Finding the representation matrix of the linear transformation $S: mathbb{R}^3longrightarrow P_2: (a,b,c)...












-1












$begingroup$


I know that the basis of $mathbb{R}^3$ is ${(1,0,0), (0,1,0),(0,0,1)}$.



Now, let $P_2(mathbb{R}$ be the set of polynomials with degree less or equal to 2 with coefficients in $mathbb{R}$ (i.e. $P_2(mathbb{R}):={p(x)in mathbb{R}[x] | text{degree}(p(x))leq 2}$.



I also know that the basis of $P_2(mathbb{R})$ is ${(1,x)}$ and with more simple examples I have no problem with this.



Now, let $$S: mathbb{R}^3longrightarrow P_2(mathbb{R}): (a,b,c) mapsto (a+b+c)+(3b-c)x$$ How do I find the matrix of "S"?






linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Maths SX! What do you denote $P_2$?
    $endgroup$
    – Bernard
    Jan 13 at 11:27










  • $begingroup$
    The set of linear polynomials with degree less than 2, so of the form $p(x)=a+bx$?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:30
















-1












$begingroup$


I know that the basis of $mathbb{R}^3$ is ${(1,0,0), (0,1,0),(0,0,1)}$.



Now, let $P_2(mathbb{R}$ be the set of polynomials with degree less or equal to 2 with coefficients in $mathbb{R}$ (i.e. $P_2(mathbb{R}):={p(x)in mathbb{R}[x] | text{degree}(p(x))leq 2}$.



I also know that the basis of $P_2(mathbb{R})$ is ${(1,x)}$ and with more simple examples I have no problem with this.



Now, let $$S: mathbb{R}^3longrightarrow P_2(mathbb{R}): (a,b,c) mapsto (a+b+c)+(3b-c)x$$ How do I find the matrix of "S"?






linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Maths SX! What do you denote $P_2$?
    $endgroup$
    – Bernard
    Jan 13 at 11:27










  • $begingroup$
    The set of linear polynomials with degree less than 2, so of the form $p(x)=a+bx$?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:30














-1












-1








-1





$begingroup$


I know that the basis of $mathbb{R}^3$ is ${(1,0,0), (0,1,0),(0,0,1)}$.



Now, let $P_2(mathbb{R}$ be the set of polynomials with degree less or equal to 2 with coefficients in $mathbb{R}$ (i.e. $P_2(mathbb{R}):={p(x)in mathbb{R}[x] | text{degree}(p(x))leq 2}$.



I also know that the basis of $P_2(mathbb{R})$ is ${(1,x)}$ and with more simple examples I have no problem with this.



Now, let $$S: mathbb{R}^3longrightarrow P_2(mathbb{R}): (a,b,c) mapsto (a+b+c)+(3b-c)x$$ How do I find the matrix of "S"?






linear-transformations







share|cite|improve this question











$endgroup$




I know that the basis of $mathbb{R}^3$ is ${(1,0,0), (0,1,0),(0,0,1)}$.



Now, let $P_2(mathbb{R}$ be the set of polynomials with degree less or equal to 2 with coefficients in $mathbb{R}$ (i.e. $P_2(mathbb{R}):={p(x)in mathbb{R}[x] | text{degree}(p(x))leq 2}$.



I also know that the basis of $P_2(mathbb{R})$ is ${(1,x)}$ and with more simple examples I have no problem with this.



Now, let $$S: mathbb{R}^3longrightarrow P_2(mathbb{R}): (a,b,c) mapsto (a+b+c)+(3b-c)x$$ How do I find the matrix of "S"?






linear-transformations




linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 12:17









idriskameni

749321




749321










asked Jan 13 at 11:26









Andrea BrabrookAndrea Brabrook

33




33












  • $begingroup$
    Welcome to Maths SX! What do you denote $P_2$?
    $endgroup$
    – Bernard
    Jan 13 at 11:27










  • $begingroup$
    The set of linear polynomials with degree less than 2, so of the form $p(x)=a+bx$?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:30


















  • $begingroup$
    Welcome to Maths SX! What do you denote $P_2$?
    $endgroup$
    – Bernard
    Jan 13 at 11:27










  • $begingroup$
    The set of linear polynomials with degree less than 2, so of the form $p(x)=a+bx$?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:30
















$begingroup$
Welcome to Maths SX! What do you denote $P_2$?
$endgroup$
– Bernard
Jan 13 at 11:27




$begingroup$
Welcome to Maths SX! What do you denote $P_2$?
$endgroup$
– Bernard
Jan 13 at 11:27












$begingroup$
The set of linear polynomials with degree less than 2, so of the form $p(x)=a+bx$?
$endgroup$
– Andrea Brabrook
Jan 13 at 11:30




$begingroup$
The set of linear polynomials with degree less than 2, so of the form $p(x)=a+bx$?
$endgroup$
– Andrea Brabrook
Jan 13 at 11:30










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint:



The matrix of a linear map has as column vectors the coordinates, in the basis of $P_2$, of the images of the vectors of the basis used for $mathbf R^3$.



