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How to compute the Jacobian matrix of a multivariate function in a nonstandard matrix?
$begingroup$
Given a function $f:R^2rightarrow R^2$ such that $f(x,y)=(xy, cos xy)$, I need to compute the Jacobian matrix Df with respect to the basis ${(1,0), (1,1)}$. Not confident in my answer though. Please help me verify it. If it is not correct, please give me a hint. Here is my try:
With respect to the standard basis ${1,0), (0,1)}$, we could derive a Jacobian matrix
$$Df=
begin{bmatrix}
frac{partial f_1}{partial x} & frac{partial f_1}{partial y} \
frac{partial f_2}{partial x} & frac{partial f_2}{partial x}
end{bmatrix}
=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
.$$
To compute the Jacobian matrix with respect to the nonstandard basis ${(1,0), (1,1)}$, I multiply $Df(x,y)$ by this basis and get
$$overline{Df}(x,y)=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
times
begin{bmatrix}
1 & 1\
0 & 1
end{bmatrix}
=
begin{bmatrix}
y & y+x\
-ysin xy & -ysin xy-xsin xy
end{bmatrix}
.$$
I did it in this way, because my textbook [Hoffman] says, "the columns of the matrix relative tot he new basis will be the derivative $Df(x,y)$ applied to the new basis in $R^2$ with this image vector expressed in the new basis in $R^2$."
Anyone disagrees or agrees with my answer?
multivariable-calculus derivatives partial-derivative change-of-basis jacobian
edited Jan 23 at 0:49
Namaste
1
asked Jan 23 at 0:30
James WangJames Wang
49828
$endgroup$
add a comment |
$begingroup$
Given a function $f:R^2rightarrow R^2$ such that $f(x,y)=(xy, cos xy)$, I need to compute the Jacobian matrix Df with respect to the basis ${(1,0), (1,1)}$. Not confident in my answer though. Please help me verify it. If it is not correct, please give me a hint. Here is my try:
With respect to the standard basis ${1,0), (0,1)}$, we could derive a Jacobian matrix
$$Df=
begin{bmatrix}
frac{partial f_1}{partial x} & frac{partial f_1}{partial y} \
frac{partial f_2}{partial x} & frac{partial f_2}{partial x}
end{bmatrix}
=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
.$$
To compute the Jacobian matrix with respect to the nonstandard basis ${(1,0), (1,1)}$, I multiply $Df(x,y)$ by this basis and get
$$overline{Df}(x,y)=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
times
begin{bmatrix}
1 & 1\
0 & 1
end{bmatrix}
=
begin{bmatrix}
y & y+x\
-ysin xy & -ysin xy-xsin xy
end{bmatrix}
.$$
I did it in this way, because my textbook [Hoffman] says, "the columns of the matrix relative tot he new basis will be the derivative $Df(x,y)$ applied to the new basis in $R^2$ with this image vector expressed in the new basis in $R^2$."
Anyone disagrees or agrees with my answer?
multivariable-calculus derivatives partial-derivative change-of-basis jacobian
edited Jan 23 at 0:49
Namaste
1
asked Jan 23 at 0:30
James WangJames Wang
49828
$endgroup$
$begingroup$
In what basis is the input data represented? The output data?
$endgroup$
– David Kraemer
Jan 23 at 1:15
$begingroup$
My calculation does not agree with yours. See my answer.
$endgroup$
– Matematleta
Jan 23 at 1:22
add a comment |
$begingroup$
Given a function $f:R^2rightarrow R^2$ such that $f(x,y)=(xy, cos xy)$, I need to compute the Jacobian matrix Df with respect to the basis ${(1,0), (1,1)}$. Not confident in my answer though. Please help me verify it. If it is not correct, please give me a hint. Here is my try:
With respect to the standard basis ${1,0), (0,1)}$, we could derive a Jacobian matrix
$$Df=
begin{bmatrix}
frac{partial f_1}{partial x} & frac{partial f_1}{partial y} \
frac{partial f_2}{partial x} & frac{partial f_2}{partial x}
end{bmatrix}
=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
.$$
To compute the Jacobian matrix with respect to the nonstandard basis ${(1,0), (1,1)}$, I multiply $Df(x,y)$ by this basis and get
$$overline{Df}(x,y)=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
times
begin{bmatrix}
1 & 1\
0 & 1
end{bmatrix}
=
begin{bmatrix}
y & y+x\
-ysin xy & -ysin xy-xsin xy
end{bmatrix}
.$$
I did it in this way, because my textbook [Hoffman] says, "the columns of the matrix relative tot he new basis will be the derivative $Df(x,y)$ applied to the new basis in $R^2$ with this image vector expressed in the new basis in $R^2$."
