Singular value decomposition

Let \(A\in M_{m\times n}(\CC)\). The singular value decomposition says that there exist unitary matrices \[ U\in U(m),\qquad V\in U(n) \] and an \(m\times n\) diagonal matrix \[ \Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_r,0,\ldots), \qquad \sigma_i\ge0, \] such that \[ A=U\Sigma V^*. \] The numbers \(\sigma_i\) are the square roots of the nonzero eigenvalues of \(A^*A\).

Since \(A^*A\) is Hermitian positive semidefinite, the spectral theorem gives an orthonormal basis \[ v_1,\ldots,v_n \] of eigenvectors with \[ A^*Av_i=\sigma_i^2v_i, \qquad \sigma_i\ge0. \] For each \(\sigma_i>0\), define \[ u_i=\frac{Av_i}{\sigma_i}. \] Then \[ \langle u_i,u_j\rangle =\frac1{\sigma_i\sigma_j}\langle v_i,A^*Av_j\rangle =\delta_{ij}. \] Thus the \(u_i\) form an orthonormal set. Extend them to an orthonormal basis of \(\CC^m\).

Let \(V\) be the unitary matrix with columns \(v_i\) and \(U\) the unitary matrix with columns \(u_i\). Then \[ Av_i=\sigma_i u_i \] for the nonzero singular values, while \(Av_i=0\) when \(\sigma_i=0\). Therefore \[ AV=U\Sigma, \] so \[ A=U\Sigma V^*. \]

Over \(\RR\), the same proof gives orthogonal \(U,V\).