Mixing
Linear operator and rotation matrices
$begingroup$
I have come across a question involving a linear operator $A$ that is represented by the following matrix:
$$
begin{pmatrix}
0 & 1 & 0\
1 & 0 & 0 \
0 & 0 & 2
end{pmatrix}
$$
This matrix represents the operator in $mathbb{R^3}$. Now if I need to find the matrix representation after a rotation around the z-axis by 90 degrees, I would have thought that the matrix would simply be $AR$ where A is the matrix above and $R$ is the matrix representing the rotation. However, the solution I was given was that the linear transformation $A'$ is given by $$ A' = R A R^{-1}$$. What is the explanation behind this?
linear-algebra linear-transformations rotations
edited Jan 5 at 18:34
David G. Stork
10.7k31332
asked Jan 5 at 18:33
daljit97daljit97
166111
$endgroup$
add a comment |
$begingroup$
I have come across a question involving a linear operator $A$ that is represented by the following matrix:
$$
begin{pmatrix}
0 & 1 & 0\
1 & 0 & 0 \
0 & 0 & 2
end{pmatrix}
$$
This matrix represents the operator in $mathbb{R^3}$. Now if I need to find the matrix representation after a rotation around the z-axis by 90 degrees, I would have thought that the matrix would simply be $AR$ where A is the matrix above and $R$ is the matrix representing the rotation. However, the solution I was given was that the linear transformation $A'$ is given by $$ A' = R A R^{-1}$$. What is the explanation behind this?
linear-algebra linear-transformations rotations
edited Jan 5 at 18:34
David G. Stork
10.7k31332
asked Jan 5 at 18:33
daljit97daljit97
166111
$endgroup$
$begingroup$
Do you know about similar matrices?
$endgroup$
– Thomas Shelby
Jan 5 at 18:36
$begingroup$
@ThomasShelby yes
$endgroup$
– daljit97
Jan 5 at 18:39
2
$begingroup$
I think you've misinterpreted the question. You are answering the question "What is the matrix of the transformation obtained by first performing a rotation of $90^circ$ and then the transformation represented by $A?$" But the intent of the question is, "What is the matrix of the operation represented by A if the coordinates are rotated by $90^circ?$"
$endgroup$
– saulspatz
Jan 5 at 18:41
add a comment |
$begingroup$
I have come across a question involving a linear operator $A$ that is represented by the following matrix:
$$
begin{pmatrix}
0 & 1 & 0\
1 & 0 & 0 \
0 & 0 & 2
end{pmatrix}
$$
This matrix represents the operator in $mathbb{R^3}$. Now if I need to find the matrix representation after a rotation around the z-axis by 90 degrees, I would have thought that the matrix would simply be $AR$ where A is the matrix above and $R$ is the matrix representing the rotation. However, the solution I was given was that the linear transformation $A'$ is given by $$ A' = R A R^{-1}$$. What is the explanation behind this?
linear-algebra linear-transformations rotations
edited Jan 5 at 18:34
David G. Stork
10.7k31332
asked Jan 5 at 18:33
daljit97daljit97
166111
$endgroup$
I have come across a question involving a linear operator $A$ that is represented by the following matrix:
$$
begin{pmatrix}
0 & 1 & 0\
1 & 0 & 0 \
0 & 0 & 2
end{pmatrix}
$$
This matrix represents the operator in $mathbb{R^3}$. Now if I need to find the matrix representation after a rotation around the z-axis by 90 degrees, I would have thought that the matrix would simply be $AR$ where A is the matrix above and $R$ is the matrix representing the rotation. However, the solution I was given was that the linear transformation $A'$ is given by $$ A' = R A R^{-1}$$. What is the explanation behind this?
linear-algebra linear-transformations rotations
linear-algebra linear-transformations rotations
edited Jan 5 at 18:34
David G. Stork
10.7k31332
asked Jan 5 at 18:33
daljit97daljit97
166111
edited Jan 5 at 18:34
David G. Stork
10.7k31332
asked Jan 5 at 18:33
daljit97daljit97
166111
edited Jan 5 at 18:34
David G. Stork
10.7k31332
edited Jan 5 at 18:34
David G. Stork
10.7k31332
edited Jan 5 at 18:34
David G. Stork
10.7k31332
10.7k31332
asked Jan 5 at 18:33
daljit97daljit97
166111
asked Jan 5 at 18:33
daljit97daljit97
166111
asked Jan 5 at 18:33
daljit97daljit97
166111
166111
$begingroup$
Do you know about similar matrices?
