## the SVD: overview of the proof

sometimes a proof is of comparable importance to the theorem itself. Sometimes, at the other extreme, the proof is almost trivial. In this case, the proof falls between those limits. to be specific, the decompositions of

u = (u1 u2)

and

v = (v1 v2)

are worth knowing in their own right.

let me outline the proof.

•  get the eigendecomposition of  $X^T X = v\ D^2\ v^{-1}$
•  … with orthogonal basis of eigenvectors v
•  … and the diagonal eigenvalue matrix written as  $D^2$
•  split the matrix v = (v1 v2), with all the zero eigenvalues associated with v2
•  let $\Sigma$ be the diagonal matrix of square roots of the nonzero eigenvalues
•  define $u_1 = X\ v_1\ {\Sigma}^{-1}$  by analogy with the impossible (because w is not invertible) $u = X\ v\ w^{-1}$
•  extend u1 by gram-schmidt to u.