The difference between applying a rotation matrix to a vector (points) and to a matrix (transformation)
Suppose that the rotation matrix is defined as $mathbf{R}$. Then in order to rotate a vector and a matrix, the following expressions are, respectively, used
$mathbf{u'}=mathbf{R} mathbf{u}$
and
$mathbf{U'}=mathbf{R} mathbf{U} mathbf{R}^T$,
where $mathbf{u}$ and $mathbf{U}$ are, respectively, an arbitrary vector and an arbitrary matrix.
For me, the first one is obvious since you simply multiply the rotation matrix by the vector (for example a point coordinate in 3D) and obtain the rotated vector (rotated point coordinate in 3D). However, the second one is not clear for me and why the rotation should be multiplied from both sides and how this expression is derived.
P.S. The matrix $mathbf{U}$ can be interpreted as a stretch matrix in 3D.
linear-algebra matrices vectors rotations
add a comment |
Suppose that the rotation matrix is defined as $mathbf{R}$. Then in order to rotate a vector and a matrix, the following expressions are, respectively, used
$mathbf{u'}=mathbf{R} mathbf{u}$
and
$mathbf{U'}=mathbf{R} mathbf{U} mathbf{R}^T$,
where $mathbf{u}$ and $mathbf{U}$ are, respectively, an arbitrary vector and an arbitrary matrix.
For me, the first one is obvious since you simply multiply the rotation matrix by the vector (for example a point coordinate in 3D) and obtain the rotated vector (rotated point coordinate in 3D). However, the second one is not clear for me and why the rotation should be multiplied from both sides and how this expression is derived.
P.S. The matrix $mathbf{U}$ can be interpreted as a stretch matrix in 3D.
linear-algebra matrices vectors rotations
If you have a linear algebra textbook on hand, you might find it instructive to read about changes of basis. The matrix $R$, in this context, can be nicely thought of as a change-of-basis matrix.
– Omnomnomnom
yesterday
1
The expression $RU$ also is a kind of "rotation of $U$," namely, it takes whatever transformation $U$ was going to perform and composes that with a subsequent rotation represented by $R.$ But this is actually giving us a new transformation within the old coordinate system, not rewriting the old transformation in a new coordinate system as we would get from $RUR^T.$
– David K
yesterday
When you say "$U$ is a matrix", it makes a difference whether $U$ is merely a set of points (in which case $RU$ is already the "rotation of $U$"), or itself a transformation matrix (as in this case). Can you please edit your title/body to be more clear?
– smci
yesterday
@smci Look at the P.S. note. I have mentioned that U is a stretch matrix. not a set of points.
– Msen Rezaee
21 hours ago
Yes that's more clear now
– smci
20 hours ago
add a comment |
Suppose that the rotation matrix is defined as $mathbf{R}$. Then in order to rotate a vector and a matrix, the following expressions are, respectively, used
$mathbf{u'}=mathbf{R} mathbf{u}$
and
$mathbf{U'}=mathbf{R} mathbf{U} mathbf{R}^T$,
where $mathbf{u}$ and $mathbf{U}$ are, respectively, an arbitrary vector and an arbitrary matrix.
For me, the first one is obvious since you simply multiply the rotation matrix by the vector (for example a point coordinate in 3D) and obtain the rotated vector (rotated point coordinate in 3D). However, the second one is not clear for me and why the rotation should be multiplied from both sides and how this expression is derived.
P.S. The matrix $mathbf{U}$ can be interpreted as a stretch matrix in 3D.
linear-algebra matrices vectors rotations
Suppose that the rotation matrix is defined as $mathbf{R}$. Then in order to rotate a vector and a matrix, the following expressions are, respectively, used
$mathbf{u'}=mathbf{R} mathbf{u}$
and
$mathbf{U'}=mathbf{R} mathbf{U} mathbf{R}^T$,
where $mathbf{u}$ and $mathbf{U}$ are, respectively, an arbitrary vector and an arbitrary matrix.
For me, the first one is obvious since you simply multiply the rotation matrix by the vector (for example a point coordinate in 3D) and obtain the rotated vector (rotated point coordinate in 3D). However, the second one is not clear for me and why the rotation should be multiplied from both sides and how this expression is derived.