$endgroup$
– Thomas Shelby
Jan 5 at 18:36
$begingroup$
@ThomasShelby yes
$endgroup$
– daljit97
Jan 5 at 18:39
2
$begingroup$
I think you've misinterpreted the question. You are answering the question "What is the matrix of the transformation obtained by first performing a rotation of $90^circ$ and then the transformation represented by $A?$" But the intent of the question is, "What is the matrix of the operation represented by A if the coordinates are rotated by $90^circ?$"
$endgroup$
– saulspatz
Jan 5 at 18:41
add a comment |
$begingroup$
Do you know about similar matrices?
$endgroup$
– Thomas Shelby
Jan 5 at 18:36
$begingroup$
@ThomasShelby yes
$endgroup$
– daljit97
Jan 5 at 18:39
2
$begingroup$
I think you've misinterpreted the question. You are answering the question "What is the matrix of the transformation obtained by first performing a rotation of $90^circ$ and then the transformation represented by $A?$" But the intent of the question is, "What is the matrix of the operation represented by A if the coordinates are rotated by $90^circ?$"
$endgroup$
– saulspatz
Jan 5 at 18:41
$begingroup$
Do you know about similar matrices?
$endgroup$
– Thomas Shelby
Jan 5 at 18:36
$begingroup$
Do you know about similar matrices?
$endgroup$
– Thomas Shelby
Jan 5 at 18:36
$begingroup$
@ThomasShelby yes
$endgroup$
– daljit97
Jan 5 at 18:39
$begingroup$
@ThomasShelby yes
$endgroup$
– daljit97
Jan 5 at 18:39
2
2
$begingroup$
I think you've misinterpreted the question. You are answering the question "What is the matrix of the transformation obtained by first performing a rotation of $90^circ$ and then the transformation represented by $A?$" But the intent of the question is, "What is the matrix of the operation represented by A if the coordinates are rotated by $90^circ?$"
$endgroup$
– saulspatz
Jan 5 at 18:41
$begingroup$
I think you've misinterpreted the question. You are answering the question "What is the matrix of the transformation obtained by first performing a rotation of $90^circ$ and then the transformation represented by $A?$" But the intent of the question is, "What is the matrix of the operation represented by A if the coordinates are rotated by $90^circ?$"
$endgroup$
– saulspatz
Jan 5 at 18:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$AR$ rotates a vector around the z-axis by $90^circ$ and then applies the linear transformation $A$. However, that's not what they're asking for. They are asking for the representation of $A$ in a coordinate system that has been rotated by $90^circ$.
Now, without the rotation, understanding $A$ is rather simple: It turns $x$ into $Ax$. However, with the rotation, all of the vectors are now rotated around by $90^circ$. In terms of $R$, this means all of the vectors which were $z$ before the rotation are now $Rz$. Therefore, the representation of $A$ with this rotated coordinate system can be thought of as a transformation from $Rx$ to $RAx$.
If we refer to this representation $A$ as $A'$, then we have:
$$A'(Rx)=RAx text{ for all } xin Bbb{R}^3$$
Since this equation holds for all $xin Bbb{R}^3$, the matrices on both the left and right must be equal transformations, so we get:
$$A'R=RArightarrow A'=RAR^{-1}$$
Therefore, the representation of $A$ in a coordinate system that has been transformed by $R$ is $RAR^{-1}$.
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
Why is the transformation from $Rx rightarrow RAx$ and not $Rx rightarrow ARx$?
$endgroup$
– daljit97
Jan 6 at 21:35
$begingroup$
@daljit97 The original transformation $A$ is $xrightarrow Ax$. However, after rotating the coordinates, every vector gets multiplied by $R$ on the left, so the transformation becomes $Rxrightarrow RAx$.
$endgroup$
– Noble Mushtak
Jan 7 at 23:04
add a comment |
$begingroup$
I like to view this sort of thing in terms of input and output bases: the matrix of a linear transformation $T$ eats coordinate tuples of vectors that are expressed relative to some basis $mathcal B$ and spits out coordinate tuples in some, possibly different, basis $mathcal B'$. As pointed out by others, you’re not being asked to compute the composition of a rotation and the given transformation, but the express that transformation relative to a basis obtained by rotating the standard basis.
