Reordering of matrix multiplicationProving associativity of matrix multiplicationIsolate yaw-pitch-roll from rotationMultiplication of Rotation Matrices in quaternionWhat exactly does a rotation preserve?Condensing successive matrix rotations into one matrixrotation about $x$ and $y$ axis on the Bloch sphereMatrix Representation of Rotation in $mathbbR^3$Confused about rotation matricesrotate the helix using the rotation (Rz and Rx) equationsGeometric image transforms using matrix multiplicationGet translational component of particular matrix multiplication
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Reordering of matrix multiplication
Proving associativity of matrix multiplicationIsolate yaw-pitch-roll from rotationMultiplication of Rotation Matrices in quaternionWhat exactly does a rotation preserve?Condensing successive matrix rotations into one matrixrotation about $x$ and $y$ axis on the Bloch sphereMatrix Representation of Rotation in $mathbbR^3$Confused about rotation matricesrotate the helix using the rotation (Rz and Rx) equationsGeometric image transforms using matrix multiplicationGet translational component of particular matrix multiplication
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$begingroup$
I want to compute the matrix multiplication $ABA$, where $A$ and $B$ are real and orthogonal matrices. In fact, they are specifically $3times3$ rotation matrices. However, it is much easier if I can reverse the order of $BA$ somehow, because I can then perform the multiplication much easier.
I know that matrix multiplication is not commutative, however, I am asking because both $A$ and $B$ are orthogonal matrices, and hopefully, may be there is some trick to utilize their orthogonality to reorder the product.
I tried to solve this, but got stuck here:
$$
ABA = A((BA)^-1)^-1=A(A^-1B^-1)^-1
$$
Is there a way to proceed from here?
Edit:
I also know that matrix multiplication is associative; however, I am not after associativity here. I want to multiply the $A$ matrix by $A$ (or by its inverse/transpose), then multiply the result by $B$.
Edit:
To put this into context, consider the following product of rotation matrices
$$
R_x(theta)R_z(pi)R_x(theta)
$$
where, $R_x(theta)$ is the rotation matrix about the $x$-axis by $theta$, and $R_z(pi)$ is the rotation matrix about the $z$-axis by $pi$.
This product simplifies to $R_z(pi)$. Is it possible to come to this conclusion without carrying out the matrix multiplication of the three rotation matrices? I looks that the $R_x(theta)$ got cancelled somehow.
linear-algebra matrices orthogonal-matrices
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
$endgroup$
add a comment
|
$begingroup$
I want to compute the matrix multiplication $ABA$, where $A$ and $B$ are real and orthogonal matrices. In fact, they are specifically $3times3$ rotation matrices. However, it is much easier if I can reverse the order of $BA$ somehow, because I can then perform the multiplication much easier.
I know that matrix multiplication is not commutative, however, I am asking because both $A$ and $B$ are orthogonal matrices, and hopefully, may be there is some trick to utilize their orthogonality to reorder the product.
I tried to solve this, but got stuck here:
$$
ABA = A((BA)^-1)^-1=A(A^-1B^-1)^-1
$$
Is there a way to proceed from here?
Edit:
I also know that matrix multiplication is associative; however, I am not after associativity here. I want to multiply the $A$ matrix by $A$ (or by its inverse/transpose), then multiply the result by $B$.
Edit:
To put this into context, consider the following product of rotation matrices
$$
R_x(theta)R_z(pi)R_x(theta)
$$
where, $R_x(theta)$ is the rotation matrix about the $x$-axis by $theta$, and $R_z(pi)$ is the rotation matrix about the $z$-axis by $pi$.
This product simplifies to $R_z(pi)$. Is it possible to come to this conclusion without carrying out the matrix multiplication of the three rotation matrices? I looks that the $R_x(theta)$ got cancelled somehow.
linear-algebra matrices orthogonal-matrices
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
$endgroup$
1
$begingroup$
How can the product $AB$ be easier then $BA$ ??? In any case, whatever the computation procedure, the result is $ABA$ in all cases !
$endgroup$
– Yves Daoust
Sep 14 at 12:53
$begingroup$
I did not say this, @Yves Daoust. Sorry for the misunderstanding. I said if I can reorder the $BA$, then the final product of the three matrices will be easier to compute.
$endgroup$
– user8396743
Sep 14 at 12:57
$begingroup$
How can it be if the result is the same ? There is no free lunch.
