Direct sum and orthogonal transformation Let $V$ be a Euclidean space, and $\mathscr{A}$ is an orthogonal transformation on $V$, that is, $||\mathscr{A}\alpha||=||\alpha||$, for each $\alpha\in V$. 
Now define $$V_1=\{\alpha\in V; \mathscr{A}(\alpha)=\alpha\}; V_2=\{\alpha-\mathscr{A}(\alpha);\alpha \in V\}.$$
Prove that $V=V_1\oplus V_2$.
I knew that $\dim V_1+\dim V_2=\dim ker(\mathscr{I})-\mathscr{A})+\dim im(\mathscr{I}-\mathscr{A})=n$, and it does only need to show $V_1\cap V_2=\{0\}$ or just $V=V_1+V_2$, but I could just now show it conveniently...
 A: One could show that $V_1 \perp V_2$. If $a=A(a)$, $\langle a,b-A(b) \rangle=\langle a,b \rangle - \langle a, A(b) \rangle= \langle a,b \rangle- \langle A^*(a), b\rangle= \langle a-A^*(a),b \rangle= \langle A(a)-a, A(b) \rangle=0$, for each $b\in V$.
A: If A is a real orthogonal matrix then $AA^t=A^tA=Id$. Thus, $A$ is unitary as a complex matrix. Therefore A is diagonalizable as a complex matrix. Therefore $A=RDR^{-1}$ and $Id-A=R(Id-D)R^{-1}$, where $D$ is a complex diagonal matrix. 
Now, the rank of $R(Id-D)^2R^{-1}$ is the rank of $(Id-D)^2$, which is the rank of $Id-D$ (squaring the diagonal matrix does not increase the numbers of zeros in the diagonal). Therefore the rank of $(Id-A)^2=R(Id-D)^2R^{-1}$ is the rank of $R(Id-D)R^{-1}=Id-A$.
Thus, $\dim(\ker(Id-A)^2)=\dim(\ker(Id-A))$, but $\ker(Id-A)\subset\ker(Id-A)^2$. Therefore, $\ker(Id-A)=\ker(Id-A)^2$.
So, if $A(v-Av)=(v-Av)$ then $(Id-A)(Id-A)v=0$. Therefore,  $v\in\ker(Id-A)^2=\ker(Id-A)$ and $(Id-A)v=v-Av=0$. 
Finally, if $w\in V_1\cap V_2$ then $w=v-Av$ and $A(v-Av)=v-Av$. Thus, $w=v-Av$=0.  