P.S. The matrix $mathbf{U}$ can be interpreted as a stretch matrix in 3D.
linear-algebra matrices vectors rotations
linear-algebra matrices vectors rotations
edited 21 hours ago
asked yesterday
Msen Rezaee
306312
306312
If you have a linear algebra textbook on hand, you might find it instructive to read about changes of basis. The matrix $R$, in this context, can be nicely thought of as a change-of-basis matrix.
– Omnomnomnom
yesterday
1
The expression $RU$ also is a kind of "rotation of $U$," namely, it takes whatever transformation $U$ was going to perform and composes that with a subsequent rotation represented by $R.$ But this is actually giving us a new transformation within the old coordinate system, not rewriting the old transformation in a new coordinate system as we would get from $RUR^T.$
– David K
yesterday
When you say "$U$ is a matrix", it makes a difference whether $U$ is merely a set of points (in which case $RU$ is already the "rotation of $U$"), or itself a transformation matrix (as in this case). Can you please edit your title/body to be more clear?
– smci
yesterday
@smci Look at the P.S. note. I have mentioned that U is a stretch matrix. not a set of points.
– Msen Rezaee
21 hours ago
Yes that's more clear now
– smci
20 hours ago
add a comment |
If you have a linear algebra textbook on hand, you might find it instructive to read about changes of basis. The matrix $R$, in this context, can be nicely thought of as a change-of-basis matrix.
– Omnomnomnom
yesterday
1
The expression $RU$ also is a kind of "rotation of $U$," namely, it takes whatever transformation $U$ was going to perform and composes that with a subsequent rotation represented by $R.$ But this is actually giving us a new transformation within the old coordinate system, not rewriting the old transformation in a new coordinate system as we would get from $RUR^T.$
– David K
yesterday
When you say "$U$ is a matrix", it makes a difference whether $U$ is merely a set of points (in which case $RU$ is already the "rotation of $U$"), or itself a transformation matrix (as in this case). Can you please edit your title/body to be more clear?
– smci
yesterday
@smci Look at the P.S. note. I have mentioned that U is a stretch matrix. not a set of points.
– Msen Rezaee
21 hours ago
Yes that's more clear now
– smci
20 hours ago
If you have a linear algebra textbook on hand, you might find it instructive to read about changes of basis. The matrix $R$, in this context, can be nicely thought of as a change-of-basis matrix.
– Omnomnomnom
yesterday
If you have a linear algebra textbook on hand, you might find it instructive to read about changes of basis. The matrix $R$, in this context, can be nicely thought of as a change-of-basis matrix.
– Omnomnomnom
yesterday
1
1
The expression $RU$ also is a kind of "rotation of $U$," namely, it takes whatever transformation $U$ was going to perform and composes that with a subsequent rotation represented by $R.$ But this is actually giving us a new transformation within the old coordinate system, not rewriting the old transformation in a new coordinate system as we would get from $RUR^T.$
– David K
yesterday
The expression $RU$ also is a kind of "rotation of $U$," namely, it takes whatever transformation $U$ was going to perform and composes that with a subsequent rotation represented by $R.$ But this is actually giving us a new transformation within the old coordinate system, not rewriting the old transformation in a new coordinate system as we would get from $RUR^T.$
– David K
yesterday
When you say "$U$ is a matrix", it makes a difference whether $U$ is merely a set of points (in which case $RU$ is already the "rotation of $U$"), or itself a transformation matrix (as in this case). Can you please edit your title/body to be more clear?
– smci
yesterday
When you say "$U$ is a matrix", it makes a difference whether $U$ is merely a set of points (in which case $RU$ is already the "rotation of $U$"), or itself a transformation matrix (as in this case). Can you please edit your title/body to be more clear?
– smci
yesterday
@smci Look at the P.S. note. I have mentioned that U is a stretch matrix. not a set of points.
– Msen Rezaee
21 hours ago
@smci Look at the P.S. note. I have mentioned that U is a stretch matrix. not a set of points.
– Msen Rezaee
21 hours ago
Yes that's more clear now
– smci
20 hours ago
Yes that's more clear now
– smci
20 hours ago
add a comment |
4 Answers
4
active
oldest
votes
Here the matrix $U$ is considered not as a bunch of column vectors, but as a (matrix of the) linear map $Fcolon {Bbb R}^nto {Bbb R}^n$
$$
y=F(x)=Ux.
$$
What happens if we rotate both $y$ and $x$ by $R$? We get (since $R^TR=I$ for rotations)
$$
y=UxquadRightarrowquad Ry=RUxquadRightarrowquad Ry=underbrace{RUR^T}_{U'}RxquadRightarrowquad y'=U'x'.
$$
Thus the matrix $U'=RUR^T$ corresponds to the same linear map $F$ in the new coordinates after rotation ($x'mapsto y'$).