$endgroup$
– Yves Daoust
Sep 14 at 13:06
$begingroup$
@Yves Daoust please read the context example I put at the end. It may give you some insight of what I am looking for. Also, read David K 's answer to have more insight about the example.
$endgroup$
– user8396743
Sep 14 at 13:10
$begingroup$
The product of three axial rotations is easy, as those matrices have a majority of zeroes/one. And it perfectly illustrates my comment: the final result is the same, whatever the computation procedure. mathworld.wolfram.com/EulerAngles.html
$endgroup$
– Yves Daoust
Sep 14 at 13:15
add a comment
|
$begingroup$
I want to compute the matrix multiplication $ABA$, where $A$ and $B$ are real and orthogonal matrices. In fact, they are specifically $3times3$ rotation matrices. However, it is much easier if I can reverse the order of $BA$ somehow, because I can then perform the multiplication much easier.
I know that matrix multiplication is not commutative, however, I am asking because both $A$ and $B$ are orthogonal matrices, and hopefully, may be there is some trick to utilize their orthogonality to reorder the product.
I tried to solve this, but got stuck here:
$$
ABA = A((BA)^-1)^-1=A(A^-1B^-1)^-1
$$
Is there a way to proceed from here?
Edit:
I also know that matrix multiplication is associative; however, I am not after associativity here. I want to multiply the $A$ matrix by $A$ (or by its inverse/transpose), then multiply the result by $B$.
Edit:
To put this into context, consider the following product of rotation matrices
$$
R_x(theta)R_z(pi)R_x(theta)
$$
where, $R_x(theta)$ is the rotation matrix about the $x$-axis by $theta$, and $R_z(pi)$ is the rotation matrix about the $z$-axis by $pi$.
This product simplifies to $R_z(pi)$. Is it possible to come to this conclusion without carrying out the matrix multiplication of the three rotation matrices? I looks that the $R_x(theta)$ got cancelled somehow.
linear-algebra matrices orthogonal-matrices
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
$endgroup$
I want to compute the matrix multiplication $ABA$, where $A$ and $B$ are real and orthogonal matrices. In fact, they are specifically $3times3$ rotation matrices. However, it is much easier if I can reverse the order of $BA$ somehow, because I can then perform the multiplication much easier.
I know that matrix multiplication is not commutative, however, I am asking because both $A$ and $B$ are orthogonal matrices, and hopefully, may be there is some trick to utilize their orthogonality to reorder the product.
I tried to solve this, but got stuck here:
$$
ABA = A((BA)^-1)^-1=A(A^-1B^-1)^-1
$$
Is there a way to proceed from here?
Edit:
I also know that matrix multiplication is associative; however, I am not after associativity here. I want to multiply the $A$ matrix by $A$ (or by its inverse/transpose), then multiply the result by $B$.
Edit:
To put this into context, consider the following product of rotation matrices
$$
R_x(theta)R_z(pi)R_x(theta)
$$
where, $R_x(theta)$ is the rotation matrix about the $x$-axis by $theta$, and $R_z(pi)$ is the rotation matrix about the $z$-axis by $pi$.
This product simplifies to $R_z(pi)$. Is it possible to come to this conclusion without carrying out the matrix multiplication of the three rotation matrices? I looks that the $R_x(theta)$ got cancelled somehow.
linear-algebra matrices orthogonal-matrices
linear-algebra matrices orthogonal-matrices
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
edited Sep 14 at 12:47
user8396743
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
asked Sep 14 at 12:19
user8396743user8396743
635 bronze badges
635 bronze badges
1
$begingroup$
How can the product $AB$ be easier then $BA$ ??? In any case, whatever the computation procedure, the result is $ABA$ in all cases !
$endgroup$
– Yves Daoust
Sep 14 at 12:53
$begingroup$
I did not say this, @Yves Daoust. Sorry for the misunderstanding. I said if I can reorder the $BA$, then the final product of the three matrices will be easier to compute.
$endgroup$
– user8396743
Sep 14 at 12:57
$begingroup$
How can it be if the result is the same ? There is no free lunch.
$endgroup$
– Yves Daoust
Sep 14 at 13:06
$begingroup$
@Yves Daoust please read the context example I put at the end. It may give you some insight of what I am looking for. Also, read David K 's answer to have more insight about the example.
