Mixing
Determining a matrix representation of a linear transformation relative to a basis
T: $R^4$ to P2 defined by T(a1, a2, a3, a4)=(a1+a2)+(a2+a3)x+(a3+a4)$x^2$
B={ (1,1,1,1), (1,1,1,0), (1,1,0,0), (1,0,0,0) }
C=( 1, 1+x, 1+$x^2$ )
Question: Find the matrix representation relative to the basis B and basis C
Attempt:
T(1,1,1,1)=2+2x+2$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,1,0)=2+2x+$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,0,0)=2+x=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,0,0,0)=1=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
After solving for a,b,c,d I get the matrix
$begin{bmatrix}-2 & -1 & 1\ 2 & 2 & 1\ 2 & 1 & 0end{bmatrix}$
matrices linear-transformations
asked Apr 14 '18 at 23:38
EssieEssie
535
|
show 3 more comments
T: $R^4$ to P2 defined by T(a1, a2, a3, a4)=(a1+a2)+(a2+a3)x+(a3+a4)$x^2$
B={ (1,1,1,1), (1,1,1,0), (1,1,0,0), (1,0,0,0) }
C=( 1, 1+x, 1+$x^2$ )
Question: Find the matrix representation relative to the basis B and basis C
Attempt:
T(1,1,1,1)=2+2x+2$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,1,0)=2+2x+$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,0,0)=2+x=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,0,0,0)=1=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
After solving for a,b,c,d I get the matrix
$begin{bmatrix}-2 & -1 & 1\ 2 & 2 & 1\ 2 & 1 & 0end{bmatrix}$
matrices linear-transformations
asked Apr 14 '18 at 23:38
EssieEssie
535
Where is $x$ defined? How can you have a quadratic term in a linear transformation?
– saulspatz
Apr 14 '18 at 23:42
1
A linear transformation from a 4-dimensional vector space to a 3-dimensional space will have a $3times4$ matrix, not a $3times3$.
– Gerry Myerson
Apr 15 '18 at 1:36
1
To get you started: $2+2x+2x^2=(-2)(1)+(2)(1+x)+(2)(1+x^2)$ So your first column is certainly correct. You have just left out the 4th column.
– Gerry Myerson
Apr 15 '18 at 1:37
1
So, can you get the full answer now, Essie?
– Gerry Myerson
Apr 16 '18 at 2:31
1
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Apr 17 '18 at 5:38
|
show 3 more comments
T: $R^4$ to P2 defined by T(a1, a2, a3, a4)=(a1+a2)+(a2+a3)x+(a3+a4)$x^2$
B={ (1,1,1,1), (1,1,1,0), (1,1,0,0), (1,0,0,0) }
C=( 1, 1+x, 1+$x^2$ )
Question: Find the matrix representation relative to the basis B and basis C
Attempt:
T(1,1,1,1)=2+2x+2$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,1,0)=2+2x+$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,0,0)=2+x=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,0,0,0)=1=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
After solving for a,b,c,d I get the matrix
$begin{bmatrix}-2 & -1 & 1\ 2 & 2 & 1\ 2 & 1 & 0end{bmatrix}$
matrices linear-transformations
asked Apr 14 '18 at 23:38
EssieEssie
535
T: $R^4$ to P2 defined by T(a1, a2, a3, a4)=(a1+a2)+(a2+a3)x+(a3+a4)$x^2$
B={ (1,1,1,1), (1,1,1,0), (1,1,0,0), (1,0,0,0) }
C=( 1, 1+x, 1+$x^2$ )
Question: Find the matrix representation relative to the basis B and basis C
Attempt:
T(1,1,1,1)=2+2x+2$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,1,0)=2+2x+$x^2$=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,1,0,0)=2+x=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
T(1,0,0,0)=1=a(1,1,1,1)+b(1,1,1,0)+c(1,1,0,0)+d(1,0,0,0)
After solving for a,b,c,d I get the matrix
$begin{bmatrix}-2 & -1 & 1\ 2 & 2 & 1\ 2 & 1 & 0end{bmatrix}$
matrices linear-transformations
matrices linear-transformations
asked Apr 14 '18 at 23:38
EssieEssie
535
asked Apr 14 '18 at 23:38
EssieEssie
535
asked Apr 14 '18 at 23:38
EssieEssie
535
asked Apr 14 '18 at 23:38
EssieEssie
535
asked Apr 14 '18 at 23:38
EssieEssie
535
535
Where is $x$ defined? How can you have a quadratic term in a linear transformation?
