Final answer:
To prove that S is a reflection, we need to show that there exists a subspace W of V such that S=R, where R is a reflection. We can decompose any vector u in V as a sum of vectors w and w', where w is the projection of u onto W and w' is the projection onto W's orthogonal complement. By showing that S(w) = w and S(w') = -w', we can conclude that S is a reflection.
Step-by-step explanation:
To prove that S is a reflection, we need to show that there exists a subspace W of V such that S=R, where R is a reflection. Since S is a self-adjoint isometry, it has eigenvalues of -1 and 1. Let's consider the eigenvalue 1. Let v be an eigenvector corresponding to the eigenvalue 1. Since S is self-adjoint, its eigenvectors corresponding to different eigenvalues are orthogonal. Therefore, v is orthogonal to S(v). Let W be the subspace spanned by v. Now, we need to show that S is equal to R. Let u be any vector in V. We can decompose u as u=w+w', where w is the projection of u onto W and w' is the projection onto W's orthogonal complement. Since v is an eigenvector of S corresponding to eigenvalue 1, S(v) = v. Therefore, S(w) = w and S(w') = -w'. This shows that S is a reflection.