$endgroup$
– user8396743
Sep 14 at 13:10
$begingroup$
The product of three axial rotations is easy, as those matrices have a majority of zeroes/one. And it perfectly illustrates my comment: the final result is the same, whatever the computation procedure. mathworld.wolfram.com/EulerAngles.html
$endgroup$
– Yves Daoust
Sep 14 at 13:15
add a comment
|
1
$begingroup$
How can the product $AB$ be easier then $BA$ ??? In any case, whatever the computation procedure, the result is $ABA$ in all cases !
$endgroup$
– Yves Daoust
Sep 14 at 12:53
$begingroup$
I did not say this, @Yves Daoust. Sorry for the misunderstanding. I said if I can reorder the $BA$, then the final product of the three matrices will be easier to compute.
$endgroup$
– user8396743
Sep 14 at 12:57
$begingroup$
How can it be if the result is the same ? There is no free lunch.
$endgroup$
– Yves Daoust
Sep 14 at 13:06
$begingroup$
@Yves Daoust please read the context example I put at the end. It may give you some insight of what I am looking for. Also, read David K 's answer to have more insight about the example.
$endgroup$
– user8396743
Sep 14 at 13:10
$begingroup$
The product of three axial rotations is easy, as those matrices have a majority of zeroes/one. And it perfectly illustrates my comment: the final result is the same, whatever the computation procedure. mathworld.wolfram.com/EulerAngles.html
$endgroup$
– Yves Daoust
Sep 14 at 13:15
1
1
$begingroup$
How can the product $AB$ be easier then $BA$ ??? In any case, whatever the computation procedure, the result is $ABA$ in all cases !
$endgroup$
– Yves Daoust
Sep 14 at 12:53
$begingroup$
How can the product $AB$ be easier then $BA$ ??? In any case, whatever the computation procedure, the result is $ABA$ in all cases !
$endgroup$
– Yves Daoust
Sep 14 at 12:53
$begingroup$
I did not say this, @Yves Daoust. Sorry for the misunderstanding. I said if I can reorder the $BA$, then the final product of the three matrices will be easier to compute.
$endgroup$
– user8396743
Sep 14 at 12:57
$begingroup$
I did not say this, @Yves Daoust. Sorry for the misunderstanding. I said if I can reorder the $BA$, then the final product of the three matrices will be easier to compute.
$endgroup$
– user8396743
Sep 14 at 12:57
$begingroup$
How can it be if the result is the same ? There is no free lunch.
$endgroup$
– Yves Daoust
Sep 14 at 13:06
$begingroup$
How can it be if the result is the same ? There is no free lunch.
$endgroup$
– Yves Daoust
Sep 14 at 13:06
$begingroup$
@Yves Daoust please read the context example I put at the end. It may give you some insight of what I am looking for. Also, read David K 's answer to have more insight about the example.
$endgroup$
– user8396743
Sep 14 at 13:10
$begingroup$
@Yves Daoust please read the context example I put at the end. It may give you some insight of what I am looking for. Also, read David K 's answer to have more insight about the example.
$endgroup$
– user8396743
Sep 14 at 13:10
$begingroup$
The product of three axial rotations is easy, as those matrices have a majority of zeroes/one. And it perfectly illustrates my comment: the final result is the same, whatever the computation procedure. mathworld.wolfram.com/EulerAngles.html
$endgroup$
– Yves Daoust
Sep 14 at 13:15
$begingroup$
The product of three axial rotations is easy, as those matrices have a majority of zeroes/one. And it perfectly illustrates my comment: the final result is the same, whatever the computation procedure. mathworld.wolfram.com/EulerAngles.html
$endgroup$
– Yves Daoust
Sep 14 at 13:15
add a comment
|
3 Answers
3
active
oldest
votes
$begingroup$
Sadly, no. As @TobyMak points out, the associativity
$$
A(BA) = (AB)A
$$
means you can choose to evaluate either $AB$ or $BA$ but there's no trick that will allow you to work with $A^2$.
You can probably show that with an example in which $A^2 = I$.
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
$endgroup$
add a comment
|
$begingroup$
I take it that what you're trying to get at is that you have some matrix $A$ that you will use several times with several different other matrices in the place of $B.$
You would like some way to precompute a matrix $M_A$ by doing some operations on $A$
so that for any matrix $B$ you have $ABA = M_A B.$
First of all, if this is true in general it is true when $B=I,$ and therefore
$AIA = M_AI$ implies $M_A = A^2.$
But then for the formula to work in general you need $ABA = A^2 B$,
which implies $BA = AB,$ that is, the matrices have to commute.
