Show that $L_A$ acts on by orthogonal transformation and in particular rotation.

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.

MY TRY::

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 $$\det A=1$$.
• 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? – user596656 Dec 9 '18 at 15:19
• 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. – José Carlos Santos Dec 9 '18 at 15:22
• Are you including the vector $v$ along with the basis B to form a basis of $L_A$ – user596656 Dec 9 '18 at 15:33