Anyone disagrees or agrees with my answer?
multivariable-calculus derivatives partial-derivative change-of-basis jacobian
edited Jan 23 at 0:49
Namaste
1
asked Jan 23 at 0:30
James WangJames Wang
49828
$endgroup$
Given a function $f:R^2rightarrow R^2$ such that $f(x,y)=(xy, cos xy)$, I need to compute the Jacobian matrix Df with respect to the basis ${(1,0), (1,1)}$. Not confident in my answer though. Please help me verify it. If it is not correct, please give me a hint. Here is my try:
With respect to the standard basis ${1,0), (0,1)}$, we could derive a Jacobian matrix
$$Df=
begin{bmatrix}
frac{partial f_1}{partial x} & frac{partial f_1}{partial y} \
frac{partial f_2}{partial x} & frac{partial f_2}{partial x}
end{bmatrix}
=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
.$$
To compute the Jacobian matrix with respect to the nonstandard basis ${(1,0), (1,1)}$, I multiply $Df(x,y)$ by this basis and get
$$overline{Df}(x,y)=
begin{bmatrix}
y & x\
-ysin xy & -xsin xy
end{bmatrix}
times
begin{bmatrix}
1 & 1\
0 & 1
end{bmatrix}
=
begin{bmatrix}
y & y+x\
-ysin xy & -ysin xy-xsin xy
end{bmatrix}
.$$
I did it in this way, because my textbook [Hoffman] says, "the columns of the matrix relative tot he new basis will be the derivative $Df(x,y)$ applied to the new basis in $R^2$ with this image vector expressed in the new basis in $R^2$."
Anyone disagrees or agrees with my answer?
multivariable-calculus derivatives partial-derivative change-of-basis jacobian
multivariable-calculus derivatives partial-derivative change-of-basis jacobian
edited Jan 23 at 0:49
Namaste
1
asked Jan 23 at 0:30
James WangJames Wang
49828
edited Jan 23 at 0:49
Namaste
1
asked Jan 23 at 0:30
James WangJames Wang
49828
edited Jan 23 at 0:49
Namaste
1
edited Jan 23 at 0:49
Namaste
1
edited Jan 23 at 0:49
Namaste
1
1
asked Jan 23 at 0:30
James WangJames Wang
49828
asked Jan 23 at 0:30
James WangJames Wang
49828
asked Jan 23 at 0:30
James WangJames Wang
49828
49828
$begingroup$
In what basis is the input data represented? The output data?
$endgroup$
– David Kraemer
Jan 23 at 1:15
$begingroup$
My calculation does not agree with yours. See my answer.
$endgroup$
– Matematleta
Jan 23 at 1:22
add a comment |
$begingroup$
In what basis is the input data represented? The output data?
$endgroup$
– David Kraemer
Jan 23 at 1:15
$begingroup$
My calculation does not agree with yours. See my answer.
$endgroup$
– Matematleta
Jan 23 at 1:22
$begingroup$
In what basis is the input data represented? The output data?
$endgroup$
– David Kraemer
Jan 23 at 1:15
$begingroup$
In what basis is the input data represented? The output data?
$endgroup$
– David Kraemer
Jan 23 at 1:15
$begingroup$
My calculation does not agree with yours. See my answer.
$endgroup$
– Matematleta
Jan 23 at 1:22
$begingroup$
My calculation does not agree with yours. See my answer.
$endgroup$
– Matematleta
Jan 23 at 1:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can check this by going back to the definitions and properties of linear transformations.
Suppose $A:mathbb R^2to mathbb R^2$ is a linear transformation such that $A((1,0))=a(1,0)+b(0,1)$ and $A((0,1))=c(1,0)+d(0,1)$. This determines $A$ uniquely and the matrix with respect to the basis $(1,0),(0,1)$ is
begin{bmatrix}
a& c\
b & d
end{bmatrix}
Now,
$A((1,0))=a(1,0)+b(0,1)=a(1,0)+b((1,1)-(1,0))=(a-b)(1,0)+b(1,1)$
and
$A((1,1))=A((1,0))+A((0,1))=(a+c)(1,0)+(b+d)(0,1)=(a+c)(1,0)+(b+d)((1,1)-(1,0))=((a+c)-(b+d))(1,0)+(b+d)(1,1)$
Then, the matrix with respect to the basis $(1,0), (1,1)$ is
begin{bmatrix}
a-b& (a+c)-(b+d)\
b & b+d
end{bmatrix}
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
You can check this by going back to the definitions and properties of linear transformations.