– saulspatz
Apr 14 '18 at 23:42
1
A linear transformation from a 4-dimensional vector space to a 3-dimensional space will have a $3times4$ matrix, not a $3times3$.
– Gerry Myerson
Apr 15 '18 at 1:36
1
To get you started: $2+2x+2x^2=(-2)(1)+(2)(1+x)+(2)(1+x^2)$ So your first column is certainly correct. You have just left out the 4th column.
– Gerry Myerson
Apr 15 '18 at 1:37
1
So, can you get the full answer now, Essie?
– Gerry Myerson
Apr 16 '18 at 2:31
1
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Apr 17 '18 at 5:38
|
show 3 more comments
Where is $x$ defined? How can you have a quadratic term in a linear transformation?
– saulspatz
Apr 14 '18 at 23:42
1
A linear transformation from a 4-dimensional vector space to a 3-dimensional space will have a $3times4$ matrix, not a $3times3$.
– Gerry Myerson
Apr 15 '18 at 1:36
1
To get you started: $2+2x+2x^2=(-2)(1)+(2)(1+x)+(2)(1+x^2)$ So your first column is certainly correct. You have just left out the 4th column.
– Gerry Myerson
Apr 15 '18 at 1:37
1
So, can you get the full answer now, Essie?
– Gerry Myerson
Apr 16 '18 at 2:31
1
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Apr 17 '18 at 5:38
Where is $x$ defined? How can you have a quadratic term in a linear transformation?
– saulspatz
Apr 14 '18 at 23:42
Where is $x$ defined? How can you have a quadratic term in a linear transformation?
– saulspatz
Apr 14 '18 at 23:42
1
1
A linear transformation from a 4-dimensional vector space to a 3-dimensional space will have a $3times4$ matrix, not a $3times3$.
– Gerry Myerson
Apr 15 '18 at 1:36
A linear transformation from a 4-dimensional vector space to a 3-dimensional space will have a $3times4$ matrix, not a $3times3$.
– Gerry Myerson
Apr 15 '18 at 1:36
1
1
To get you started: $2+2x+2x^2=(-2)(1)+(2)(1+x)+(2)(1+x^2)$ So your first column is certainly correct. You have just left out the 4th column.
– Gerry Myerson
Apr 15 '18 at 1:37
To get you started: $2+2x+2x^2=(-2)(1)+(2)(1+x)+(2)(1+x^2)$ So your first column is certainly correct. You have just left out the 4th column.
– Gerry Myerson
Apr 15 '18 at 1:37
1
1
So, can you get the full answer now, Essie?
– Gerry Myerson
Apr 16 '18 at 2:31
So, can you get the full answer now, Essie?
– Gerry Myerson
Apr 16 '18 at 2:31
1
1
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Apr 17 '18 at 5:38
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Apr 17 '18 at 5:38
|
show 3 more comments
1 Answer
1
active
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You should express the results of your four transformations in terms of the ordered basis $mathscr C,$ not $mathscr B$ (it appears that you did that to get a partially-correct answer, but your attempt suggests some incomplete understanding):
$$begin{align}
T(1,1,1,1) & = 2 + 2x + 2x^2\
& = -2(1) + 2(1+x) + 2(1+x^2)\\
T(1,1,1,0) & = 2 + 2x + x^2\
& = -1(1) + 2(1+x) + 1(1+x^2)\\
T(1,1,0,0) & = 2 + x\
& = 1(1) + 1(1+x) + 0(1+x^2)\\
T(1,0,0,0) & = 1\
& = 1(1) + 0(1+x) + 0(1+x^2).end{align}$$
The coefficients of the elements of $mathscr C$ are the columns of the matrix of $T$ relative to $mathscr B$ and $mathscr C:$
$$begin{bmatrix}
-2 & -1 & 1 & 1\
2 & 2 & 1 & 0\
2 & 1 & 0 & 0end{bmatrix}.$$
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
add a comment |
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1 Answer
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You should express the results of your four transformations in terms of the ordered basis $mathscr C,$ not $mathscr B$ (it appears that you did that to get a partially-correct answer, but your attempt suggests some incomplete understanding):
