Orthogonal transformations preserve length

[lys04]

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?

 

[pasmith]

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 $$\|v\| = \|Av\| = |\lambda| \|v\| \quad\Rightarrow\quad |\lambda| = 1.$$