Let $A$ be a $3\times 3$ orthogonal matrix with determinant $=1$. Let $v$ be an eigen vector corresponding to $1$ of $A$.Let $W=\text{span}\{v\}$. Show that $L_A$ preserves $W^\perp$ and it acts on it by orthogonal transformation and in particular rotation.


Given $Av=v\implies v=A^{-1}v$

Also $L_A:\Bbb R^3\to \Bbb R^3$ defined by $L_A(x)=Ax$.

Let $w\in W^\perp\implies \langle w,v\rangle =0$

To show $L_A(w)\in W^\perp$.

Now $L_A(w)=Aw$

Also $\langle Aw,v\rangle =\langle w,A^Tv\rangle= \langle w,A^{-1}v\rangle=\langle w,v\rangle =0$

Thus $L_A(w)\in W^\perp$.-------------(Proved)

hence we can consider the restriction $L_A:W^\perp\to W^\perp$

But how can I show that it is an orthogonal transformation and in aprticular a rotation?


Since $A.W^\perp\subset W^\perp$ and since $A$ is orthogonal, $A$ induces an orthogonal map from $W^\perp$ into itself. Suppose that it is not a rotation. Fix an orthonormal basis $B$ of $W^\perp$. Then the matrix of $A|_{W^\perp}$ with respect to $B$ is an orthogonal matrix with determinant $-1$. But then $\det A=1\times(-1)=-1$. This is impossible, since we're assuming that $\det A=1$.

  • $\begingroup$ My problem is I dont understand how to show that restriction of A to $W^\perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why? $\endgroup$ – user596656 Dec 9 '18 at 15:19
  • $\begingroup$ How to prove it? $\endgroup$ – user596656 Dec 9 '18 at 15:19
  • $\begingroup$ Since $A$ is orthogonal, then$$(\forall v,w\in\mathbb{R}^3):\langle A.v,A.w\rangle=\langle v,w\rangle.$$In particular,$$(\forall v,w\in W^\perp):\langle A.v,A.w\rangle=\langle v,w\rangle,$$which means that $A|_{W^\perp}$ is orthogonal. $\endgroup$ – José Carlos Santos Dec 9 '18 at 15:22
  • $\begingroup$ Are you including the vector $v$ along with the basis B to form a basis of $L_A$ $\endgroup$ – user596656 Dec 9 '18 at 15:33
  • $\begingroup$ Otherwise how can you include the eigen value -1 $\endgroup$ – user596656 Dec 9 '18 at 15:33

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.