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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?

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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 $\det A=1$.

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  • $\begingroup$ 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

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