Complement of a subspace












0














So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.






vector-spaces







share|cite|improve this question



























    0














    So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.






    vector-spaces







    share|cite|improve this question

























      0












      0








      0







      So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.






      vector-spaces







      share|cite|improve this question













      So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.






      vector-spaces




      vector-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 at 14:31









      Rakoon Berry

      12




      12






















          2 Answers
          2






          active

          oldest

          votes


















          2














          This is not an answer but it provides the steps which will lead to a solution



          Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.



          Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.



          Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.






          share|cite|improve this answer





















          • Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
            – Rakoon Berry
            Dec 9 at 15:39












          • @RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
            – Yanko
            Dec 9 at 17:03












          • Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
            – Rakoon Berry
            Dec 9 at 17:22












          • So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
            – Rakoon Berry
            Dec 9 at 18:06










          • About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
            – Yanko
            Dec 10 at 13:58



















          0














          There are lots of ways to do this. One way is the following:



          First you should find a basis for $U$ so you know what dimension it is.



          After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.



          This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.



          Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.






          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%2f3032428%2fcomplement-of-a-subspace%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









            2














            This is not an answer but it provides the steps which will lead to a solution



            Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.



            Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.



            Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.






            share|cite|improve this answer





















            • Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
              – Rakoon Berry
              Dec 9 at 15:39












            • @RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
              – Yanko
              Dec 9 at 17:03












            • Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
              – Rakoon Berry
              Dec 9 at 17:22












            • So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
              – Rakoon Berry
              Dec 9 at 18:06










            • About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
              – Yanko
              Dec 10 at 13:58
















            2














            This is not an answer but it provides the steps which will lead to a solution



            Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.



            Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.



            Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.






            share|cite|improve this answer





















            • Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
              – Rakoon Berry
              Dec 9 at 15:39












            • @RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
              – Yanko
              Dec 9 at 17:03












            • Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
              – Rakoon Berry
              Dec 9 at 17:22












            • So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
              – Rakoon Berry
              Dec 9 at 18:06










            • About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
              – Yanko
              Dec 10 at 13:58














            2












            2








            2






            This is not an answer but it provides the steps which will lead to a solution



            Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.



            Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.



            Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.






            share|cite|improve this answer












            This is not an answer but it provides the steps which will lead to a solution



            Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.



            Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.



            Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 at 14:38









            Yanko

            6,104724




            6,104724












            • Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
              – Rakoon Berry
              Dec 9 at 15:39












            • @RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
              – Yanko
              Dec 9 at 17:03












            • Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
              – Rakoon Berry
              Dec 9 at 17:22












            • So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
              – Rakoon Berry
              Dec 9 at 18:06










            • About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
              – Yanko
              Dec 10 at 13:58


















            • Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
              – Rakoon Berry
              Dec 9 at 15:39












            • @RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
              – Yanko
              Dec 9 at 17:03












            • Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
              – Rakoon Berry
              Dec 9 at 17:22












            • So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
              – Rakoon Berry
              Dec 9 at 18:06










            • About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
              – Yanko
              Dec 10 at 13:58
















            Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
            – Rakoon Berry
            Dec 9 at 15:39






            Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
            – Rakoon Berry
            Dec 9 at 15:39














            @RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
            – Yanko
            Dec 9 at 17:03






            @RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
            – Yanko
            Dec 9 at 17:03














            Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
            – Rakoon Berry
            Dec 9 at 17:22






            Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
            – Rakoon Berry
            Dec 9 at 17:22














            So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
            – Rakoon Berry
            Dec 9 at 18:06




            So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
            – Rakoon Berry
            Dec 9 at 18:06












            About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
            – Yanko
            Dec 10 at 13:58




            About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
            – Yanko
            Dec 10 at 13:58











            0














            There are lots of ways to do this. One way is the following:



            First you should find a basis for $U$ so you know what dimension it is.



            After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.



            This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.



            Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.






            share|cite|improve this answer


























              0














              There are lots of ways to do this. One way is the following:



              First you should find a basis for $U$ so you know what dimension it is.



              After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.



              This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.



              Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.






              share|cite|improve this answer
























                0












                0








                0






                There are lots of ways to do this. One way is the following:



                First you should find a basis for $U$ so you know what dimension it is.



                After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.



                This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.



                Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.






                share|cite|improve this answer












                There are lots of ways to do this. One way is the following:



                First you should find a basis for $U$ so you know what dimension it is.



                After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.



                This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.



                Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 at 14:44









                Ben

                2,488616




                2,488616






























                    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%2f3032428%2fcomplement-of-a-subspace%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