(Incidentally, there is no such thing as THE basis of a vector space: $mathbf R^3$ has an infinite number of bases. The basis you mention is usually considered as the canonical basis of $mathbf R^3$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, yes i should have said that in this case they were in respect to the standard bases. It's the images of the vectors in the basis for $R^3$ that's causing me the problem!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:43










  • $begingroup$
    Can't you calculate $S(1,0,0)$, $S(0,1,0)$ and $S(0,0,1)$?
    $endgroup$
    – Bernard
    Jan 13 at 11:49










  • $begingroup$
    Well, what i'm getting is $(3b-c,0,0), (0,3b-c,0),(0,0,3b-c)$ but i'm pretty sure this is completely wrong and don't know how to proceed then to find the coordinates in the basis of $P_2$
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:03










  • $begingroup$
    But there can be no $a,b$ or $c$ in the answer: they should represent the coordinates of vectors in the basis!
    $endgroup$
    – Bernard
    Jan 13 at 12:09










  • $begingroup$
    Which is why I was sure it was wrong! So, in answer to can i not calculate the images, no i can't!! Would I replace the a,b,c with x,y,z?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:14



















0












$begingroup$

Do you understand that a matrix represents a given matrix only in a given choice of basis for both domain and range spaces? What bases do you want to use? A natural choice would be {(1,0,0), (0,1,0), (0,0,1)} for the domain space and {1, x} for the range space. Apply the transformation to each of the domain space basis vectors in turn and write the result in terms of the range space basis. Each gives a column of the matrix representation.



This transformation maps (1, 0, 0) into 1+ 0x so first column of the matrix is (1, 0). The transformation maps (0, 1, 0) into 1+ 3x so the second column of the matrix is (1, 3). The transformation maps (0, 0, 1) into 1- x so the third column of the matrix is (1, -1). The matrix representation of the linear transformation using these choices of basis is $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Huge help, thank you..I was sort of getting there with the previous help but not quite! And yes, i do understand all that, thanks! struggling mature student here!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:46












Your Answer








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%2f3071903%2ffinding-the-representation-matrix-of-the-linear-transformation-s-mathbbr3%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









0












$begingroup$

Hint:



The matrix of a linear map has as column vectors the coordinates, in the basis of $P_2$, of the images of the vectors of the basis used for $mathbf R^3$.



(Incidentally, there is no such thing as THE basis of a vector space: $mathbf R^3$ has an infinite number of bases. The basis you mention is usually considered as the canonical basis of $mathbf R^3$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, yes i should have said that in this case they were in respect to the standard bases. It's the images of the vectors in the basis for $R^3$ that's causing me the problem!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:43










  • $begingroup$
    Can't you calculate $S(1,0,0)$, $S(0,1,0)$ and $S(0,0,1)$?
    $endgroup$
    – Bernard
    Jan 13 at 11:49










  • $begingroup$
    Well, what i'm getting is $(3b-c,0,0), (0,3b-c,0),(0,0,3b-c)$ but i'm pretty sure this is completely wrong and don't know how to proceed then to find the coordinates in the basis of $P_2$
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:03










  • $begingroup$
    But there can be no $a,b$ or $c$ in the answer: they should represent the coordinates of vectors in the basis!
    $endgroup$
    – Bernard
    Jan 13 at 12:09










  • $begingroup$
    Which is why I was sure it was wrong! So, in answer to can i not calculate the images, no i can't!! Would I replace the a,b,c with x,y,z?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:14
















0












$begingroup$

Hint:



The matrix of a linear map has as column vectors the coordinates, in the basis of $P_2$, of the images of the vectors of the basis used for $mathbf R^3$.



(Incidentally, there is no such thing as THE basis of a vector space: $mathbf R^3$ has an infinite number of bases. The basis you mention is usually considered as the canonical basis of $mathbf R^3$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, yes i should have said that in this case they were in respect to the standard bases. It's the images of the vectors in the basis for $R^3$ that's causing me the problem!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:43










  • $begingroup$
    Can't you calculate $S(1,0,0)$, $S(0,1,0)$ and $S(0,0,1)$?
    $endgroup$
    – Bernard
    Jan 13 at 11:49










  • $begingroup$
    Well, what i'm getting is $(3b-c,0,0), (0,3b-c,0),(0,0,3b-c)$ but i'm pretty sure this is completely wrong and don't know how to proceed then to find the coordinates in the basis of $P_2$
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:03










  • $begingroup$
    But there can be no $a,b$ or $c$ in the answer: they should represent the coordinates of vectors in the basis!
    $endgroup$
    – Bernard
    Jan 13 at 12:09










  • $begingroup$
    Which is why I was sure it was wrong! So, in answer to can i not calculate the images, no i can't!! Would I replace the a,b,c with x,y,z?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:14














0












0








0





$begingroup$

Hint:



The matrix of a linear map has as column vectors the coordinates, in the basis of $P_2$, of the images of the vectors of the basis used for $mathbf R^3$.