To put the task a slightly different way, you need to find a matrix that will eat and spit out vectors expressed relative to the rotated basis. You have a matrix that eats and spits out vectors in the standard basis, so a way to accomplish this is to first convert the vector to the standard basis, apply the matrix that you have, and then convert its output to the rotated basis. If $R$ is the matrix that converts coordinates from the standard basis to the rotated basis, then it should be clear that $R^{-1}$ performs the conversion in the other direction. Putting this all together, you get $RAR^{-1}$.
(You didn’t ask why its $RAR^{-1}$ instead of $R^{-1}AR$, which can also be a bit confusing. That’s covered in many places: look for discussions of passive vs. active transformations.)
answered Jan 5 at 21:05
amdamd
29.7k21050
$endgroup$
$begingroup$
That's a very helpful way of thinking about, but yeah my concern is about why $RAR^{-1}$ instead of $R^{-1} AR$. Is there any reference you could recommend for this topic?
$endgroup$
– daljit97
Jan 7 at 0:35
$begingroup$
@daljit97 That’s not what you asked in your question, though. There you asked why $RAR^{-1}$ instead of $AR$. For this new question, you first need to define what you mean by $R$ because either expression could be the correct one depending on what $R$ means.
$endgroup$
– amd
Jan 7 at 0:37
$begingroup$
Oh yes, I meant that I am now wondering that. As for $R$ in my question I meant a rotation matrix but more generally $R$ should be any matrix that represents a coordinate transformation (linear).
$endgroup$
– daljit97
Jan 7 at 0:41
$begingroup$
@daljit97 Yes, I understand that $R$ is a rotation matrix, but does it represent the transformation that the coordinate axes undergo or does it represent the corresponding change of coordinates of a point? That distinction is crucial.
$endgroup$
– amd
Jan 7 at 0:43
$begingroup$
The question I got it from used the following wording: "$R(e3, pi/2)$ is a matrix representing a rotation around the 3-axis through 90 degrees".
$endgroup$
– daljit97
Jan 7 at 0:46
add a comment |
$begingroup$
If a linear operator $A$ maps each vector from $Bbb R^n$ to $Bbb R^n$ and is invertible, then a rotation applied on $A$ resulting a new operator $A'$, should map the rotated vectors in $Bbb R^n$ to rotated counterparts in $Bbb R^n$, say, if $$u=Avquad,quad u,vin Bbb R^n$$then for some rotation matrix $R$ we should have $$Ru=A'(Rv)$$which means that if $A$ maps $v$ to $u$ bijectively, then similarly $A$ maps $Rv$ to $Ru$ bijectively. The above equation means that $$u=R^{-1}A'Rv=Av$$therefore $$R^{-1}A'R=A$$or equivalently $$A'=RAR^{-1}$$
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$AR$ rotates a vector around the z-axis by $90^circ$ and then applies the linear transformation $A$. However, that's not what they're asking for. They are asking for the representation of $A$ in a coordinate system that has been rotated by $90^circ$.
Now, without the rotation, understanding $A$ is rather simple: It turns $x$ into $Ax$. However, with the rotation, all of the vectors are now rotated around by $90^circ$. In terms of $R$, this means all of the vectors which were $z$ before the rotation are now $Rz$. Therefore, the representation of $A$ with this rotated coordinate system can be thought of as a transformation from $Rx$ to $RAx$.
If we refer to this representation $A$ as $A'$, then we have:
$$A'(Rx)=RAx text{ for all } xin Bbb{R}^3$$
Since this equation holds for all $xin Bbb{R}^3$, the matrices on both the left and right must be equal transformations, so we get:
$$A'R=RArightarrow A'=RAR^{-1}$$
Therefore, the representation of $A$ in a coordinate system that has been transformed by $R$ is $RAR^{-1}$.
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
Why is the transformation from $Rx rightarrow RAx$ and not $Rx rightarrow ARx$?
$endgroup$
– daljit97
Jan 6 at 21:35
$begingroup$
@daljit97 The original transformation $A$ is $xrightarrow Ax$. However, after rotating the coordinates, every vector gets multiplied by $R$ on the left, so the transformation becomes $Rxrightarrow RAx$.
