titles     Numbered titles     A titles     B titles     C titles     D titles     E titles     F titles     G titles     H titles     I titles     J titles     K titles     L titles     M titles     N titles     O titles     P titles     Q titles     R titles     S titles     T titles     U titles     V titles     W titles     X titles     Y titles     Z titles     Unique cheats     Buried secrets     Fixed generic cheats     Generic cheats     Secret passwords     Password generators     List of games with no cheats found    Unpublished Game Boy cheats    Unpublished Game Boy Color cheats Misc Stuff    Pirate Market    BONZA    FORTUNA    Decrypted SmyNES ROMs    NES inside (rus)    FCEUMM project Blog    cah4e3.wordpress.com Mail me:    cah4e3 at gmail dot c...
                      Read more

                      Dobbiaco

                      Image
                      Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
                      Read more