Orthogonal transformations preserve length

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SUMMARY

Orthogonal transformations, represented by orthogonal matrices such as Q, preserve the length of vectors. The transformation of a vector (p,q) using the orthogonal matrix results in a magnitude of sqrt(p^2+q^2), confirming that the length remains unchanged. This property arises from the orthogonality of the column vectors v1 and v2, which have magnitudes of 1 and are perpendicular. Additionally, it is established that the eigenvalues of orthogonal matrices have a magnitude of 1, as demonstrated by the relationship |λ| = 1 for eigenvectors.

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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?
 
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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.
 
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