$endgroup$
– Noble Mushtak
Jan 7 at 23:04
add a comment |
$begingroup$
$AR$ rotates a vector around the z-axis by $90^circ$ and then applies the linear transformation $A$. However, that's not what they're asking for. They are asking for the representation of $A$ in a coordinate system that has been rotated by $90^circ$.
Now, without the rotation, understanding $A$ is rather simple: It turns $x$ into $Ax$. However, with the rotation, all of the vectors are now rotated around by $90^circ$. In terms of $R$, this means all of the vectors which were $z$ before the rotation are now $Rz$. Therefore, the representation of $A$ with this rotated coordinate system can be thought of as a transformation from $Rx$ to $RAx$.
If we refer to this representation $A$ as $A'$, then we have:
$$A'(Rx)=RAx text{ for all } xin Bbb{R}^3$$
Since this equation holds for all $xin Bbb{R}^3$, the matrices on both the left and right must be equal transformations, so we get:
$$A'R=RArightarrow A'=RAR^{-1}$$
Therefore, the representation of $A$ in a coordinate system that has been transformed by $R$ is $RAR^{-1}$.
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
Why is the transformation from $Rx rightarrow RAx$ and not $Rx rightarrow ARx$?
$endgroup$
– daljit97
Jan 6 at 21:35
$begingroup$
@daljit97 The original transformation $A$ is $xrightarrow Ax$. However, after rotating the coordinates, every vector gets multiplied by $R$ on the left, so the transformation becomes $Rxrightarrow RAx$.
$endgroup$
– Noble Mushtak
Jan 7 at 23:04
add a comment |
$begingroup$
$AR$ rotates a vector around the z-axis by $90^circ$ and then applies the linear transformation $A$. However, that's not what they're asking for. They are asking for the representation of $A$ in a coordinate system that has been rotated by $90^circ$.
Now, without the rotation, understanding $A$ is rather simple: It turns $x$ into $Ax$. However, with the rotation, all of the vectors are now rotated around by $90^circ$. In terms of $R$, this means all of the vectors which were $z$ before the rotation are now $Rz$. Therefore, the representation of $A$ with this rotated coordinate system can be thought of as a transformation from $Rx$ to $RAx$.
If we refer to this representation $A$ as $A'$, then we have:
$$A'(Rx)=RAx text{ for all } xin Bbb{R}^3$$
Since this equation holds for all $xin Bbb{R}^3$, the matrices on both the left and right must be equal transformations, so we get:
$$A'R=RArightarrow A'=RAR^{-1}$$
Therefore, the representation of $A$ in a coordinate system that has been transformed by $R$ is $RAR^{-1}$.
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$AR$ rotates a vector around the z-axis by $90^circ$ and then applies the linear transformation $A$. However, that's not what they're asking for. They are asking for the representation of $A$ in a coordinate system that has been rotated by $90^circ$.
Now, without the rotation, understanding $A$ is rather simple: It turns $x$ into $Ax$. However, with the rotation, all of the vectors are now rotated around by $90^circ$. In terms of $R$, this means all of the vectors which were $z$ before the rotation are now $Rz$. Therefore, the representation of $A$ with this rotated coordinate system can be thought of as a transformation from $Rx$ to $RAx$.
If we refer to this representation $A$ as $A'$, then we have:
$$A'(Rx)=RAx text{ for all } xin Bbb{R}^3$$
Since this equation holds for all $xin Bbb{R}^3$, the matrices on both the left and right must be equal transformations, so we get:
$$A'R=RArightarrow A'=RAR^{-1}$$
Therefore, the representation of $A$ in a coordinate system that has been transformed by $R$ is $RAR^{-1}$.
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 5 at 18:42
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
$begingroup$
Why is the transformation from $Rx rightarrow RAx$ and not $Rx rightarrow ARx$?
$endgroup$
– daljit97
Jan 6 at 21:35
$begingroup$
@daljit97 The original transformation $A$ is $xrightarrow Ax$. However, after rotating the coordinates, every vector gets multiplied by $R$ on the left, so the transformation becomes $Rxrightarrow RAx$.
$endgroup$
– Noble Mushtak
Jan 7 at 23:04
add a comment |
$begingroup$
Why is the transformation from $Rx rightarrow RAx$ and not $Rx rightarrow ARx$?
