Find the coordinates of a polynomial relative to a given basis.Find a subset that forms a basis for the span of a setfind the standard basis of a linear transformationChange of basis matrix exercise, find the basis given the matrix.Change of basis to find coordinatesFinding a basis for the nullspaceFind $x$ in $mathbbR^2$ whose cordinate vector relative to the basis $B$Find the coordinates of p(x) relative to the basisFind Coordinate Vector with Respect to Given BasisFind matrix of polynomial linear transformation relative to basiscoordinate vector relative to basis for $P_2$
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Find the coordinates of a polynomial relative to a given basis.
Find a subset that forms a basis for the span of a setfind the standard basis of a linear transformationChange of basis matrix exercise, find the basis given the matrix.Change of basis to find coordinatesFinding a basis for the nullspaceFind $x$ in $mathbbR^2$ whose cordinate vector relative to the basis $B$Find the coordinates of p(x) relative to the basisFind Coordinate Vector with Respect to Given BasisFind matrix of polynomial linear transformation relative to basiscoordinate vector relative to basis for $P_2$
$begingroup$
The set $B=−4−3x^2, −16−3x−12x^2, −35−9x−27x^2$ is a basis for $P_3$ (or $Bbb R_3[x]$, if you wish.) Find the coordinates of $p(x)=−160−39x−123x^2$ relative to the basis $B$.
What I did to try to solve this problem was:
I let $x = 1$ making the set $B$ into a matrix:
beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrix
Then I transposed it:
beginbmatrix-4&-16&-3\0&-3&-12\-3&-12&-27endbmatrix
Then I made a matrix out of $p(x)$:
beginbmatrix-160\-39\-123endbmatrix
Then I multiplied the two matrices and received:
beginbmatrix5569\1224\4269endbmatrix
But this answer is wrong!
What am I doing wrong?
linear-algebra
edited Mar 15 at 16:16
Rócherz
2,9863821
asked Mar 15 at 16:10
Scott MScott M
32
$endgroup$
add a comment |
$begingroup$
The set $B=−4−3x^2, −16−3x−12x^2, −35−9x−27x^2$ is a basis for $P_3$ (or $Bbb R_3[x]$, if you wish.) Find the coordinates of $p(x)=−160−39x−123x^2$ relative to the basis $B$.
What I did to try to solve this problem was:
I let $x = 1$ making the set $B$ into a matrix:
beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrix
Then I transposed it:
beginbmatrix-4&-16&-3\0&-3&-12\-3&-12&-27endbmatrix
Then I made a matrix out of $p(x)$:
beginbmatrix-160\-39\-123endbmatrix
Then I multiplied the two matrices and received:
beginbmatrix5569\1224\4269endbmatrix
But this answer is wrong!
What am I doing wrong?
linear-algebra
edited Mar 15 at 16:16
Rócherz
2,9863821
asked Mar 15 at 16:10
Scott MScott M
32
$endgroup$
$begingroup$
You need to take the inverse of the matrix before you multiply it by your vector. What you want is a mix of the three basis polynomials that gives you $p(x)$; instead, what you've done is treated the coefficients of $p(x)$ as a mix, and computed what polynomial that gives you.
$endgroup$
– Brian Tung
Mar 15 at 16:15
add a comment |
$begingroup$
The set $B=−4−3x^2, −16−3x−12x^2, −35−9x−27x^2$ is a basis for $P_3$ (or $Bbb R_3[x]$, if you wish.) Find the coordinates of $p(x)=−160−39x−123x^2$ relative to the basis $B$.
What I did to try to solve this problem was:
I let $x = 1$ making the set $B$ into a matrix:
beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrix
Then I transposed it:
beginbmatrix-4&-16&-3\0&-3&-12\-3&-12&-27endbmatrix
Then I made a matrix out of $p(x)$:
beginbmatrix-160\-39\-123endbmatrix
Then I multiplied the two matrices and received:
beginbmatrix5569\1224\4269endbmatrix
But this answer is wrong!
What am I doing wrong?
linear-algebra
edited Mar 15 at 16:16
Rócherz
2,9863821
asked Mar 15 at 16:10
Scott MScott M
32
$endgroup$
The set $B=−4−3x^2, −16−3x−12x^2, −35−9x−27x^2$ is a basis for $P_3$ (or $Bbb R_3[x]$, if you wish.) Find the coordinates of $p(x)=−160−39x−123x^2$ relative to the basis $B$.
