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I was wondering whether people can comment on the problem of finding the derivative of the singular values and the singular vector w.r.t. the entries of that matrix.

Any help is appreciated.

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The singular values of $A$ are the square roots of the nonzero eigenvalues of $M = A^* A$. Let $P(t) = \det(M - t I)$ be the characteristic polynomial of $M$. If $\lambda$ is one of the eigenvalues and $P'(\lambda) \ne 0$, implicit differentiation gives $\frac{\partial \lambda}{\partial m_{ij}} = - \frac{1}{P'(\lambda)} \text{adj}(M - \lambda I)_{ji}$, where $\text{adj}(M - \lambda I)$ is the adjugate or classical adjoint of $M - \lambda I$. Now use the chain rule with $\frac{\partial m_{ii}}{\partial a_{ki}} = 2 a_{ki}$, $\frac{\partial m_{ij}}{\partial a_{ki}} = a_{kj}$ and $\frac{\partial m_{ij}}{\partial a_{kj}} = a_{ki}$ if $i \ne j$, $\frac{\partial m_{ij}}{\partial a_{kl}} = 0$ otherwise

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What is $P^'(\lambda)$? $\frac{\mbox{d} P}{\mbox{d} \lambda}$? – Shiyu Jun 18 '11 at 4:39

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