Self adjoint operators and trace class property












2












$begingroup$


This is a variation of the problem questioned some time ago.



For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.



Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?



The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$



The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.



For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$



Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?






functional-analysis operator-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
    $endgroup$
    – Martin Argerami
    Dec 22 '18 at 16:31










  • $begingroup$
    @MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
    $endgroup$
    – hyperkahler
    Dec 22 '18 at 16:35












  • $begingroup$
    It's not immediately obvious to me that the answer to my question answers yours.
    $endgroup$
    – Martin Argerami
    Dec 25 '18 at 2:53










  • $begingroup$
    @MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
    $endgroup$
    – hyperkahler
    Jan 5 at 0:37










  • $begingroup$
    hyperkahler: that answer is very wrong.
    $endgroup$
    – Martin Argerami
    Jan 5 at 3:10


















2












$begingroup$


This is a variation of the problem questioned some time ago.



For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.



Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?



The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$



The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.



For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$



Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?






functional-analysis operator-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
    $endgroup$
    – Martin Argerami
    Dec 22 '18 at 16:31










  • $begingroup$
    @MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
    $endgroup$
    – hyperkahler
    Dec 22 '18 at 16:35












  • $begingroup$
    It's not immediately obvious to me that the answer to my question answers yours.
    $endgroup$
    – Martin Argerami
    Dec 25 '18 at 2:53










  • $begingroup$
    @MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
    $endgroup$
    – hyperkahler
    Jan 5 at 0:37










  • $begingroup$
    hyperkahler: that answer is very wrong.
    $endgroup$
    – Martin Argerami
    Jan 5 at 3:10
















2












2








2


1



$begingroup$


This is a variation of the problem questioned some time ago.



For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.



Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?



The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$



The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.



For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$



Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?






functional-analysis operator-theory







share|cite|improve this question











$endgroup$




This is a variation of the problem questioned some time ago.



For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.



Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?



The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$



The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.



For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$



Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?






functional-analysis operator-theory




functional-analysis operator-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 16:33







hyperkahler

















asked Dec 22 '18 at 15:57









hyperkahlerhyperkahler

1,471714




1,471714








  • 1




    $begingroup$
    As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
    $endgroup$
    – Martin Argerami
    Dec 22 '18 at 16:31










  • $begingroup$
    @MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
    $endgroup$
    – hyperkahler
    Dec 22 '18 at 16:35












  • $begingroup$
    It's not immediately obvious to me that the answer to my question answers yours.
    $endgroup$
    – Martin Argerami
    Dec 25 '18 at 2:53










  • $begingroup$
    @MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
    $endgroup$
    – hyperkahler
    Jan 5 at 0:37










  • $begingroup$
    hyperkahler: that answer is very wrong.
    $endgroup$
    – Martin Argerami
    Jan 5 at 3:10
















  • 1




    $begingroup$
    As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
    $endgroup$
    – Martin Argerami
    Dec 22 '18 at 16:31










  • $begingroup$
    @MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
    $endgroup$
    – hyperkahler
    Dec 22 '18 at 16:35












  • $begingroup$
    It's not immediately obvious to me that the answer to my question answers yours.
    $endgroup$
    – Martin Argerami
    Dec 25 '18 at 2:53










  • $begingroup$
    @MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
    $endgroup$
    – hyperkahler
    Jan 5 at 0:37










  • $begingroup$
    hyperkahler: that answer is very wrong.
    $endgroup$
    – Martin Argerami
    Jan 5 at 3:10










1




1




$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31




$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31












$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35






$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35














$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53




$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53












$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37




$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37












$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10






$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10












1 Answer
1






active

oldest

votes


















1












$begingroup$

Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.



Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.



If $T$ is not trace-class, then
$$
operatorname{Tr}(T)=sum_j|lambda_j|=infty.
$$
If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
$$
sum_jlangle Te_j,e_jrangle
$$
cannot converge absolutely.






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%2f3049581%2fself-adjoint-operators-and-trace-class-property%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












    $begingroup$

    Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.



    Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.



    If $T$ is not trace-class, then
    $$
    operatorname{Tr}(T)=sum_j|lambda_j|=infty.
    $$
    If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
    $$
    sum_jlangle Te_j,e_jrangle
    $$
    cannot converge absolutely.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.



      Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.



      If $T$ is not trace-class, then
      $$
      operatorname{Tr}(T)=sum_j|lambda_j|=infty.
      $$
      If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
      $$
      sum_jlangle Te_j,e_jrangle
      $$
      cannot converge absolutely.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.



        Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.



        If $T$ is not trace-class, then
        $$
        operatorname{Tr}(T)=sum_j|lambda_j|=infty.
        $$
        If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
        $$
        sum_jlangle Te_j,e_jrangle
        $$
        cannot converge absolutely.






        share|cite|improve this answer











        $endgroup$



        Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.



        Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.



        If $T$ is not trace-class, then
        $$
        operatorname{Tr}(T)=sum_j|lambda_j|=infty.
        $$
        If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
        $$
        sum_jlangle Te_j,e_jrangle
        $$
        cannot converge absolutely.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 16:56

























        answered Dec 22 '18 at 19:20









        Martin ArgeramiMartin Argerami

        126k1182181




        126k1182181






























            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%2f3049581%2fself-adjoint-operators-and-trace-class-property%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more