What I did to try to solve this problem was:
I let $x = 1$ making the set $B$ into a matrix:
beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrix
Then I transposed it:
beginbmatrix-4&-16&-3\0&-3&-12\-3&-12&-27endbmatrix
Then I made a matrix out of $p(x)$:
beginbmatrix-160\-39\-123endbmatrix
Then I multiplied the two matrices and received:
beginbmatrix5569\1224\4269endbmatrix
But this answer is wrong!
What am I doing wrong?
linear-algebra
linear-algebra
edited Mar 15 at 16:16
Rócherz
2,9863821
asked Mar 15 at 16:10
Scott MScott M
32
edited Mar 15 at 16:16
Rócherz
2,9863821
asked Mar 15 at 16:10
Scott MScott M
32
edited Mar 15 at 16:16
Rócherz
2,9863821
edited Mar 15 at 16:16
Rócherz
2,9863821
edited Mar 15 at 16:16
Rócherz
2,9863821
2,9863821
asked Mar 15 at 16:10
Scott MScott M
32
asked Mar 15 at 16:10
Scott MScott M
32
asked Mar 15 at 16:10
Scott MScott M
32
32
$begingroup$
You need to take the inverse of the matrix before you multiply it by your vector. What you want is a mix of the three basis polynomials that gives you $p(x)$; instead, what you've done is treated the coefficients of $p(x)$ as a mix, and computed what polynomial that gives you.
$endgroup$
– Brian Tung
Mar 15 at 16:15
add a comment |
$begingroup$
You need to take the inverse of the matrix before you multiply it by your vector. What you want is a mix of the three basis polynomials that gives you $p(x)$; instead, what you've done is treated the coefficients of $p(x)$ as a mix, and computed what polynomial that gives you.
$endgroup$
– Brian Tung
Mar 15 at 16:15
$begingroup$
You need to take the inverse of the matrix before you multiply it by your vector. What you want is a mix of the three basis polynomials that gives you $p(x)$; instead, what you've done is treated the coefficients of $p(x)$ as a mix, and computed what polynomial that gives you.
$endgroup$
– Brian Tung
Mar 15 at 16:15
$begingroup$
You need to take the inverse of the matrix before you multiply it by your vector. What you want is a mix of the three basis polynomials that gives you $p(x)$; instead, what you've done is treated the coefficients of $p(x)$ as a mix, and computed what polynomial that gives you.
$endgroup$
– Brian Tung
Mar 15 at 16:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have a lot of the right ingredients here, but it's worth seeing why these
ingredients are correct.
We have the basis
$$
beta=f_1, f_2, f_3
$$
of $Bbb R_2[x]$ where $f_1, f_2, f_3inBbb R_2[x]$ are given by
beginalign*
f_1(x) &= -3 , x^2 - 4 & f_2(x) &= -12 , x^2 - 3 , x - 16 & f_3(x) &= -27 , x^2 - 9 , x - 35
endalign*
We wish to compute the vector $[p]_beta$ where $pinBbb R_2[x]$ is given by
$p(x)=-123 , x^2 - 39 , x - 160$.
Our desired vector $[p]_beta$ is defined as $[p]_beta=langle c_1, c_2, c_3rangle$
where $c_1,c_2,c_3inBbb R$ satisfy the equation
$$
c_1cdot f_1+c_2cdot f_2+c_3cdot f_3 = p
$$
To find these scalars, note that this equation takes the form
$$
c_1cdot(-3 , x^2 - 4)
+c_2cdot(-12 , x^2 - 3 , x - 16)
+c_3cdot(-27 , x^2 - 9 , x - 35)
=
-123 , x^2 - 39 , x - 160
$$
Rearranging this equation gives
$$
(-3 , c_1 - 12 , c_2 - 27 , c_3)cdot x^2
+(-3 , c_2 - 9 , c_3)cdot x
+(-4 , c_1 - 16 , c_2 - 35 , c_3)
=
-123 , x^2 - 39 , x - 160
$$
Comparing coefficients on both sides of our equation then gives the system
$$
beginarrayrcrcrcr
-3,c_1 &-& 12,c_2 &-& 27,c_3 &=& -123 \
& & -3,c_2 &-& 9,c_3 &=& -39 \
-4,c_1 &-& 16,c_2 &-& 35,c_3 &=& -160
endarray
$$
This system is represented by the augmented matrix
$$
left[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]
$$
Row reducing then gives
$$
operatornamerrefleft[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]=left[beginarrayrrr
1 & 0 & 0 & 1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 4
endarrayright]
$$
This gives our desired vector $[p]_beta=langle 1, 1, 4rangle$.