$endgroup$
– daljit97
Jan 6 at 21:35
$begingroup$
@daljit97 The original transformation $A$ is $xrightarrow Ax$. However, after rotating the coordinates, every vector gets multiplied by $R$ on the left, so the transformation becomes $Rxrightarrow RAx$.
$endgroup$
– Noble Mushtak
Jan 7 at 23:04
$begingroup$
Why is the transformation from $Rx rightarrow RAx$ and not $Rx rightarrow ARx$?
$endgroup$
– daljit97
Jan 6 at 21:35
$begingroup$
Why is the transformation from $Rx rightarrow RAx$ and not $Rx rightarrow ARx$?
$endgroup$
– daljit97
Jan 6 at 21:35
$begingroup$
@daljit97 The original transformation $A$ is $xrightarrow Ax$. However, after rotating the coordinates, every vector gets multiplied by $R$ on the left, so the transformation becomes $Rxrightarrow RAx$.
$endgroup$
– Noble Mushtak
Jan 7 at 23:04
$begingroup$
@daljit97 The original transformation $A$ is $xrightarrow Ax$. However, after rotating the coordinates, every vector gets multiplied by $R$ on the left, so the transformation becomes $Rxrightarrow RAx$.
$endgroup$
– Noble Mushtak
Jan 7 at 23:04
add a comment |
$begingroup$
I like to view this sort of thing in terms of input and output bases: the matrix of a linear transformation $T$ eats coordinate tuples of vectors that are expressed relative to some basis $mathcal B$ and spits out coordinate tuples in some, possibly different, basis $mathcal B'$. As pointed out by others, you’re not being asked to compute the composition of a rotation and the given transformation, but the express that transformation relative to a basis obtained by rotating the standard basis.
To put the task a slightly different way, you need to find a matrix that will eat and spit out vectors expressed relative to the rotated basis. You have a matrix that eats and spits out vectors in the standard basis, so a way to accomplish this is to first convert the vector to the standard basis, apply the matrix that you have, and then convert its output to the rotated basis. If $R$ is the matrix that converts coordinates from the standard basis to the rotated basis, then it should be clear that $R^{-1}$ performs the conversion in the other direction. Putting this all together, you get $RAR^{-1}$.
(You didn’t ask why its $RAR^{-1}$ instead of $R^{-1}AR$, which can also be a bit confusing. That’s covered in many places: look for discussions of passive vs. active transformations.)
answered Jan 5 at 21:05
amdamd
29.7k21050
$endgroup$
$begingroup$
That's a very helpful way of thinking about, but yeah my concern is about why $RAR^{-1}$ instead of $R^{-1} AR$. Is there any reference you could recommend for this topic?
$endgroup$
– daljit97
Jan 7 at 0:35
$begingroup$
@daljit97 That’s not what you asked in your question, though. There you asked why $RAR^{-1}$ instead of $AR$. For this new question, you first need to define what you mean by $R$ because either expression could be the correct one depending on what $R$ means.
$endgroup$
– amd
Jan 7 at 0:37
$begingroup$
Oh yes, I meant that I am now wondering that. As for $R$ in my question I meant a rotation matrix but more generally $R$ should be any matrix that represents a coordinate transformation (linear).
$endgroup$
– daljit97
Jan 7 at 0:41
$begingroup$
@daljit97 Yes, I understand that $R$ is a rotation matrix, but does it represent the transformation that the coordinate axes undergo or does it represent the corresponding change of coordinates of a point? That distinction is crucial.
$endgroup$
– amd
Jan 7 at 0:43
$begingroup$
The question I got it from used the following wording: "$R(e3, pi/2)$ is a matrix representing a rotation around the 3-axis through 90 degrees".
$endgroup$
– daljit97
Jan 7 at 0:46
add a comment |
$begingroup$
I like to view this sort of thing in terms of input and output bases: the matrix of a linear transformation $T$ eats coordinate tuples of vectors that are expressed relative to some basis $mathcal B$ and spits out coordinate tuples in some, possibly different, basis $mathcal B'$. As pointed out by others, you’re not being asked to compute the composition of a rotation and the given transformation, but the express that transformation relative to a basis obtained by rotating the standard basis.
