Mixing
What does it mean for the cross product to be invariant under orthogonal transformation?
$begingroup$
This article claims $Ax times Ay = det(A)(xtimes y)$, but using the vectors $(3,2,1)$ and $(5,4,6)$ and the orthogonal transformation with determinant $1$:
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)$$
does not hold.
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
3 \
2 \
1
end{matrix}right)=left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)$$
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
5 \
4 \
6
end{matrix}right)=left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)$$
and
$$left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)times left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)=left(begin{matrix}
frac{13*sqrt{3}+8}{2} \
frac{8*sqrt{3}-13}{2} \
2
end{matrix}right)$$
but $(3,2,1)times (5,4,6)=(8,-13,2)$.
So, what does it mean for the cross product to be invariant under orthogonal transformation?
linear-algebra cross-product
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
$endgroup$
add a comment |
$begingroup$
This article claims $Ax times Ay = det(A)(xtimes y)$, but using the vectors $(3,2,1)$ and $(5,4,6)$ and the orthogonal transformation with determinant $1$:
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)$$
does not hold.
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
3 \
2 \
1
end{matrix}right)=left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)$$
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
5 \
4 \
6
end{matrix}right)=left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)$$
and
$$left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)times left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)=left(begin{matrix}
frac{13*sqrt{3}+8}{2} \
frac{8*sqrt{3}-13}{2} \
2
end{matrix}right)$$
but $(3,2,1)times (5,4,6)=(8,-13,2)$.
So, what does it mean for the cross product to be invariant under orthogonal transformation?
linear-algebra cross-product
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
$endgroup$
1
$begingroup$
The claim in the linked note is obviously wrong. I think it is correct if you multiply the cross-product on the right by A.
$endgroup$
– Lukas Geyer
Jan 28 at 19:53
1
$begingroup$
You say "this article" : first of all, it is not an article, it is a page page written by anybody anywhere in the Realm of nowhere... It is not because something is written that it is true...
$endgroup$
– Jean Marie
Jan 28 at 20:41
add a comment |
$begingroup$
This article claims $Ax times Ay = det(A)(xtimes y)$, but using the vectors $(3,2,1)$ and $(5,4,6)$ and the orthogonal transformation with determinant $1$:
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)$$
does not hold.
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
3 \
2 \
1
end{matrix}right)=left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)$$
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
5 \
4 \
6
end{matrix}right)=left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)$$
and
$$left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)times left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)=left(begin{matrix}
frac{13*sqrt{3}+8}{2} \
frac{8*sqrt{3}-13}{2} \
2
end{matrix}right)$$
but $(3,2,1)times (5,4,6)=(8,-13,2)$.
So, what does it mean for the cross product to be invariant under orthogonal transformation?
linear-algebra cross-product
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
$endgroup$
This article claims $Ax times Ay = det(A)(xtimes y)$, but using the vectors $(3,2,1)$ and $(5,4,6)$ and the orthogonal transformation with determinant $1$:
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)$$
does not hold.
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
3 \
2 \
1
end{matrix}right)=left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)$$
$$left(begin{matrix}
frac{1}{2} & frac{-sqrt{3}}{2} & 0 \
frac{sqrt{3}}{2} & frac{1}{2} & 0 \
0 & 0 & 1
end{matrix}right)left(begin{matrix}
5 \
4 \
6
end{matrix}right)=left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)$$
and
$$left(begin{matrix}
frac{-2*sqrt{3}+3}{2} \
frac{3*sqrt{3}+2}{2} \
1
end{matrix}right)times left(begin{matrix}
frac{-4*sqrt{3}+5}{2} \
frac{5*sqrt{3}+4}{2} \
6
end{matrix}right)=left(begin{matrix}
frac{13*sqrt{3}+8}{2} \
frac{8*sqrt{3}-13}{2} \
2
end{matrix}right)$$
but $(3,2,1)times (5,4,6)=(8,-13,2)$.