In your particular example, the rotation $R_z(pi)$ maps the axis of $R_x(theta)$ onto itself, but reversed. Hence the initial rotation by $theta$ becomes a rotation by $-theta$. You will not get such a nice result for any $B$ that is not a rotation by a multiple of $pi$, and even when $B = R_z(pi)$ you will not get such a nice result for any non-trivial rotation around an axis that is not either the $z$ axis or an axis in the $x,y$ plane.
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
$endgroup$
$begingroup$
I might add that rotations in $mathbb R^3$ that have a common axis do commute.
$endgroup$
– amd
Sep 14 at 19:15
add a comment
|
$begingroup$
$$ABA = (AB)A$$
since matrix multiplication is associative. One proof of this is on this other post on Math SE.
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
$endgroup$
$begingroup$
Thanks for the answer, but this is actually not what I am after. By reordering I meant changing the order from $ABA$ to something like $AAB$ or $AA^-1B$ or something similar. Let me edit my question to reflect this fact.
$endgroup$
– user8396743
Sep 14 at 12:26
1
$begingroup$
That's fine. Could you include more context about the orthogonal matrices? I still find it hard to understand which particular forms you would like.
$endgroup$
– Toby Mak
Sep 14 at 12:29
$begingroup$
Sure, I'm on it.
$endgroup$
– user8396743
Sep 14 at 12:33
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
$begingroup$
Sadly, no. As @TobyMak points out, the associativity
$$
A(BA) = (AB)A
$$
means you can choose to evaluate either $AB$ or $BA$ but there's no trick that will allow you to work with $A^2$.
You can probably show that with an example in which $A^2 = I$.
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
$endgroup$
add a comment
|
$begingroup$
Sadly, no. As @TobyMak points out, the associativity
$$
A(BA) = (AB)A
$$
means you can choose to evaluate either $AB$ or $BA$ but there's no trick that will allow you to work with $A^2$.
You can probably show that with an example in which $A^2 = I$.
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
$endgroup$
add a comment
|
$begingroup$
Sadly, no. As @TobyMak points out, the associativity
$$
A(BA) = (AB)A
$$
means you can choose to evaluate either $AB$ or $BA$ but there's no trick that will allow you to work with $A^2$.
You can probably show that with an example in which $A^2 = I$.
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
$endgroup$
Sadly, no. As @TobyMak points out, the associativity
$$
A(BA) = (AB)A
$$
means you can choose to evaluate either $AB$ or $BA$ but there's no trick that will allow you to work with $A^2$.
You can probably show that with an example in which $A^2 = I$.
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
answered Sep 14 at 12:30
Ethan BolkerEthan Bolker
59.9k5 gold badges67 silver badges143 bronze badges
59.9k5 gold badges67 silver badges143 bronze badges
add a comment
|
add a comment
|
$begingroup$
I take it that what you're trying to get at is that you have some matrix $A$ that you will use several times with several different other matrices in the place of $B.$
You would like some way to precompute a matrix $M_A$ by doing some operations on $A$
so that for any matrix $B$ you have $ABA = M_A B.$
First of all, if this is true in general it is true when $B=I,$ and therefore
$AIA = M_AI$ implies $M_A = A^2.$
But then for the formula to work in general you need $ABA = A^2 B$,
which implies $BA = AB,$ that is, the matrices have to commute.
In your particular example, the rotation $R_z(pi)$ maps the axis of $R_x(theta)$ onto itself, but reversed. Hence the initial rotation by $theta$ becomes a rotation by $-theta$. You will not get such a nice result for any $B$ that is not a rotation by a multiple of $pi$, and even when $B = R_z(pi)$ you will not get such a nice result for any non-trivial rotation around an axis that is not either the $z$ axis or an axis in the $x,y$ plane.
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
$endgroup$
$begingroup$
I might add that rotations in $mathbb R^3$ that have a common axis do commute.
$endgroup$
– amd
Sep 14 at 19:15
add a comment
|
$begingroup$
I take it that what you're trying to get at is that you have some matrix $A$ that you will use several times with several different other matrices in the place of $B.$
You would like some way to precompute a matrix $M_A$ by doing some operations on $A$
so that for any matrix $B$ you have $ABA = M_A B.$
First of all, if this is true in general it is true when $B=I,$ and therefore
$AIA = M_AI$ implies $M_A = A^2.$
But then for the formula to work in general you need $ABA = A^2 B$,
which implies $BA = AB,$ that is, the matrices have to commute.