To put the task a slightly different way, you need to find a matrix that will eat and spit out vectors expressed relative to the rotated basis. You have a matrix that eats and spits out vectors in the standard basis, so a way to accomplish this is to first convert the vector to the standard basis, apply the matrix that you have, and then convert its output to the rotated basis. If $R$ is the matrix that converts coordinates from the standard basis to the rotated basis, then it should be clear that $R^{-1}$ performs the conversion in the other direction. Putting this all together, you get $RAR^{-1}$.
(You didn’t ask why its $RAR^{-1}$ instead of $R^{-1}AR$, which can also be a bit confusing. That’s covered in many places: look for discussions of passive vs. active transformations.)
answered Jan 5 at 21:05
amdamd
29.7k21050
$endgroup$
$begingroup$
That's a very helpful way of thinking about, but yeah my concern is about why $RAR^{-1}$ instead of $R^{-1} AR$. Is there any reference you could recommend for this topic?
$endgroup$
– daljit97
Jan 7 at 0:35
$begingroup$
@daljit97 That’s not what you asked in your question, though. There you asked why $RAR^{-1}$ instead of $AR$. For this new question, you first need to define what you mean by $R$ because either expression could be the correct one depending on what $R$ means.
$endgroup$
– amd
Jan 7 at 0:37
$begingroup$
Oh yes, I meant that I am now wondering that. As for $R$ in my question I meant a rotation matrix but more generally $R$ should be any matrix that represents a coordinate transformation (linear).
$endgroup$
– daljit97
Jan 7 at 0:41
$begingroup$
@daljit97 Yes, I understand that $R$ is a rotation matrix, but does it represent the transformation that the coordinate axes undergo or does it represent the corresponding change of coordinates of a point? That distinction is crucial.
$endgroup$
– amd
Jan 7 at 0:43
$begingroup$
The question I got it from used the following wording: "$R(e3, pi/2)$ is a matrix representing a rotation around the 3-axis through 90 degrees".
$endgroup$
– daljit97
Jan 7 at 0:46
add a comment |
$begingroup$
I like to view this sort of thing in terms of input and output bases: the matrix of a linear transformation $T$ eats coordinate tuples of vectors that are expressed relative to some basis $mathcal B$ and spits out coordinate tuples in some, possibly different, basis $mathcal B'$. As pointed out by others, you’re not being asked to compute the composition of a rotation and the given transformation, but the express that transformation relative to a basis obtained by rotating the standard basis.
To put the task a slightly different way, you need to find a matrix that will eat and spit out vectors expressed relative to the rotated basis. You have a matrix that eats and spits out vectors in the standard basis, so a way to accomplish this is to first convert the vector to the standard basis, apply the matrix that you have, and then convert its output to the rotated basis. If $R$ is the matrix that converts coordinates from the standard basis to the rotated basis, then it should be clear that $R^{-1}$ performs the conversion in the other direction. Putting this all together, you get $RAR^{-1}$.
(You didn’t ask why its $RAR^{-1}$ instead of $R^{-1}AR$, which can also be a bit confusing. That’s covered in many places: look for discussions of passive vs. active transformations.)
answered Jan 5 at 21:05
amdamd
29.7k21050
$endgroup$
I like to view this sort of thing in terms of input and output bases: the matrix of a linear transformation $T$ eats coordinate tuples of vectors that are expressed relative to some basis $mathcal B$ and spits out coordinate tuples in some, possibly different, basis $mathcal B'$. As pointed out by others, you’re not being asked to compute the composition of a rotation and the given transformation, but the express that transformation relative to a basis obtained by rotating the standard basis.
To put the task a slightly different way, you need to find a matrix that will eat and spit out vectors expressed relative to the rotated basis. You have a matrix that eats and spits out vectors in the standard basis, so a way to accomplish this is to first convert the vector to the standard basis, apply the matrix that you have, and then convert its output to the rotated basis. If $R$ is the matrix that converts coordinates from the standard basis to the rotated basis, then it should be clear that $R^{-1}$ performs the conversion in the other direction. Putting this all together, you get $RAR^{-1}$.
(You didn’t ask why its $RAR^{-1}$ instead of $R^{-1}AR$, which can also be a bit confusing. That’s covered in many places: look for discussions of passive vs. active transformations.)
answered Jan 5 at 21:05
amdamd
29.7k21050
answered Jan 5 at 21:05
amdamd
29.7k21050
answered Jan 5 at 21:05
amdamd
29.7k21050
answered Jan 5 at 21:05
amdamd
29.7k21050
29.7k21050
$begingroup$
That's a very helpful way of thinking about, but yeah my concern is about why $RAR^{-1}$ instead of $R^{-1} AR$. Is there any reference you could recommend for this topic?