(Incidentally, there is no such thing as THE basis of a vector space: $mathbf R^3$ has an infinite number of bases. The basis you mention is usually considered as the canonical basis of $mathbf R^3$).






share|cite|improve this answer









$endgroup$



Hint:



The matrix of a linear map has as column vectors the coordinates, in the basis of $P_2$, of the images of the vectors of the basis used for $mathbf R^3$.



(Incidentally, there is no such thing as THE basis of a vector space: $mathbf R^3$ has an infinite number of bases. The basis you mention is usually considered as the canonical basis of $mathbf R^3$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 11:35









BernardBernard

124k742117




124k742117












  • $begingroup$
    Thanks, yes i should have said that in this case they were in respect to the standard bases. It's the images of the vectors in the basis for $R^3$ that's causing me the problem!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:43










  • $begingroup$
    Can't you calculate $S(1,0,0)$, $S(0,1,0)$ and $S(0,0,1)$?
    $endgroup$
    – Bernard
    Jan 13 at 11:49










  • $begingroup$
    Well, what i'm getting is $(3b-c,0,0), (0,3b-c,0),(0,0,3b-c)$ but i'm pretty sure this is completely wrong and don't know how to proceed then to find the coordinates in the basis of $P_2$
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:03










  • $begingroup$
    But there can be no $a,b$ or $c$ in the answer: they should represent the coordinates of vectors in the basis!
    $endgroup$
    – Bernard
    Jan 13 at 12:09










  • $begingroup$
    Which is why I was sure it was wrong! So, in answer to can i not calculate the images, no i can't!! Would I replace the a,b,c with x,y,z?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:14


















  • $begingroup$
    Thanks, yes i should have said that in this case they were in respect to the standard bases. It's the images of the vectors in the basis for $R^3$ that's causing me the problem!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 11:43










  • $begingroup$
    Can't you calculate $S(1,0,0)$, $S(0,1,0)$ and $S(0,0,1)$?
    $endgroup$
    – Bernard
    Jan 13 at 11:49










  • $begingroup$
    Well, what i'm getting is $(3b-c,0,0), (0,3b-c,0),(0,0,3b-c)$ but i'm pretty sure this is completely wrong and don't know how to proceed then to find the coordinates in the basis of $P_2$
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:03










  • $begingroup$
    But there can be no $a,b$ or $c$ in the answer: they should represent the coordinates of vectors in the basis!
    $endgroup$
    – Bernard
    Jan 13 at 12:09










  • $begingroup$
    Which is why I was sure it was wrong! So, in answer to can i not calculate the images, no i can't!! Would I replace the a,b,c with x,y,z?
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:14
















$begingroup$
Thanks, yes i should have said that in this case they were in respect to the standard bases. It's the images of the vectors in the basis for $R^3$ that's causing me the problem!
$endgroup$
– Andrea Brabrook
Jan 13 at 11:43




$begingroup$
Thanks, yes i should have said that in this case they were in respect to the standard bases. It's the images of the vectors in the basis for $R^3$ that's causing me the problem!
$endgroup$
– Andrea Brabrook
Jan 13 at 11:43












$begingroup$
Can't you calculate $S(1,0,0)$, $S(0,1,0)$ and $S(0,0,1)$?
$endgroup$
– Bernard
Jan 13 at 11:49




$begingroup$
Can't you calculate $S(1,0,0)$, $S(0,1,0)$ and $S(0,0,1)$?
$endgroup$
– Bernard
Jan 13 at 11:49












$begingroup$
Well, what i'm getting is $(3b-c,0,0), (0,3b-c,0),(0,0,3b-c)$ but i'm pretty sure this is completely wrong and don't know how to proceed then to find the coordinates in the basis of $P_2$
$endgroup$
– Andrea Brabrook
Jan 13 at 12:03




$begingroup$
Well, what i'm getting is $(3b-c,0,0), (0,3b-c,0),(0,0,3b-c)$ but i'm pretty sure this is completely wrong and don't know how to proceed then to find the coordinates in the basis of $P_2$
$endgroup$
– Andrea Brabrook
Jan 13 at 12:03












$begingroup$
But there can be no $a,b$ or $c$ in the answer: they should represent the coordinates of vectors in the basis!
$endgroup$
– Bernard
Jan 13 at 12:09