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
add a comment |
$begingroup$
You should inverse the matrix, not transpose it.
But it is more efficient to solve the linear system.
$$beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrixbeginpmatrixa\b\cendpmatrix=beginpmatrix-160\-39\-123endpmatrix
$$
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Actually, the answer was correct when I transposed it, took the inverse of it, then multiplied it be the 3x1 matrix, p(x).
$endgroup$
– Scott M
Mar 15 at 16:20
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have a lot of the right ingredients here, but it's worth seeing why these
ingredients are correct.
We have the basis
$$
beta=f_1, f_2, f_3
$$
of $Bbb R_2[x]$ where $f_1, f_2, f_3inBbb R_2[x]$ are given by
beginalign*
f_1(x) &= -3 , x^2 - 4 & f_2(x) &= -12 , x^2 - 3 , x - 16 & f_3(x) &= -27 , x^2 - 9 , x - 35
endalign*
We wish to compute the vector $[p]_beta$ where $pinBbb R_2[x]$ is given by
$p(x)=-123 , x^2 - 39 , x - 160$.
Our desired vector $[p]_beta$ is defined as $[p]_beta=langle c_1, c_2, c_3rangle$
where $c_1,c_2,c_3inBbb R$ satisfy the equation
$$
c_1cdot f_1+c_2cdot f_2+c_3cdot f_3 = p
$$
To find these scalars, note that this equation takes the form
$$
c_1cdot(-3 , x^2 - 4)
+c_2cdot(-12 , x^2 - 3 , x - 16)
+c_3cdot(-27 , x^2 - 9 , x - 35)
=
-123 , x^2 - 39 , x - 160
$$
Rearranging this equation gives
$$
(-3 , c_1 - 12 , c_2 - 27 , c_3)cdot x^2
+(-3 , c_2 - 9 , c_3)cdot x
+(-4 , c_1 - 16 , c_2 - 35 , c_3)
=
-123 , x^2 - 39 , x - 160
$$
Comparing coefficients on both sides of our equation then gives the system
$$
beginarrayrcrcrcr
-3,c_1 &-& 12,c_2 &-& 27,c_3 &=& -123 \
& & -3,c_2 &-& 9,c_3 &=& -39 \
-4,c_1 &-& 16,c_2 &-& 35,c_3 &=& -160
endarray
$$
This system is represented by the augmented matrix
$$
left[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]
$$
Row reducing then gives
$$
operatornamerrefleft[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]=left[beginarrayrrr
1 & 0 & 0 & 1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 4
endarrayright]
$$
This gives our desired vector $[p]_beta=langle 1, 1, 4rangle$.
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
add a comment |
$begingroup$
You have a lot of the right ingredients here, but it's worth seeing why these
ingredients are correct.
We have the basis
$$
beta=f_1, f_2, f_3
$$
of $Bbb R_2[x]$ where $f_1, f_2, f_3inBbb R_2[x]$ are given by
beginalign*
f_1(x) &= -3 , x^2 - 4 & f_2(x) &= -12 , x^2 - 3 , x - 16 & f_3(x) &= -27 , x^2 - 9 , x - 35
endalign*
We wish to compute the vector $[p]_beta$ where $pinBbb R_2[x]$ is given by
$p(x)=-123 , x^2 - 39 , x - 160$.
