Self-adjoint and isometric then $A^2=Id$

Question

If $E$ is a real vector space and $A:E\to E$ a linear operator self-adjoint and isometric (preserves inner product) then $A^2=Id$.

I think this problem is easy, but I think I'm missing something, could someone help me?

Answer 1

Let $(\cdot, \cdot )$ the inner product. For $x \in E$ we have

$(A^2x-x, A^2x-x)=(A^2x,A^2x)-2(A^2x,x)+(x,x)=(Ax,Ax)-2(Ax,Ax)+(x,x)=-(Ax,Ax)+(x,x)=-(x,x)+(x,x)=0.$