In general, for any change of the basis $x'=Sx$, $y'=Sy$ the corresponding change of the matrix $U$ is
$$
Sy=underbrace{SUS^{-1}}_{U'}SxquadRightarrowquad y'=U'x'.
$$
It means that the class of all similar matrices ${SUS^{-1}colon Stext{ invertible}}$ is exactly the class of all matrices that describe the same linear map in different bases.
add a comment |
Using your example where $U$ is a stretch matrix in 3D, if you want to "rotate" this matrix, you essentially want this stretch action to occur in a different direction / axis. Suppose you have some shape aligned to this new axis. You want to know what the $U'$ is that stretches the shape parallel to this axis. To do this, you use $R^T$ to rotate everything back to the original orientation. Then you do the original stretch transformation $U$. Then you rotate this back using $R$. So $U'=RUR^T$.
Thanks for your answer. But could you please give me a clearer example. I understood your point. However, I couldn't understand your example.
– Msen Rezaee
yesterday
@MsenRezaee I added a picture to help illustrate the point. It is in 2D rather than 3D, but it should be simple to see how this generalises
– John Doe
yesterday
add a comment |
One thing that may be instructive is to recall that every matrix can be represented as the linear combination of a series of dyadic/outer products between two vectors, $U = sum_i a_i otimes b_i = sum_i a_i b_i^T$ where $a_i$ and $b_i$ are a sequence of column vectors.
When changing the basis of the matrix, we are in effect applying the vector rule for changing bases to both sequences of vectors:
$$U’ = sum_i a_i’ otimes b_i’ = sum_i Ra_i (Rb_i)^T = sum_iRa_ib_i^T R^T = RUR^T$$
Hope this helps!
2
This is not language I'm familiar with, but wouldn't $ab^T$ always be rank $1$, so only rank $1$ matrices can be represented this way?
– Callus
yesterday
2
@Callus I guess it was meant that any matrix could be represented as a span of dyadic matrices.
– A.Γ.
yesterday
Correct, thank you for spotting that @Callus and @A.Γ.! Will update answer shortly.
– aghostinthefigures
yesterday
add a comment |
The columns of $mathbf{U}$ tell you what happens to the coordinate vectors $hat{e}_1,hat{e}_2,hat{e}_3$. For example, if the first column is $[a,b,c]^T$, then $mathbf{U}hat{e}_1 = ahat{e}_1 + bhat{e}_2 + chat{e}_3$.
The matrix $mathbf{U}'$ is the matrix that behaves the same way on the rotated coordinate basis $hat{f}_i = mathbf{R}hat{e}_i$. This is because $mathbf{R}^T = mathbf{R}^{-1}$ so for example
$$begin{align*}
mathbf{U}'hat{f}_1 &= mathbf{R}mathbf{U}mathbf{R}^Tmathbf{R}hat{e}_1 \
&= mathbf{R}left(mathbf{U}hat{e}_1right) \
&= mathbf{R}(ahat{e}_1 + bhat{e}_2 + chat{e}_3) \
&= amathbf{R}hat{e}_1 + bmathbf{R}hat{e}_2 + cmathbf{R}hat{e}_3 \
&= ahat{f}_1 + bhat{f}_2 + chat{f}_3
end{align*}
$$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here the matrix $U$ is considered not as a bunch of column vectors, but as a (matrix of the) linear map $Fcolon {Bbb R}^nto {Bbb R}^n$
$$
y=F(x)=Ux.
$$
What happens if we rotate both $y$ and $x$ by $R$? We get (since $R^TR=I$ for rotations)
$$
y=UxquadRightarrowquad Ry=RUxquadRightarrowquad Ry=underbrace{RUR^T}_{U'}RxquadRightarrowquad y'=U'x'.
$$
Thus the matrix $U'=RUR^T$ corresponds to the same linear map $F$ in the new coordinates after rotation ($x'mapsto y'$).
