finding the dimension of eigenspace with characteristic and minimal polynomial












0














I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
How can I calculate it?



For example, suppose you're given a $ 6 times 6 $
matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
and that the i-eigenspace is 1-dimensional.



Thanks for your help






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question





























    0














    I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
    How can I calculate it?



    For example, suppose you're given a $ 6 times 6 $
    matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
    its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
    and that the i-eigenspace is 1-dimensional.



    Thanks for your help






    linear-algebra matrices eigenvalues-eigenvectors







    share|cite|improve this question



























      0












      0








      0







      I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
      How can I calculate it?



      For example, suppose you're given a $ 6 times 6 $
      matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
      its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
      and that the i-eigenspace is 1-dimensional.



      Thanks for your help






      linear-algebra matrices eigenvalues-eigenvectors







      share|cite|improve this question















      I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
      How can I calculate it?



      For example, suppose you're given a $ 6 times 6 $
      matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
      its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
      and that the i-eigenspace is 1-dimensional.



      Thanks for your help






      linear-algebra matrices eigenvalues-eigenvectors




      linear-algebra matrices eigenvalues-eigenvectors






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 at 9:27









      José Carlos Santos

      148k22117218




      148k22117218










      asked Jun 13 '17 at 9:46









      fateme jl

      8710




      8710






















          3 Answers
          3






          active

          oldest

          votes


















          1














          Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



          $$JB_{lambda=3}=begin{pmatrix}
          color{red}3&color{red}1&0&0\
          color{red}0&color{red}3&0&0\
          0&0&color{green}3&0\
          0&0&0&color{blue}3end{pmatrix}$$



          Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






          share|cite|improve this answer





















          • Isn't two Jordan blocks of size 2 a possibility?
            – Rab
            Jun 13 '17 at 10:05










          • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
            – Rab
            Jun 13 '17 at 10:10










          • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
            – DonAntonio
            Jun 13 '17 at 11:20








          • 1




            Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
            – DonAntonio
            Jun 13 '17 at 11:22












          • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
            – Peter Melech
            Jun 16 '17 at 7:52



















          1














          That implies its Jordan normal form is:



          $begin{pmatrix}3&1&0&0&0&0\
          0&3&0&0&0&0\
          0&0&3&0&0&0\
          0&0&0&3&0&0\
          0&0&0&0&i&1\
          0&0&0&0&0&iend{pmatrix}$






          share|cite|improve this answer





















          • why we don't have two 2*2 blocks for eigenvalue 3?
            – fateme jl
            Jun 15 '17 at 12:05










          • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
            – Peter Melech
            Jun 16 '17 at 7:48



















          1














          The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






          share|cite|improve this answer





















            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%2f2320813%2ffinding-the-dimension-of-eigenspace-with-characteristic-and-minimal-polynomial%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






            share|cite|improve this answer





















            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52
















            1














            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






            share|cite|improve this answer





















            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52














            1












            1








            1






            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






            share|cite|improve this answer












            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 13 '17 at 10:02









            DonAntonio

            176k1491225




            176k1491225












            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52


















            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52
















            Isn't two Jordan blocks of size 2 a possibility?
            – Rab
            Jun 13 '17 at 10:05




            Isn't two Jordan blocks of size 2 a possibility?
            – Rab
            Jun 13 '17 at 10:05












            Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
            – Rab
            Jun 13 '17 at 10:10




            Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
            – Rab
            Jun 13 '17 at 10:10












            @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
            – DonAntonio
            Jun 13 '17 at 11:20






            @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
            – DonAntonio
            Jun 13 '17 at 11:20






            1




            1




            Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
            – DonAntonio
            Jun 13 '17 at 11:22






            Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
            – DonAntonio
            Jun 13 '17 at 11:22














            @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
            – Peter Melech
            Jun 16 '17 at 7:52




            @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
            – Peter Melech
            Jun 16 '17 at 7:52











            1














            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$






            share|cite|improve this answer





















            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48
















            1














            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$






            share|cite|improve this answer





















            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48














            1












            1








            1






            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$






            share|cite|improve this answer












            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 13 '17 at 10:06









            Peter Melech

            2,529813




            2,529813












            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48


















            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48
















            why we don't have two 2*2 blocks for eigenvalue 3?
            – fateme jl
            Jun 15 '17 at 12:05




            why we don't have two 2*2 blocks for eigenvalue 3?
            – fateme jl
            Jun 15 '17 at 12:05












            Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
            – Peter Melech
            Jun 16 '17 at 7:48




            Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
            – Peter Melech
            Jun 16 '17 at 7:48











            1














            The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






            share|cite|improve this answer


























              1














              The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






              share|cite|improve this answer
























                1












                1








                1






                The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






                share|cite|improve this answer












                The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jun 13 '17 at 10:08









                José Carlos Santos

                148k22117218




                148k22117218






























                    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%2f2320813%2ffinding-the-dimension-of-eigenspace-with-characteristic-and-minimal-polynomial%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