Prove that every reflection matrix is diagonalizableSolution

Prove that every reflection matrix is diagonalizable.

Solution

If T is a projection, that means there\'s a subspace W onto which it projects. It maps every vector in W to itself. Therefore every vector in W is an eigenvector with eigenvalue 11. Every vector not in WWis mapped to a vector in W. Take any vector v and write

v=Tv+(vTv),

so the first term Tv is in W. It is easy to see that the second term, vTv, is in the kernel of T: the first term is mapped to Tv, and the second is mapped to TvT2v. But since Tv is in W, it must be fixed by T, so T2v=Tv; thus T(vTV)=0. In this way, every vector vv is written as the sum of a vector in WW, which is an eigenvector with eigenvalue 11, and a vector in the kernel of TT, which is an eigenvector with eigenvalue 00. So form a basis of the whole space by taking the union of a basis of WW and a basis of the kernel of TT, and the matrix of TT with respect to that

(and all off-diagonal entries are 00) where the number of 11s is the dimension of WW and the number of 00s is the dimension of the kernel of TT.

hence every reflection matrix is diagonaizable.