In your particular example, the rotation $R_z(pi)$ maps the axis of $R_x(theta)$ onto itself, but reversed. Hence the initial rotation by $theta$ becomes a rotation by $-theta$. You will not get such a nice result for any $B$ that is not a rotation by a multiple of $pi$, and even when $B = R_z(pi)$ you will not get such a nice result for any non-trivial rotation around an axis that is not either the $z$ axis or an axis in the $x,y$ plane.
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
$endgroup$
$begingroup$
I might add that rotations in $mathbb R^3$ that have a common axis do commute.
$endgroup$
– amd
Sep 14 at 19:15
add a comment
|
$begingroup$
I take it that what you're trying to get at is that you have some matrix $A$ that you will use several times with several different other matrices in the place of $B.$
You would like some way to precompute a matrix $M_A$ by doing some operations on $A$
so that for any matrix $B$ you have $ABA = M_A B.$
First of all, if this is true in general it is true when $B=I,$ and therefore
$AIA = M_AI$ implies $M_A = A^2.$
But then for the formula to work in general you need $ABA = A^2 B$,
which implies $BA = AB,$ that is, the matrices have to commute.
In your particular example, the rotation $R_z(pi)$ maps the axis of $R_x(theta)$ onto itself, but reversed. Hence the initial rotation by $theta$ becomes a rotation by $-theta$. You will not get such a nice result for any $B$ that is not a rotation by a multiple of $pi$, and even when $B = R_z(pi)$ you will not get such a nice result for any non-trivial rotation around an axis that is not either the $z$ axis or an axis in the $x,y$ plane.
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
$endgroup$
I take it that what you're trying to get at is that you have some matrix $A$ that you will use several times with several different other matrices in the place of $B.$
You would like some way to precompute a matrix $M_A$ by doing some operations on $A$
so that for any matrix $B$ you have $ABA = M_A B.$
First of all, if this is true in general it is true when $B=I,$ and therefore
$AIA = M_AI$ implies $M_A = A^2.$
But then for the formula to work in general you need $ABA = A^2 B$,
which implies $BA = AB,$ that is, the matrices have to commute.
In your particular example, the rotation $R_z(pi)$ maps the axis of $R_x(theta)$ onto itself, but reversed. Hence the initial rotation by $theta$ becomes a rotation by $-theta$. You will not get such a nice result for any $B$ that is not a rotation by a multiple of $pi$, and even when $B = R_z(pi)$ you will not get such a nice result for any non-trivial rotation around an axis that is not either the $z$ axis or an axis in the $x,y$ plane.
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
edited Sep 14 at 19:23
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
answered Sep 14 at 12:59
David KDavid K
62.1k4 gold badges49 silver badges141 bronze badges
62.1k4 gold badges49 silver badges141 bronze badges
$begingroup$
I might add that rotations in $mathbb R^3$ that have a common axis do commute.
$endgroup$
– amd
Sep 14 at 19:15
add a comment
|
$begingroup$
I might add that rotations in $mathbb R^3$ that have a common axis do commute.
$endgroup$
– amd
Sep 14 at 19:15
$begingroup$
I might add that rotations in $mathbb R^3$ that have a common axis do commute.
$endgroup$
– amd
Sep 14 at 19:15
$begingroup$
I might add that rotations in $mathbb R^3$ that have a common axis do commute.
$endgroup$
– amd
Sep 14 at 19:15
add a comment
|
$begingroup$
$$ABA = (AB)A$$
since matrix multiplication is associative. One proof of this is on this other post on Math SE.
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
$endgroup$
$begingroup$
Thanks for the answer, but this is actually not what I am after. By reordering I meant changing the order from $ABA$ to something like $AAB$ or $AA^-1B$ or something similar. Let me edit my question to reflect this fact.
$endgroup$
– user8396743
Sep 14 at 12:26
1
$begingroup$
That's fine. Could you include more context about the orthogonal matrices? I still find it hard to understand which particular forms you would like.
$endgroup$
– Toby Mak
Sep 14 at 12:29
$begingroup$
Sure, I'm on it.
$endgroup$
– user8396743
Sep 14 at 12:33
add a comment
|
$begingroup$
$$ABA = (AB)A$$
since matrix multiplication is associative. One proof of this is on this other post on Math SE.
