Mixing
Finding the representation matrix of the linear transformation $S: mathbb{R}^3longrightarrow P_2: (a,b,c)...
$begingroup$
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
edited Jan 13 at 12:17
idriskameni
641319
asked Jan 13 at 11:26
Andrea BrabrookAndrea Brabrook
33
$endgroup$
add a comment |
$begingroup$
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
edited Jan 13 at 12:17
idriskameni
641319
asked Jan 13 at 11:26
Andrea BrabrookAndrea Brabrook
33
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 13 at 12:17
idriskameni
641319
asked Jan 13 at 11:26
Andrea BrabrookAndrea Brabrook
33
$endgroup$
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
edited Jan 13 at 12:17
idriskameni
641319
asked Jan 13 at 11:26
Andrea BrabrookAndrea Brabrook
33
edited Jan 13 at 12:17
idriskameni
641319
asked Jan 13 at 11:26
Andrea BrabrookAndrea Brabrook
33
edited Jan 13 at 12:17
idriskameni
641319
edited Jan 13 at 12:17
idriskameni
641319
edited Jan 13 at 12:17
idriskameni
641319
641319
asked Jan 13 at 11:26
Andrea BrabrookAndrea Brabrook
33
asked Jan 13 at 11:26
Andrea BrabrookAndrea Brabrook
33
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $mathbf R^3$).
answered Jan 13 at 11:35
BernardBernard
121k740116
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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 $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$
answered Jan 13 at 12:40
user247327user247327
11.1k1515
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
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oldest
votes
$begingroup$
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 $mathbf R^3$).
answered Jan 13 at 11:35
BernardBernard
121k740116
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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 $mathbf R^3$).
answered Jan 13 at 11:35
BernardBernard
121k740116
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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 $mathbf R^3$).
answered Jan 13 at 11:35
BernardBernard
121k740116
$endgroup$
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 $mathbf R^3$).
answered Jan 13 at 11:35
BernardBernard
121k740116
answered Jan 13 at 11:35
BernardBernard
121k740116
answered Jan 13 at 11:35
BernardBernard
121k740116
answered Jan 13 at 11:35
BernardBernard
121k740116
121k740116
$begingroup$
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
$begingroup$
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 $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$
answered Jan 13 at 12:40
user247327user247327
11.1k1515
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$
answered Jan 13 at 12:40
user247327user247327
11.1k1515
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$
answered Jan 13 at 12:40
user247327user247327
11.1k1515
$endgroup$
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 $begin{bmatrix}1 & 1 & 1 \ 0 & 3 & -1end{bmatrix}$
answered Jan 13 at 12:40
user247327user247327
11.1k1515
answered Jan 13 at 12:40
user247327user247327
11.1k1515
answered Jan 13 at 12:40
user247327user247327
11.1k1515
answered Jan 13 at 12:40
user247327user247327
11.1k1515
11.1k1515
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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