Our desired vector $[p]_beta$ is defined as $[p]_beta=langle c_1, c_2, c_3rangle$
where $c_1,c_2,c_3inBbb R$ satisfy the equation
$$
c_1cdot f_1+c_2cdot f_2+c_3cdot f_3 = p
$$
To find these scalars, note that this equation takes the form
$$
c_1cdot(-3 , x^2 - 4)
+c_2cdot(-12 , x^2 - 3 , x - 16)
+c_3cdot(-27 , x^2 - 9 , x - 35)
=
-123 , x^2 - 39 , x - 160
$$
Rearranging this equation gives
$$
(-3 , c_1 - 12 , c_2 - 27 , c_3)cdot x^2
+(-3 , c_2 - 9 , c_3)cdot x
+(-4 , c_1 - 16 , c_2 - 35 , c_3)
=
-123 , x^2 - 39 , x - 160
$$
Comparing coefficients on both sides of our equation then gives the system
$$
beginarrayrcrcrcr
-3,c_1 &-& 12,c_2 &-& 27,c_3 &=& -123 \
& & -3,c_2 &-& 9,c_3 &=& -39 \
-4,c_1 &-& 16,c_2 &-& 35,c_3 &=& -160
endarray
$$
This system is represented by the augmented matrix
$$
left[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]
$$
Row reducing then gives
$$
operatornamerrefleft[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]=left[beginarrayrrr
1 & 0 & 0 & 1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 4
endarrayright]
$$
This gives our desired vector $[p]_beta=langle 1, 1, 4rangle$.
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
add a comment |
$begingroup$
You have a lot of the right ingredients here, but it's worth seeing why these
ingredients are correct.
We have the basis
$$
beta=f_1, f_2, f_3
$$
of $Bbb R_2[x]$ where $f_1, f_2, f_3inBbb R_2[x]$ are given by
beginalign*
f_1(x) &= -3 , x^2 - 4 & f_2(x) &= -12 , x^2 - 3 , x - 16 & f_3(x) &= -27 , x^2 - 9 , x - 35
endalign*
We wish to compute the vector $[p]_beta$ where $pinBbb R_2[x]$ is given by
$p(x)=-123 , x^2 - 39 , x - 160$.
Our desired vector $[p]_beta$ is defined as $[p]_beta=langle c_1, c_2, c_3rangle$
where $c_1,c_2,c_3inBbb R$ satisfy the equation
$$
c_1cdot f_1+c_2cdot f_2+c_3cdot f_3 = p
$$
To find these scalars, note that this equation takes the form
$$
c_1cdot(-3 , x^2 - 4)
+c_2cdot(-12 , x^2 - 3 , x - 16)
+c_3cdot(-27 , x^2 - 9 , x - 35)
=
-123 , x^2 - 39 , x - 160
$$
Rearranging this equation gives
$$
(-3 , c_1 - 12 , c_2 - 27 , c_3)cdot x^2
+(-3 , c_2 - 9 , c_3)cdot x
+(-4 , c_1 - 16 , c_2 - 35 , c_3)
=
-123 , x^2 - 39 , x - 160
$$
Comparing coefficients on both sides of our equation then gives the system
$$
beginarrayrcrcrcr
-3,c_1 &-& 12,c_2 &-& 27,c_3 &=& -123 \
& & -3,c_2 &-& 9,c_3 &=& -39 \
-4,c_1 &-& 16,c_2 &-& 35,c_3 &=& -160
endarray
$$
This system is represented by the augmented matrix
$$
left[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]
$$
Row reducing then gives
$$
operatornamerrefleft[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]=left[beginarrayrrr
1 & 0 & 0 & 1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 4
endarrayright]
$$
This gives our desired vector $[p]_beta=langle 1, 1, 4rangle$.
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
You have a lot of the right ingredients here, but it's worth seeing why these
ingredients are correct.
We have the basis
$$
beta=f_1, f_2, f_3
$$
of $Bbb R_2[x]$ where $f_1, f_2, f_3inBbb R_2[x]$ are given by
beginalign*
f_1(x) &= -3 , x^2 - 4 & f_2(x) &= -12 , x^2 - 3 , x - 16 & f_3(x) &= -27 , x^2 - 9 , x - 35
endalign*
We wish to compute the vector $[p]_beta$ where $pinBbb R_2[x]$ is given by
$p(x)=-123 , x^2 - 39 , x - 160$.