$$begin{align}
T(1,1,1,1) & = 2 + 2x + 2x^2\
& = -2(1) + 2(1+x) + 2(1+x^2)\\
T(1,1,1,0) & = 2 + 2x + x^2\
& = -1(1) + 2(1+x) + 1(1+x^2)\\
T(1,1,0,0) & = 2 + x\
& = 1(1) + 1(1+x) + 0(1+x^2)\\
T(1,0,0,0) & = 1\
& = 1(1) + 0(1+x) + 0(1+x^2).end{align}$$
The coefficients of the elements of $mathscr C$ are the columns of the matrix of $T$ relative to $mathscr B$ and $mathscr C:$
$$begin{bmatrix}
-2 & -1 & 1 & 1\
2 & 2 & 1 & 0\
2 & 1 & 0 & 0end{bmatrix}.$$
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
add a comment |
You should express the results of your four transformations in terms of the ordered basis $mathscr C,$ not $mathscr B$ (it appears that you did that to get a partially-correct answer, but your attempt suggests some incomplete understanding):
$$begin{align}
T(1,1,1,1) & = 2 + 2x + 2x^2\
& = -2(1) + 2(1+x) + 2(1+x^2)\\
T(1,1,1,0) & = 2 + 2x + x^2\
& = -1(1) + 2(1+x) + 1(1+x^2)\\
T(1,1,0,0) & = 2 + x\
& = 1(1) + 1(1+x) + 0(1+x^2)\\
T(1,0,0,0) & = 1\
& = 1(1) + 0(1+x) + 0(1+x^2).end{align}$$
The coefficients of the elements of $mathscr C$ are the columns of the matrix of $T$ relative to $mathscr B$ and $mathscr C:$
$$begin{bmatrix}
-2 & -1 & 1 & 1\
2 & 2 & 1 & 0\
2 & 1 & 0 & 0end{bmatrix}.$$
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
add a comment |
You should express the results of your four transformations in terms of the ordered basis $mathscr C,$ not $mathscr B$ (it appears that you did that to get a partially-correct answer, but your attempt suggests some incomplete understanding):
$$begin{align}
T(1,1,1,1) & = 2 + 2x + 2x^2\
& = -2(1) + 2(1+x) + 2(1+x^2)\\
T(1,1,1,0) & = 2 + 2x + x^2\
& = -1(1) + 2(1+x) + 1(1+x^2)\\
T(1,1,0,0) & = 2 + x\
& = 1(1) + 1(1+x) + 0(1+x^2)\\
T(1,0,0,0) & = 1\
& = 1(1) + 0(1+x) + 0(1+x^2).end{align}$$
The coefficients of the elements of $mathscr C$ are the columns of the matrix of $T$ relative to $mathscr B$ and $mathscr C:$
$$begin{bmatrix}
-2 & -1 & 1 & 1\
2 & 2 & 1 & 0\
2 & 1 & 0 & 0end{bmatrix}.$$
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
You should express the results of your four transformations in terms of the ordered basis $mathscr C,$ not $mathscr B$ (it appears that you did that to get a partially-correct answer, but your attempt suggests some incomplete understanding):
$$begin{align}
T(1,1,1,1) & = 2 + 2x + 2x^2\
& = -2(1) + 2(1+x) + 2(1+x^2)\\
T(1,1,1,0) & = 2 + 2x + x^2\
& = -1(1) + 2(1+x) + 1(1+x^2)\\
T(1,1,0,0) & = 2 + x\
& = 1(1) + 1(1+x) + 0(1+x^2)\\
T(1,0,0,0) & = 1\
& = 1(1) + 0(1+x) + 0(1+x^2).end{align}$$
The coefficients of the elements of $mathscr C$ are the columns of the matrix of $T$ relative to $mathscr B$ and $mathscr C:$
$$begin{bmatrix}
-2 & -1 & 1 & 1\
2 & 2 & 1 & 0\
2 & 1 & 0 & 0end{bmatrix}.$$
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
answered Nov 21 '18 at 17:52
Maurice PMaurice P
1,3901732
1,3901732
add a comment |
add a comment |
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Where is $x$ defined? How can you have a quadratic term in a linear transformation?
– saulspatz
Apr 14 '18 at 23:42
1
A linear transformation from a 4-dimensional vector space to a 3-dimensional space will have a $3times4$ matrix, not a $3times3$.
– Gerry Myerson
Apr 15 '18 at 1:36
1
To get you started: $2+2x+2x^2=(-2)(1)+(2)(1+x)+(2)(1+x^2)$ So your first column is certainly correct. You have just left out the 4th column.
– Gerry Myerson
Apr 15 '18 at 1:37
1
So, can you get the full answer now, Essie?
– Gerry Myerson
Apr 16 '18 at 2:31
1
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Apr 17 '18 at 5:38