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
$endgroup$
$begingroup$
Thanks for the answer, but this is actually not what I am after. By reordering I meant changing the order from $ABA$ to something like $AAB$ or $AA^-1B$ or something similar. Let me edit my question to reflect this fact.
$endgroup$
– user8396743
Sep 14 at 12:26
1
$begingroup$
That's fine. Could you include more context about the orthogonal matrices? I still find it hard to understand which particular forms you would like.
$endgroup$
– Toby Mak
Sep 14 at 12:29
$begingroup$
Sure, I'm on it.
$endgroup$
– user8396743
Sep 14 at 12:33
add a comment
|
$begingroup$
$$ABA = (AB)A$$
since matrix multiplication is associative. One proof of this is on this other post on Math SE.
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
$endgroup$
$$ABA = (AB)A$$
since matrix multiplication is associative. One proof of this is on this other post on Math SE.
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
answered Sep 14 at 12:22
Toby MakToby Mak
6,3412 gold badges15 silver badges30 bronze badges
6,3412 gold badges15 silver badges30 bronze badges
$begingroup$
Thanks for the answer, but this is actually not what I am after. By reordering I meant changing the order from $ABA$ to something like $AAB$ or $AA^-1B$ or something similar. Let me edit my question to reflect this fact.
$endgroup$
– user8396743
Sep 14 at 12:26
1
$begingroup$
That's fine. Could you include more context about the orthogonal matrices? I still find it hard to understand which particular forms you would like.
$endgroup$
– Toby Mak
Sep 14 at 12:29
$begingroup$
Sure, I'm on it.
$endgroup$
– user8396743
Sep 14 at 12:33
add a comment
|
$begingroup$
Thanks for the answer, but this is actually not what I am after. By reordering I meant changing the order from $ABA$ to something like $AAB$ or $AA^-1B$ or something similar. Let me edit my question to reflect this fact.
$endgroup$
– user8396743
Sep 14 at 12:26
1
$begingroup$
That's fine. Could you include more context about the orthogonal matrices? I still find it hard to understand which particular forms you would like.
$endgroup$
– Toby Mak
Sep 14 at 12:29
$begingroup$
Sure, I'm on it.
$endgroup$
– user8396743
Sep 14 at 12:33
$begingroup$
Thanks for the answer, but this is actually not what I am after. By reordering I meant changing the order from $ABA$ to something like $AAB$ or $AA^-1B$ or something similar. Let me edit my question to reflect this fact.
$endgroup$
– user8396743
Sep 14 at 12:26
$begingroup$
Thanks for the answer, but this is actually not what I am after. By reordering I meant changing the order from $ABA$ to something like $AAB$ or $AA^-1B$ or something similar. Let me edit my question to reflect this fact.
$endgroup$
– user8396743
Sep 14 at 12:26
1
1
$begingroup$
That's fine. Could you include more context about the orthogonal matrices? I still find it hard to understand which particular forms you would like.
$endgroup$
– Toby Mak
Sep 14 at 12:29
$begingroup$
That's fine. Could you include more context about the orthogonal matrices? I still find it hard to understand which particular forms you would like.
$endgroup$
– Toby Mak
Sep 14 at 12:29
$begingroup$
Sure, I'm on it.
$endgroup$
– user8396743
Sep 14 at 12:33
$begingroup$
Sure, I'm on it.
$endgroup$
– user8396743
Sep 14 at 12:33
add a comment
|
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How can the product $AB$ be easier then $BA$ ??? In any case, whatever the computation procedure, the result is $ABA$ in all cases !
$endgroup$
– Yves Daoust
Sep 14 at 12:53
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I did not say this, @Yves Daoust. Sorry for the misunderstanding. I said if I can reorder the $BA$, then the final product of the three matrices will be easier to compute.
$endgroup$
– user8396743
Sep 14 at 12:57
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How can it be if the result is the same ? There is no free lunch.
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– Yves Daoust
Sep 14 at 13:06
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@Yves Daoust please read the context example I put at the end. It may give you some insight of what I am looking for. Also, read David K 's answer to have more insight about the example.
$endgroup$
– user8396743
Sep 14 at 13:10
$begingroup$
The product of three axial rotations is easy, as those matrices have a majority of zeroes/one. And it perfectly illustrates my comment: the final result is the same, whatever the computation procedure. mathworld.wolfram.com/EulerAngles.html
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– Yves Daoust
Sep 14 at 13:15