Our desired vector $[p]_beta$ is defined as $[p]_beta=langle c_1, c_2, c_3rangle$
where $c_1,c_2,c_3inBbb R$ satisfy the equation
$$
c_1cdot f_1+c_2cdot f_2+c_3cdot f_3 = p
$$
To find these scalars, note that this equation takes the form
$$
c_1cdot(-3 , x^2 - 4)
+c_2cdot(-12 , x^2 - 3 , x - 16)
+c_3cdot(-27 , x^2 - 9 , x - 35)
=
-123 , x^2 - 39 , x - 160
$$
Rearranging this equation gives
$$
(-3 , c_1 - 12 , c_2 - 27 , c_3)cdot x^2
+(-3 , c_2 - 9 , c_3)cdot x
+(-4 , c_1 - 16 , c_2 - 35 , c_3)
=
-123 , x^2 - 39 , x - 160
$$
Comparing coefficients on both sides of our equation then gives the system
$$
beginarrayrcrcrcr
-3,c_1 &-& 12,c_2 &-& 27,c_3 &=& -123 \
& & -3,c_2 &-& 9,c_3 &=& -39 \
-4,c_1 &-& 16,c_2 &-& 35,c_3 &=& -160
endarray
$$
This system is represented by the augmented matrix
$$
left[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]
$$
Row reducing then gives
$$
operatornamerrefleft[beginarrayrrr
-3 & -12 & -27 & -123 \
0 & -3 & -9 & -39 \
-4 & -16 & -35 & -160
endarrayright]=left[beginarrayrrr
1 & 0 & 0 & 1 \
0 & 1 & 0 & 1 \
0 & 0 & 1 & 4
endarrayright]
$$
This gives our desired vector $[p]_beta=langle 1, 1, 4rangle$.
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Mar 15 at 18:29
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
21.8k42959
add a comment |
add a comment |
$begingroup$
You should inverse the matrix, not transpose it.
But it is more efficient to solve the linear system.
$$beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrixbeginpmatrixa\b\cendpmatrix=beginpmatrix-160\-39\-123endpmatrix
$$
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Actually, the answer was correct when I transposed it, took the inverse of it, then multiplied it be the 3x1 matrix, p(x).
$endgroup$
– Scott M
Mar 15 at 16:20
add a comment |
$begingroup$
You should inverse the matrix, not transpose it.
But it is more efficient to solve the linear system.
$$beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrixbeginpmatrixa\b\cendpmatrix=beginpmatrix-160\-39\-123endpmatrix
$$
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Actually, the answer was correct when I transposed it, took the inverse of it, then multiplied it be the 3x1 matrix, p(x).
$endgroup$
– Scott M
Mar 15 at 16:20
add a comment |
$begingroup$
You should inverse the matrix, not transpose it.
But it is more efficient to solve the linear system.
$$beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrixbeginpmatrixa\b\cendpmatrix=beginpmatrix-160\-39\-123endpmatrix
$$
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
$endgroup$
You should inverse the matrix, not transpose it.
But it is more efficient to solve the linear system.
$$beginbmatrix-4&0&-3\-16&-3&-12\-35&-9&-27endbmatrixbeginpmatrixa\b\cendpmatrix=beginpmatrix-160\-39\-123endpmatrix
$$
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
answered Mar 15 at 16:19
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
Actually, the answer was correct when I transposed it, took the inverse of it, then multiplied it be the 3x1 matrix, p(x).
$endgroup$
– Scott M
Mar 15 at 16:20
add a comment |
$begingroup$
Actually, the answer was correct when I transposed it, took the inverse of it, then multiplied it be the 3x1 matrix, p(x).
$endgroup$
– Scott M
Mar 15 at 16:20
$begingroup$
Actually, the answer was correct when I transposed it, took the inverse of it, then multiplied it be the 3x1 matrix, p(x).
$endgroup$
– Scott M
Mar 15 at 16:20
$begingroup$
Actually, the answer was correct when I transposed it, took the inverse of it, then multiplied it be the 3x1 matrix, p(x).
$endgroup$
– Scott M
Mar 15 at 16:20
add a comment |
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$begingroup$
You need to take the inverse of the matrix before you multiply it by your vector. What you want is a mix of the three basis polynomials that gives you $p(x)$; instead, what you've done is treated the coefficients of $p(x)$ as a mix, and computed what polynomial that gives you.
$endgroup$
– Brian Tung
Mar 15 at 16:15