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







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$endgroup$












  • $begingroup$
    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






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share|cite|improve this question













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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












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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


















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Cabo Verde

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  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)
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Gyllenstierna

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