This tells you . But you already know that , so after canceling the you get which is just what you wanted. If you haven't seen the spectral theorem, then skip this proof. Given and an orthonormal basis of , iff iff iff are all eigenvectors of A . The desired basis is guaranteed by spectral theorem since is symmetric.
Every matrix has a singular value decomposition.
The theorem above almost gives you the SVD for any matrix. To complete the decomposition, let be an orthonormal basis for the three dimensional subspace of perpendicular to . An extra row of zeros has been added to the stretcher matrix to produce the dimensions required for the multiplication. In either case, the dimensions of the stretcher matrix will always match the dimensions of A . The decomposition shows that the action of every matrix can be described as a rotation followed by a stretch followed by another rotation.