$endgroup$
– daljit97
Jan 7 at 0:35
$begingroup$
@daljit97 That’s not what you asked in your question, though. There you asked why $RAR^{-1}$ instead of $AR$. For this new question, you first need to define what you mean by $R$ because either expression could be the correct one depending on what $R$ means.
$endgroup$
– amd
Jan 7 at 0:37
$begingroup$
Oh yes, I meant that I am now wondering that. As for $R$ in my question I meant a rotation matrix but more generally $R$ should be any matrix that represents a coordinate transformation (linear).
$endgroup$
– daljit97
Jan 7 at 0:41
$begingroup$
@daljit97 Yes, I understand that $R$ is a rotation matrix, but does it represent the transformation that the coordinate axes undergo or does it represent the corresponding change of coordinates of a point? That distinction is crucial.
$endgroup$
– amd
Jan 7 at 0:43
$begingroup$
The question I got it from used the following wording: "$R(e3, pi/2)$ is a matrix representing a rotation around the 3-axis through 90 degrees".
$endgroup$
– daljit97
Jan 7 at 0:46
add a comment |
$begingroup$
That's a very helpful way of thinking about, but yeah my concern is about why $RAR^{-1}$ instead of $R^{-1} AR$. Is there any reference you could recommend for this topic?
$endgroup$
– daljit97
Jan 7 at 0:35
$begingroup$
@daljit97 That’s not what you asked in your question, though. There you asked why $RAR^{-1}$ instead of $AR$. For this new question, you first need to define what you mean by $R$ because either expression could be the correct one depending on what $R$ means.
$endgroup$
– amd
Jan 7 at 0:37
$begingroup$
Oh yes, I meant that I am now wondering that. As for $R$ in my question I meant a rotation matrix but more generally $R$ should be any matrix that represents a coordinate transformation (linear).
$endgroup$
– daljit97
Jan 7 at 0:41
$begingroup$
@daljit97 Yes, I understand that $R$ is a rotation matrix, but does it represent the transformation that the coordinate axes undergo or does it represent the corresponding change of coordinates of a point? That distinction is crucial.
$endgroup$
– amd
Jan 7 at 0:43
$begingroup$
The question I got it from used the following wording: "$R(e3, pi/2)$ is a matrix representing a rotation around the 3-axis through 90 degrees".
$endgroup$
– daljit97
Jan 7 at 0:46
$begingroup$
That's a very helpful way of thinking about, but yeah my concern is about why $RAR^{-1}$ instead of $R^{-1} AR$. Is there any reference you could recommend for this topic?
$endgroup$
– daljit97
Jan 7 at 0:35
$begingroup$
That's a very helpful way of thinking about, but yeah my concern is about why $RAR^{-1}$ instead of $R^{-1} AR$. Is there any reference you could recommend for this topic?
$endgroup$
– daljit97
Jan 7 at 0:35
$begingroup$
@daljit97 That’s not what you asked in your question, though. There you asked why $RAR^{-1}$ instead of $AR$. For this new question, you first need to define what you mean by $R$ because either expression could be the correct one depending on what $R$ means.
$endgroup$
– amd
Jan 7 at 0:37
$begingroup$
@daljit97 That’s not what you asked in your question, though. There you asked why $RAR^{-1}$ instead of $AR$. For this new question, you first need to define what you mean by $R$ because either expression could be the correct one depending on what $R$ means.
$endgroup$
– amd
Jan 7 at 0:37
$begingroup$
Oh yes, I meant that I am now wondering that. As for $R$ in my question I meant a rotation matrix but more generally $R$ should be any matrix that represents a coordinate transformation (linear).
$endgroup$
– daljit97
Jan 7 at 0:41
$begingroup$
Oh yes, I meant that I am now wondering that. As for $R$ in my question I meant a rotation matrix but more generally $R$ should be any matrix that represents a coordinate transformation (linear).
$endgroup$
– daljit97
Jan 7 at 0:41
$begingroup$
@daljit97 Yes, I understand that $R$ is a rotation matrix, but does it represent the transformation that the coordinate axes undergo or does it represent the corresponding change of coordinates of a point? That distinction is crucial.