So, what does it mean for the cross product to be invariant under orthogonal transformation?
linear-algebra cross-product
linear-algebra cross-product
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
edited Jan 28 at 19:58
José Carlos Santos
171k23132240
171k23132240
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
asked Jan 28 at 19:45
Al JebrAl Jebr
4,37943378
4,37943378
1
$begingroup$
The claim in the linked note is obviously wrong. I think it is correct if you multiply the cross-product on the right by A.
$endgroup$
– Lukas Geyer
Jan 28 at 19:53
1
$begingroup$
You say "this article" : first of all, it is not an article, it is a page page written by anybody anywhere in the Realm of nowhere... It is not because something is written that it is true...
$endgroup$
– Jean Marie
Jan 28 at 20:41
add a comment |
1
$begingroup$
The claim in the linked note is obviously wrong. I think it is correct if you multiply the cross-product on the right by A.
$endgroup$
– Lukas Geyer
Jan 28 at 19:53
1
$begingroup$
You say "this article" : first of all, it is not an article, it is a page page written by anybody anywhere in the Realm of nowhere... It is not because something is written that it is true...
$endgroup$
– Jean Marie
Jan 28 at 20:41
1
1
$begingroup$
The claim in the linked note is obviously wrong. I think it is correct if you multiply the cross-product on the right by A.
$endgroup$
– Lukas Geyer
Jan 28 at 19:53
$begingroup$
The claim in the linked note is obviously wrong. I think it is correct if you multiply the cross-product on the right by A.
$endgroup$
– Lukas Geyer
Jan 28 at 19:53
1
1
$begingroup$
You say "this article" : first of all, it is not an article, it is a page page written by anybody anywhere in the Realm of nowhere... It is not because something is written that it is true...
$endgroup$
– Jean Marie
Jan 28 at 20:41
$begingroup$
You say "this article" : first of all, it is not an article, it is a page page written by anybody anywhere in the Realm of nowhere... It is not because something is written that it is true...
$endgroup$
– Jean Marie
Jan 28 at 20:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $v=(3,2,1)$ and $w=(5,4,6)$, then, indeed $vtimes w=(8,-13,2)$. And$$A.(vtimes w)=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).$$It turns out thatbegin{align}(Av)times(Aw)&=left(frac{3}{2}-sqrt{3},1+frac{3 sqrt{3}}{2},1right)timesleft(frac{5}{2}-2 sqrt{3},2+frac{5 sqrt{3}}{2},6right)\&=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).end{align}As you can see, $(Av)times(Aw)=det(A).A(vtimes w)$ in this case. Actually, it's true for every orthogonal matrix $A$.
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Any hints on how to prove this fact that $Av times Aw=det(A)A(vtimes w)$ when $A$ is orthogonal?
$endgroup$
– Al Jebr
Jan 28 at 19:58
$begingroup$
You will find a proof here.
$endgroup$
– José Carlos Santos
Jan 28 at 21:10
add a comment |
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1 Answer
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$begingroup$
If $v=(3,2,1)$ and $w=(5,4,6)$, then, indeed $vtimes w=(8,-13,2)$. And$$A.(vtimes w)=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).$$It turns out thatbegin{align}(Av)times(Aw)&=left(frac{3}{2}-sqrt{3},1+frac{3 sqrt{3}}{2},1right)timesleft(frac{5}{2}-2 sqrt{3},2+frac{5 sqrt{3}}{2},6right)\&=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).end{align}As you can see, $(Av)times(Aw)=det(A).A(vtimes w)$ in this case. Actually, it's true for every orthogonal matrix $A$.
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Any hints on how to prove this fact that $Av times Aw=det(A)A(vtimes w)$ when $A$ is orthogonal?
$endgroup$
– Al Jebr
Jan 28 at 19:58
$begingroup$
You will find a proof here.