In general, for any change of the basis $x'=Sx$, $y'=Sy$ the corresponding change of the matrix $U$ is
$$
Sy=underbrace{SUS^{-1}}_{U'}SxquadRightarrowquad y'=U'x'.
$$
It means that the class of all similar matrices ${SUS^{-1}colon Stext{ invertible}}$ is exactly the class of all matrices that describe the same linear map in different bases.
add a comment |
Here the matrix $U$ is considered not as a bunch of column vectors, but as a (matrix of the) linear map $Fcolon {Bbb R}^nto {Bbb R}^n$
$$
y=F(x)=Ux.
$$
What happens if we rotate both $y$ and $x$ by $R$? We get (since $R^TR=I$ for rotations)
$$
y=UxquadRightarrowquad Ry=RUxquadRightarrowquad Ry=underbrace{RUR^T}_{U'}RxquadRightarrowquad y'=U'x'.
$$
Thus the matrix $U'=RUR^T$ corresponds to the same linear map $F$ in the new coordinates after rotation ($x'mapsto y'$).
In general, for any change of the basis $x'=Sx$, $y'=Sy$ the corresponding change of the matrix $U$ is
$$
Sy=underbrace{SUS^{-1}}_{U'}SxquadRightarrowquad y'=U'x'.
$$
It means that the class of all similar matrices ${SUS^{-1}colon Stext{ invertible}}$ is exactly the class of all matrices that describe the same linear map in different bases.
add a comment |
Here the matrix $U$ is considered not as a bunch of column vectors, but as a (matrix of the) linear map $Fcolon {Bbb R}^nto {Bbb R}^n$
$$
y=F(x)=Ux.
$$
What happens if we rotate both $y$ and $x$ by $R$? We get (since $R^TR=I$ for rotations)
$$
y=UxquadRightarrowquad Ry=RUxquadRightarrowquad Ry=underbrace{RUR^T}_{U'}RxquadRightarrowquad y'=U'x'.
$$
Thus the matrix $U'=RUR^T$ corresponds to the same linear map $F$ in the new coordinates after rotation ($x'mapsto y'$).
In general, for any change of the basis $x'=Sx$, $y'=Sy$ the corresponding change of the matrix $U$ is
$$
Sy=underbrace{SUS^{-1}}_{U'}SxquadRightarrowquad y'=U'x'.
$$
It means that the class of all similar matrices ${SUS^{-1}colon Stext{ invertible}}$ is exactly the class of all matrices that describe the same linear map in different bases.
Here the matrix $U$ is considered not as a bunch of column vectors, but as a (matrix of the) linear map $Fcolon {Bbb R}^nto {Bbb R}^n$
$$
y=F(x)=Ux.
$$
What happens if we rotate both $y$ and $x$ by $R$? We get (since $R^TR=I$ for rotations)
$$
y=UxquadRightarrowquad Ry=RUxquadRightarrowquad Ry=underbrace{RUR^T}_{U'}RxquadRightarrowquad y'=U'x'.
$$
Thus the matrix $U'=RUR^T$ corresponds to the same linear map $F$ in the new coordinates after rotation ($x'mapsto y'$).
In general, for any change of the basis $x'=Sx$, $y'=Sy$ the corresponding change of the matrix $U$ is
$$
Sy=underbrace{SUS^{-1}}_{U'}SxquadRightarrowquad y'=U'x'.
$$
It means that the class of all similar matrices ${SUS^{-1}colon Stext{ invertible}}$ is exactly the class of all matrices that describe the same linear map in different bases.
edited yesterday
answered yesterday
A.Γ.
22.6k32656
22.6k32656
add a comment |
add a comment |
Using your example where $U$ is a stretch matrix in 3D, if you want to "rotate" this matrix, you essentially want this stretch action to occur in a different direction / axis. Suppose you have some shape aligned to this new axis. You want to know what the $U'$ is that stretches the shape parallel to this axis. To do this, you use $R^T$ to rotate everything back to the original orientation. Then you do the original stretch transformation $U$. Then you rotate this back using $R$. So $U'=RUR^T$.
Thanks for your answer. But could you please give me a clearer example. I understood your point. However, I couldn't understand your example.
– Msen Rezaee
yesterday
@MsenRezaee I added a picture to help illustrate the point. It is in 2D rather than 3D, but it should be simple to see how this generalises
– John Doe
yesterday
add a comment |
Using your example where $U$ is a stretch matrix in 3D, if you want to "rotate" this matrix, you essentially want this stretch action to occur in a different direction / axis. Suppose you have some shape aligned to this new axis. You want to know what the $U'$ is that stretches the shape parallel to this axis. To do this, you use $R^T$ to rotate everything back to the original orientation. Then you do the original stretch transformation $U$. Then you rotate this back using $R$. So $U'=RUR^T$.