$endgroup$
– amd
Jan 7 at 0:43
$begingroup$
@daljit97 Yes, I understand that $R$ is a rotation matrix, but does it represent the transformation that the coordinate axes undergo or does it represent the corresponding change of coordinates of a point? That distinction is crucial.
$endgroup$
– amd
Jan 7 at 0:43
$begingroup$
The question I got it from used the following wording: "$R(e3, pi/2)$ is a matrix representing a rotation around the 3-axis through 90 degrees".
$endgroup$
– daljit97
Jan 7 at 0:46
$begingroup$
The question I got it from used the following wording: "$R(e3, pi/2)$ is a matrix representing a rotation around the 3-axis through 90 degrees".
$endgroup$
– daljit97
Jan 7 at 0:46
add a comment |
$begingroup$
If a linear operator $A$ maps each vector from $Bbb R^n$ to $Bbb R^n$ and is invertible, then a rotation applied on $A$ resulting a new operator $A'$, should map the rotated vectors in $Bbb R^n$ to rotated counterparts in $Bbb R^n$, say, if $$u=Avquad,quad u,vin Bbb R^n$$then for some rotation matrix $R$ we should have $$Ru=A'(Rv)$$which means that if $A$ maps $v$ to $u$ bijectively, then similarly $A$ maps $Rv$ to $Ru$ bijectively. The above equation means that $$u=R^{-1}A'Rv=Av$$therefore $$R^{-1}A'R=A$$or equivalently $$A'=RAR^{-1}$$
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
If a linear operator $A$ maps each vector from $Bbb R^n$ to $Bbb R^n$ and is invertible, then a rotation applied on $A$ resulting a new operator $A'$, should map the rotated vectors in $Bbb R^n$ to rotated counterparts in $Bbb R^n$, say, if $$u=Avquad,quad u,vin Bbb R^n$$then for some rotation matrix $R$ we should have $$Ru=A'(Rv)$$which means that if $A$ maps $v$ to $u$ bijectively, then similarly $A$ maps $Rv$ to $Ru$ bijectively. The above equation means that $$u=R^{-1}A'Rv=Av$$therefore $$R^{-1}A'R=A$$or equivalently $$A'=RAR^{-1}$$
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
If a linear operator $A$ maps each vector from $Bbb R^n$ to $Bbb R^n$ and is invertible, then a rotation applied on $A$ resulting a new operator $A'$, should map the rotated vectors in $Bbb R^n$ to rotated counterparts in $Bbb R^n$, say, if $$u=Avquad,quad u,vin Bbb R^n$$then for some rotation matrix $R$ we should have $$Ru=A'(Rv)$$which means that if $A$ maps $v$ to $u$ bijectively, then similarly $A$ maps $Rv$ to $Ru$ bijectively. The above equation means that $$u=R^{-1}A'Rv=Av$$therefore $$R^{-1}A'R=A$$or equivalently $$A'=RAR^{-1}$$
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
If a linear operator $A$ maps each vector from $Bbb R^n$ to $Bbb R^n$ and is invertible, then a rotation applied on $A$ resulting a new operator $A'$, should map the rotated vectors in $Bbb R^n$ to rotated counterparts in $Bbb R^n$, say, if $$u=Avquad,quad u,vin Bbb R^n$$then for some rotation matrix $R$ we should have $$Ru=A'(Rv)$$which means that if $A$ maps $v$ to $u$ bijectively, then similarly $A$ maps $Rv$ to $Ru$ bijectively. The above equation means that $$u=R^{-1}A'Rv=Av$$therefore $$R^{-1}A'R=A$$or equivalently $$A'=RAR^{-1}$$
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 5 at 18:43
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
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$begingroup$
Do you know about similar matrices?
$endgroup$
– Thomas Shelby
Jan 5 at 18:36
$begingroup$
@ThomasShelby yes
$endgroup$
– daljit97
Jan 5 at 18:39
2
$begingroup$
I think you've misinterpreted the question. You are answering the question "What is the matrix of the transformation obtained by first performing a rotation of $90^circ$ and then the transformation represented by $A?$" But the intent of the question is, "What is the matrix of the operation represented by A if the coordinates are rotated by $90^circ?$"
$endgroup$
– saulspatz
Jan 5 at 18:41