$begingroup$
But there can be no $a,b$ or $c$ in the answer: they should represent the coordinates of vectors in the basis!
$endgroup$
– Bernard
Jan 13 at 12:09












$begingroup$
Which is why I was sure it was wrong! So, in answer to can i not calculate the images, no i can't!! Would I replace the a,b,c with x,y,z?
$endgroup$
– Andrea Brabrook
Jan 13 at 12:14




$begingroup$
Which is why I was sure it was wrong! So, in answer to can i not calculate the images, no i can't!! Would I replace the a,b,c with x,y,z?
$endgroup$
– Andrea Brabrook
Jan 13 at 12:14











0












$begingroup$

Do you understand that a matrix represents a given matrix only in a given choice of basis for both domain and range spaces? What bases do you want to use? A natural choice would be {(1,0,0), (0,1,0), (0,0,1)} for the domain space and {1, x} for the range space. Apply the transformation to each of the domain space basis vectors in turn and write the result in terms of the range space basis. Each gives a column of the matrix representation.



This transformation maps (1, 0, 0) into 1+ 0x so first column of the matrix is (1, 0). The transformation maps (0, 1, 0) into 1+ 3x so the second column of the matrix is (1, 3). The transformation maps (0, 0, 1) into 1- x so the third column of the matrix is (1, -1). The matrix representation of the linear transformation using these choices of basis is $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Huge help, thank you..I was sort of getting there with the previous help but not quite! And yes, i do understand all that, thanks! struggling mature student here!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:46
















0












$begingroup$

Do you understand that a matrix represents a given matrix only in a given choice of basis for both domain and range spaces? What bases do you want to use? A natural choice would be {(1,0,0), (0,1,0), (0,0,1)} for the domain space and {1, x} for the range space. Apply the transformation to each of the domain space basis vectors in turn and write the result in terms of the range space basis. Each gives a column of the matrix representation.



This transformation maps (1, 0, 0) into 1+ 0x so first column of the matrix is (1, 0). The transformation maps (0, 1, 0) into 1+ 3x so the second column of the matrix is (1, 3). The transformation maps (0, 0, 1) into 1- x so the third column of the matrix is (1, -1). The matrix representation of the linear transformation using these choices of basis is $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Huge help, thank you..I was sort of getting there with the previous help but not quite! And yes, i do understand all that, thanks! struggling mature student here!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:46














0












0








0





$begingroup$

Do you understand that a matrix represents a given matrix only in a given choice of basis for both domain and range spaces? What bases do you want to use? A natural choice would be {(1,0,0), (0,1,0), (0,0,1)} for the domain space and {1, x} for the range space. Apply the transformation to each of the domain space basis vectors in turn and write the result in terms of the range space basis. Each gives a column of the matrix representation.



This transformation maps (1, 0, 0) into 1+ 0x so first column of the matrix is (1, 0). The transformation maps (0, 1, 0) into 1+ 3x so the second column of the matrix is (1, 3). The transformation maps (0, 0, 1) into 1- x so the third column of the matrix is (1, -1). The matrix representation of the linear transformation using these choices of basis is $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$






share|cite|improve this answer









$endgroup$



Do you understand that a matrix represents a given matrix only in a given choice of basis for both domain and range spaces? What bases do you want to use? A natural choice would be {(1,0,0), (0,1,0), (0,0,1)} for the domain space and {1, x} for the range space. Apply the transformation to each of the domain space basis vectors in turn and write the result in terms of the range space basis. Each gives a column of the matrix representation.



This transformation maps (1, 0, 0) into 1+ 0x so first column of the matrix is (1, 0). The transformation maps (0, 1, 0) into 1+ 3x so the second column of the matrix is (1, 3). The transformation maps (0, 0, 1) into 1- x so the third column of the matrix is (1, -1). The matrix representation of the linear transformation using these choices of basis is $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 12:40









user247327user247327

11.7k1516




11.7k1516












  • $begingroup$
    Huge help, thank you..I was sort of getting there with the previous help but not quite! And yes, i do understand all that, thanks! struggling mature student here!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:46


















  • $begingroup$
    Huge help, thank you..I was sort of getting there with the previous help but not quite! And yes, i do understand all that, thanks! struggling mature student here!
    $endgroup$
    – Andrea Brabrook
    Jan 13 at 12:46
















$begingroup$
Huge help, thank you..I was sort of getting there with the previous help but not quite! And yes, i do understand all that, thanks! struggling mature student here!
$endgroup$
– Andrea Brabrook
Jan 13 at 12:46




$begingroup$
Huge help, thank you..I was sort of getting there with the previous help but not quite! And yes, i do understand all that, thanks! struggling mature student here!
$endgroup$
– Andrea Brabrook
Jan 13 at 12:46


















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%2f3071903%2ffinding-the-representation-matrix-of-the-linear-transformation-s-mathbbr3%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