Thanks for your answer. But could you please give me a clearer example. I understood your point. However, I couldn't understand your example.
– Msen Rezaee
yesterday
@MsenRezaee I added a picture to help illustrate the point. It is in 2D rather than 3D, but it should be simple to see how this generalises
– John Doe
yesterday
add a comment |
Using your example where $U$ is a stretch matrix in 3D, if you want to "rotate" this matrix, you essentially want this stretch action to occur in a different direction / axis. Suppose you have some shape aligned to this new axis. You want to know what the $U'$ is that stretches the shape parallel to this axis. To do this, you use $R^T$ to rotate everything back to the original orientation. Then you do the original stretch transformation $U$. Then you rotate this back using $R$. So $U'=RUR^T$.
Using your example where $U$ is a stretch matrix in 3D, if you want to "rotate" this matrix, you essentially want this stretch action to occur in a different direction / axis. Suppose you have some shape aligned to this new axis. You want to know what the $U'$ is that stretches the shape parallel to this axis. To do this, you use $R^T$ to rotate everything back to the original orientation. Then you do the original stretch transformation $U$. Then you rotate this back using $R$. So $U'=RUR^T$.
edited yesterday
answered yesterday
John Doe
10.4k11135
10.4k11135
Thanks for your answer. But could you please give me a clearer example. I understood your point. However, I couldn't understand your example.
– Msen Rezaee
yesterday
@MsenRezaee I added a picture to help illustrate the point. It is in 2D rather than 3D, but it should be simple to see how this generalises
– John Doe
yesterday
add a comment |
Thanks for your answer. But could you please give me a clearer example. I understood your point. However, I couldn't understand your example.
– Msen Rezaee
yesterday
@MsenRezaee I added a picture to help illustrate the point. It is in 2D rather than 3D, but it should be simple to see how this generalises
– John Doe
yesterday
Thanks for your answer. But could you please give me a clearer example. I understood your point. However, I couldn't understand your example.
– Msen Rezaee
yesterday
Thanks for your answer. But could you please give me a clearer example. I understood your point. However, I couldn't understand your example.
– Msen Rezaee
yesterday
@MsenRezaee I added a picture to help illustrate the point. It is in 2D rather than 3D, but it should be simple to see how this generalises
– John Doe
yesterday
@MsenRezaee I added a picture to help illustrate the point. It is in 2D rather than 3D, but it should be simple to see how this generalises
– John Doe
yesterday
add a comment |
One thing that may be instructive is to recall that every matrix can be represented as the linear combination of a series of dyadic/outer products between two vectors, $U = sum_i a_i otimes b_i = sum_i a_i b_i^T$ where $a_i$ and $b_i$ are a sequence of column vectors.
When changing the basis of the matrix, we are in effect applying the vector rule for changing bases to both sequences of vectors:
$$U’ = sum_i a_i’ otimes b_i’ = sum_i Ra_i (Rb_i)^T = sum_iRa_ib_i^T R^T = RUR^T$$
Hope this helps!
2
This is not language I'm familiar with, but wouldn't $ab^T$ always be rank $1$, so only rank $1$ matrices can be represented this way?
– Callus
yesterday
2
@Callus I guess it was meant that any matrix could be represented as a span of dyadic matrices.
– A.Γ.
yesterday
Correct, thank you for spotting that @Callus and @A.Γ.! Will update answer shortly.
– aghostinthefigures
yesterday
add a comment |
One thing that may be instructive is to recall that every matrix can be represented as the linear combination of a series of dyadic/outer products between two vectors, $U = sum_i a_i otimes b_i = sum_i a_i b_i^T$ where $a_i$ and $b_i$ are a sequence of column vectors.
When changing the basis of the matrix, we are in effect applying the vector rule for changing bases to both sequences of vectors:
$$U’ = sum_i a_i’ otimes b_i’ = sum_i Ra_i (Rb_i)^T = sum_iRa_ib_i^T R^T = RUR^T$$
Hope this helps!
2
This is not language I'm familiar with, but wouldn't $ab^T$ always be rank $1$, so only rank $1$ matrices can be represented this way?