$endgroup$
– José Carlos Santos
Jan 28 at 21:10
add a comment |
$begingroup$
If $v=(3,2,1)$ and $w=(5,4,6)$, then, indeed $vtimes w=(8,-13,2)$. And$$A.(vtimes w)=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).$$It turns out thatbegin{align}(Av)times(Aw)&=left(frac{3}{2}-sqrt{3},1+frac{3 sqrt{3}}{2},1right)timesleft(frac{5}{2}-2 sqrt{3},2+frac{5 sqrt{3}}{2},6right)\&=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).end{align}As you can see, $(Av)times(Aw)=det(A).A(vtimes w)$ in this case. Actually, it's true for every orthogonal matrix $A$.
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Any hints on how to prove this fact that $Av times Aw=det(A)A(vtimes w)$ when $A$ is orthogonal?
$endgroup$
– Al Jebr
Jan 28 at 19:58
$begingroup$
You will find a proof here.
$endgroup$
– José Carlos Santos
Jan 28 at 21:10
add a comment |
$begingroup$
If $v=(3,2,1)$ and $w=(5,4,6)$, then, indeed $vtimes w=(8,-13,2)$. And$$A.(vtimes w)=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).$$It turns out thatbegin{align}(Av)times(Aw)&=left(frac{3}{2}-sqrt{3},1+frac{3 sqrt{3}}{2},1right)timesleft(frac{5}{2}-2 sqrt{3},2+frac{5 sqrt{3}}{2},6right)\&=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).end{align}As you can see, $(Av)times(Aw)=det(A).A(vtimes w)$ in this case. Actually, it's true for every orthogonal matrix $A$.
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
If $v=(3,2,1)$ and $w=(5,4,6)$, then, indeed $vtimes w=(8,-13,2)$. And$$A.(vtimes w)=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).$$It turns out thatbegin{align}(Av)times(Aw)&=left(frac{3}{2}-sqrt{3},1+frac{3 sqrt{3}}{2},1right)timesleft(frac{5}{2}-2 sqrt{3},2+frac{5 sqrt{3}}{2},6right)\&=left(4+frac{13sqrt{3}}2,4sqrt{3}-frac{13}2,2right).end{align}As you can see, $(Av)times(Aw)=det(A).A(vtimes w)$ in this case. Actually, it's true for every orthogonal matrix $A$.
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 19:55
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
Any hints on how to prove this fact that $Av times Aw=det(A)A(vtimes w)$ when $A$ is orthogonal?
$endgroup$
– Al Jebr
Jan 28 at 19:58
$begingroup$
You will find a proof here.
$endgroup$
– José Carlos Santos
Jan 28 at 21:10
add a comment |
$begingroup$
Any hints on how to prove this fact that $Av times Aw=det(A)A(vtimes w)$ when $A$ is orthogonal?
$endgroup$
– Al Jebr
Jan 28 at 19:58
$begingroup$
You will find a proof here.
$endgroup$
– José Carlos Santos
Jan 28 at 21:10
$begingroup$
Any hints on how to prove this fact that $Av times Aw=det(A)A(vtimes w)$ when $A$ is orthogonal?
$endgroup$
– Al Jebr
Jan 28 at 19:58
$begingroup$
Any hints on how to prove this fact that $Av times Aw=det(A)A(vtimes w)$ when $A$ is orthogonal?
$endgroup$
– Al Jebr
Jan 28 at 19:58
$begingroup$
You will find a proof here.
$endgroup$
– José Carlos Santos
Jan 28 at 21:10
$begingroup$
You will find a proof here.
$endgroup$
– José Carlos Santos
Jan 28 at 21:10
add a comment |
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1
$begingroup$
The claim in the linked note is obviously wrong. I think it is correct if you multiply the cross-product on the right by A.
$endgroup$
– Lukas Geyer
Jan 28 at 19:53
1
$begingroup$
You say "this article" : first of all, it is not an article, it is a page page written by anybody anywhere in the Realm of nowhere... It is not because something is written that it is true...
$endgroup$
– Jean Marie
Jan 28 at 20:41