Tangent Space of orthogonal matrix
I want to prove that
TIO(n,R)={X∈Rn×n:X=−XT}
My only trouble is with that whole dropping second order terms business.
So I know that I want to define a curve through the identity in such a way that the tangent to the curve is X
, that is
γ={A(t)=I+tX:t∈(−ϵ,ϵ)}
Now A(0)=I
and
˙
A
(0)=X
so X∈TIO(n,R)
. Moreover
I=ATA=(I+tX)T(I+tX)=I+tX+tXT+t2XTX
Now if you drop t2XTX
you have the answer, but why do you drop it? What's the rational?
Is it because
I+tX+tXT+t2XTX→I+tX+tXT
as ϵ→0
? Is that all?
EDIT: I'm also not sure about the converse. Why can you always build γ
from a skew-symmetric matrix X
?
TIO(n,R)={X∈Rn×n:X=−XT}
My only trouble is with that whole dropping second order terms business.
So I know that I want to define a curve through the identity in such a way that the tangent to the curve is X
, that is
γ={A(t)=I+tX:t∈(−ϵ,ϵ)}
Now A(0)=I
and
˙
A
(0)=X
so X∈TIO(n,R)
. Moreover
I=ATA=(I+tX)T(I+tX)=I+tX+tXT+t2XTX
Now if you drop t2XTX
you have the answer, but why do you drop it? What's the rational?
Is it because
I+tX+tXT+t2XTX→I+tX+tXT
as ϵ→0
? Is that all?
EDIT: I'm also not sure about the converse. Why can you always build γ
from a skew-symmetric matrix X
?
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