– Callus
yesterday
2
@Callus I guess it was meant that any matrix could be represented as a span of dyadic matrices.
– A.Γ.
yesterday
Correct, thank you for spotting that @Callus and @A.Γ.! Will update answer shortly.
– aghostinthefigures
yesterday
add a comment |
One thing that may be instructive is to recall that every matrix can be represented as the linear combination of a series of dyadic/outer products between two vectors, $U = sum_i a_i otimes b_i = sum_i a_i b_i^T$ where $a_i$ and $b_i$ are a sequence of column vectors.
When changing the basis of the matrix, we are in effect applying the vector rule for changing bases to both sequences of vectors:
$$U’ = sum_i a_i’ otimes b_i’ = sum_i Ra_i (Rb_i)^T = sum_iRa_ib_i^T R^T = RUR^T$$
Hope this helps!
One thing that may be instructive is to recall that every matrix can be represented as the linear combination of a series of dyadic/outer products between two vectors, $U = sum_i a_i otimes b_i = sum_i a_i b_i^T$ where $a_i$ and $b_i$ are a sequence of column vectors.
When changing the basis of the matrix, we are in effect applying the vector rule for changing bases to both sequences of vectors:
$$U’ = sum_i a_i’ otimes b_i’ = sum_i Ra_i (Rb_i)^T = sum_iRa_ib_i^T R^T = RUR^T$$
Hope this helps!
edited yesterday
answered yesterday
aghostinthefigures
1,2301216
1,2301216
2
This is not language I'm familiar with, but wouldn't $ab^T$ always be rank $1$, so only rank $1$ matrices can be represented this way?
– Callus
yesterday
2
@Callus I guess it was meant that any matrix could be represented as a span of dyadic matrices.
– A.Γ.
yesterday
Correct, thank you for spotting that @Callus and @A.Γ.! Will update answer shortly.
– aghostinthefigures
yesterday
add a comment |
2
This is not language I'm familiar with, but wouldn't $ab^T$ always be rank $1$, so only rank $1$ matrices can be represented this way?
– Callus
yesterday
2
@Callus I guess it was meant that any matrix could be represented as a span of dyadic matrices.
– A.Γ.
yesterday
Correct, thank you for spotting that @Callus and @A.Γ.! Will update answer shortly.
– aghostinthefigures
yesterday
2
2
This is not language I'm familiar with, but wouldn't $ab^T$ always be rank $1$, so only rank $1$ matrices can be represented this way?
– Callus
yesterday
This is not language I'm familiar with, but wouldn't $ab^T$ always be rank $1$, so only rank $1$ matrices can be represented this way?
– Callus
yesterday
2
2
@Callus I guess it was meant that any matrix could be represented as a span of dyadic matrices.
– A.Γ.
yesterday
@Callus I guess it was meant that any matrix could be represented as a span of dyadic matrices.
– A.Γ.
yesterday
Correct, thank you for spotting that @Callus and @A.Γ.! Will update answer shortly.
– aghostinthefigures
yesterday
Correct, thank you for spotting that @Callus and @A.Γ.! Will update answer shortly.
– aghostinthefigures
yesterday
add a comment |
The columns of $mathbf{U}$ tell you what happens to the coordinate vectors $hat{e}_1,hat{e}_2,hat{e}_3$. For example, if the first column is $[a,b,c]^T$, then $mathbf{U}hat{e}_1 = ahat{e}_1 + bhat{e}_2 + chat{e}_3$.
The matrix $mathbf{U}'$ is the matrix that behaves the same way on the rotated coordinate basis $hat{f}_i = mathbf{R}hat{e}_i$. This is because $mathbf{R}^T = mathbf{R}^{-1}$ so for example
$$begin{align*}
mathbf{U}'hat{f}_1 &= mathbf{R}mathbf{U}mathbf{R}^Tmathbf{R}hat{e}_1 \
&= mathbf{R}left(mathbf{U}hat{e}_1right) \
&= mathbf{R}(ahat{e}_1 + bhat{e}_2 + chat{e}_3) \
&= amathbf{R}hat{e}_1 + bmathbf{R}hat{e}_2 + cmathbf{R}hat{e}_3 \
&= ahat{f}_1 + bhat{f}_2 + chat{f}_3
end{align*}
$$
add a comment |
The columns of $mathbf{U}$ tell you what happens to the coordinate vectors $hat{e}_1,hat{e}_2,hat{e}_3$. For example, if the first column is $[a,b,c]^T$, then $mathbf{U}hat{e}_1 = ahat{e}_1 + bhat{e}_2 + chat{e}_3$.