Suppose $A:mathbb R^2to mathbb R^2$ is a linear transformation such that $A((1,0))=a(1,0)+b(0,1)$ and $A((0,1))=c(1,0)+d(0,1)$. This determines $A$ uniquely and the matrix with respect to the basis $(1,0),(0,1)$ is
begin{bmatrix}
a& c\
b & d
end{bmatrix}
Now,
$A((1,0))=a(1,0)+b(0,1)=a(1,0)+b((1,1)-(1,0))=(a-b)(1,0)+b(1,1)$
and
$A((1,1))=A((1,0))+A((0,1))=(a+c)(1,0)+(b+d)(0,1)=(a+c)(1,0)+(b+d)((1,1)-(1,0))=((a+c)-(b+d))(1,0)+(b+d)(1,1)$
Then, the matrix with respect to the basis $(1,0), (1,1)$ is
begin{bmatrix}
a-b& (a+c)-(b+d)\
b & b+d
end{bmatrix}
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
You can check this by going back to the definitions and properties of linear transformations.
Suppose $A:mathbb R^2to mathbb R^2$ is a linear transformation such that $A((1,0))=a(1,0)+b(0,1)$ and $A((0,1))=c(1,0)+d(0,1)$. This determines $A$ uniquely and the matrix with respect to the basis $(1,0),(0,1)$ is
begin{bmatrix}
a& c\
b & d
end{bmatrix}
Now,
$A((1,0))=a(1,0)+b(0,1)=a(1,0)+b((1,1)-(1,0))=(a-b)(1,0)+b(1,1)$
and
$A((1,1))=A((1,0))+A((0,1))=(a+c)(1,0)+(b+d)(0,1)=(a+c)(1,0)+(b+d)((1,1)-(1,0))=((a+c)-(b+d))(1,0)+(b+d)(1,1)$
Then, the matrix with respect to the basis $(1,0), (1,1)$ is
begin{bmatrix}
a-b& (a+c)-(b+d)\
b & b+d
end{bmatrix}
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
You can check this by going back to the definitions and properties of linear transformations.
Suppose $A:mathbb R^2to mathbb R^2$ is a linear transformation such that $A((1,0))=a(1,0)+b(0,1)$ and $A((0,1))=c(1,0)+d(0,1)$. This determines $A$ uniquely and the matrix with respect to the basis $(1,0),(0,1)$ is
begin{bmatrix}
a& c\
b & d
end{bmatrix}
Now,
$A((1,0))=a(1,0)+b(0,1)=a(1,0)+b((1,1)-(1,0))=(a-b)(1,0)+b(1,1)$
and
$A((1,1))=A((1,0))+A((0,1))=(a+c)(1,0)+(b+d)(0,1)=(a+c)(1,0)+(b+d)((1,1)-(1,0))=((a+c)-(b+d))(1,0)+(b+d)(1,1)$
Then, the matrix with respect to the basis $(1,0), (1,1)$ is
begin{bmatrix}
a-b& (a+c)-(b+d)\
b & b+d
end{bmatrix}
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
$endgroup$
You can check this by going back to the definitions and properties of linear transformations.
Suppose $A:mathbb R^2to mathbb R^2$ is a linear transformation such that $A((1,0))=a(1,0)+b(0,1)$ and $A((0,1))=c(1,0)+d(0,1)$. This determines $A$ uniquely and the matrix with respect to the basis $(1,0),(0,1)$ is
begin{bmatrix}
a& c\
b & d
end{bmatrix}
Now,
$A((1,0))=a(1,0)+b(0,1)=a(1,0)+b((1,1)-(1,0))=(a-b)(1,0)+b(1,1)$
and
$A((1,1))=A((1,0))+A((0,1))=(a+c)(1,0)+(b+d)(0,1)=(a+c)(1,0)+(b+d)((1,1)-(1,0))=((a+c)-(b+d))(1,0)+(b+d)(1,1)$
Then, the matrix with respect to the basis $(1,0), (1,1)$ is
begin{bmatrix}
a-b& (a+c)-(b+d)\
b & b+d
end{bmatrix}
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
answered Jan 23 at 1:22
MatematletaMatematleta
11.5k2920
11.5k2920
add a comment |
add a comment |
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$begingroup$
In what basis is the input data represented? The output data?
$endgroup$
– David Kraemer
Jan 23 at 1:15
$begingroup$
My calculation does not agree with yours. See my answer.
$endgroup$
– Matematleta
Jan 23 at 1:22