If A^T A V = V S is the spectral decomposition of A^T A with S diagonal and V orthogonal, we have that (A V)^T (A V)= S. Hence, the nonzero columns of AV, which is associated with the nonzero columns of A, is orthogonal with all other nonzero columns of AV. Hence, is possible to find an orthogonal matrix U such that A V = U sqrt(S) -- note that the diagonal of S give us the squared norm of each column of AV, which means this is completely fine. Now, this means A=U sqrt(S) V^T. Completing U, if necessary, we have the SVD decomposition. In practice, after the calculation of the spectral decomposition of A^T A, we would calculate the Kern of (AV)^T -- orthogonal to Im(AV) -- and find an orthogonal basis to it, and of course, all the singular values associated with this process would be zero.####[]