I Orthogonal transformations preserve length

lys04
Messages
144
Reaction score
5
Let Q be an orthogonal matrix, and I want to transform (p,q), the magnitude of this vector is sqrt(p^2+q^2) using pythag. T((p,q))=pv1+qv_2 where v1 and v2 are the column vectors of Q. Since the column vectors of Q have magnitude of 1, this means pv1 has magnitude of p and qv2 has magnitude of q. v1 and v2 are also perpendicular, using the fact that ||a+b||^2=(a+b).(a+b)=a.(a+b)+b.(a+b)=a.a+a.b+b.a+b.b=a.a+0+0+b.b=||a|^2+||b||^2, so the magnitude of pv1+qv_2 is also sqrt(p^2+q^2). Hence length is preserved?

Also does this explain why orthogonal transformations have eigenvalues of magnitude of 1?
 
Last edited:
Physics news on Phys.org
By definition, an orthogonal matrix satisfies A^T = A^{-1}. An immediate consequence is that for any v,
\|Av\|^2 = (Av) \cdot (Av) = (Av)^T(Av) = v^TA^TAv = v^Tv = v \cdot v = \|v\|^2 so that \|Av\| = \|v\|. If v is an eigenvector with eigenvalue \lambda, then <br /> \|v\| = \|Av\| = |\lambda| \|v\| \quad\Rightarrow\quad |\lambda| = 1.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
2
Views
446
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K