The matrix $mathbf{U}'$ is the matrix that behaves the same way on the rotated coordinate basis $hat{f}_i = mathbf{R}hat{e}_i$. This is because $mathbf{R}^T = mathbf{R}^{-1}$ so for example
$$begin{align*}
mathbf{U}'hat{f}_1 &= mathbf{R}mathbf{U}mathbf{R}^Tmathbf{R}hat{e}_1 \
&= mathbf{R}left(mathbf{U}hat{e}_1right) \
&= mathbf{R}(ahat{e}_1 + bhat{e}_2 + chat{e}_3) \
&= amathbf{R}hat{e}_1 + bmathbf{R}hat{e}_2 + cmathbf{R}hat{e}_3 \
&= ahat{f}_1 + bhat{f}_2 + chat{f}_3
end{align*}
$$
add a comment |
The columns of $mathbf{U}$ tell you what happens to the coordinate vectors $hat{e}_1,hat{e}_2,hat{e}_3$. For example, if the first column is $[a,b,c]^T$, then $mathbf{U}hat{e}_1 = ahat{e}_1 + bhat{e}_2 + chat{e}_3$.
The matrix $mathbf{U}'$ is the matrix that behaves the same way on the rotated coordinate basis $hat{f}_i = mathbf{R}hat{e}_i$. This is because $mathbf{R}^T = mathbf{R}^{-1}$ so for example
$$begin{align*}
mathbf{U}'hat{f}_1 &= mathbf{R}mathbf{U}mathbf{R}^Tmathbf{R}hat{e}_1 \
&= mathbf{R}left(mathbf{U}hat{e}_1right) \
&= mathbf{R}(ahat{e}_1 + bhat{e}_2 + chat{e}_3) \
&= amathbf{R}hat{e}_1 + bmathbf{R}hat{e}_2 + cmathbf{R}hat{e}_3 \
&= ahat{f}_1 + bhat{f}_2 + chat{f}_3
end{align*}
$$
The columns of $mathbf{U}$ tell you what happens to the coordinate vectors $hat{e}_1,hat{e}_2,hat{e}_3$. For example, if the first column is $[a,b,c]^T$, then $mathbf{U}hat{e}_1 = ahat{e}_1 + bhat{e}_2 + chat{e}_3$.
The matrix $mathbf{U}'$ is the matrix that behaves the same way on the rotated coordinate basis $hat{f}_i = mathbf{R}hat{e}_i$. This is because $mathbf{R}^T = mathbf{R}^{-1}$ so for example
$$begin{align*}
mathbf{U}'hat{f}_1 &= mathbf{R}mathbf{U}mathbf{R}^Tmathbf{R}hat{e}_1 \
&= mathbf{R}left(mathbf{U}hat{e}_1right) \
&= mathbf{R}(ahat{e}_1 + bhat{e}_2 + chat{e}_3) \
&= amathbf{R}hat{e}_1 + bmathbf{R}hat{e}_2 + cmathbf{R}hat{e}_3 \
&= ahat{f}_1 + bhat{f}_2 + chat{f}_3
end{align*}
$$
answered yesterday
Callus
4,403922
4,403922
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If you have a linear algebra textbook on hand, you might find it instructive to read about changes of basis. The matrix $R$, in this context, can be nicely thought of as a change-of-basis matrix.
– Omnomnomnom
yesterday
1
The expression $RU$ also is a kind of "rotation of $U$," namely, it takes whatever transformation $U$ was going to perform and composes that with a subsequent rotation represented by $R.$ But this is actually giving us a new transformation within the old coordinate system, not rewriting the old transformation in a new coordinate system as we would get from $RUR^T.$
– David K
yesterday
When you say "$U$ is a matrix", it makes a difference whether $U$ is merely a set of points (in which case $RU$ is already the "rotation of $U$"), or itself a transformation matrix (as in this case). Can you please edit your title/body to be more clear?
– smci
yesterday
@smci Look at the P.S. note. I have mentioned that U is a stretch matrix. not a set of points.
– Msen Rezaee
21 hours ago
Yes that's more clear now
– smci
20 hours ago