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Hi All,

Suppose I've a symmetric matrix $A_{N \times N} = (A_{ij})$ which has a eigen value decomposition $A = UDU'$. I would like to know under what conditions $\frac{\partial U}{\partial A_{ij}}$ exists for all $i,j = 1,2, \ldots, N$. I found the following paper which talks about estimating the Jacobian of the SVD transformation

But its not very clear regarding the conditions that the matrix $A$ would need to satisfy. Any help is much appreciated.

Thanks Ashin

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2 Answers 2

up vote 5 down vote accepted

To make Igor's more precise, Kato's book tells us that

  1. if an eigenvalue of a matrix $A$ is simple, then it extends as an analytic function $M\mapsto\lambda(M)$ defined in a neighbourhood of $A$, such that $\lambda(M)$ is an eigenvalue of $M$.
  2. if $s\mapsto A(s)$ is an analytic, one-parameter, family of real symmetric matrices, their eigenvalues $\lambda_1(s),\ldots,\lambda_n(s)$ can be arranged so that they are analytic functions. Mind that they are not in increasing order in general, because their order can change at values of $s$ for which $A(s)$ has a multiple eigenvalue.
  3. The previous result becomes false when the family depends upon several parameters. A typical example with two parameters is $$A(s,t)=\begin{pmatrix} s & t \\\\ t & -s \end{pmatrix},$$ for which the eigenvalues $\pm\sqrt{s^2+t^2}$ are even not $C^1$-functions.
  4. However, Weyl inequalities tell us that for real symmetric matrices, ${\rm dist}({\rm Sp}(B),{\rm Sp}(A))\le\|B-A\|$. Hence the eigenvalues are Lipschitz function, with unit Lipschitz constant.
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But my question is more about the singular vectors. Are they also going to admit partial derivatives w.r.t. the entries of the matrix almost everywhere. Best – Ashin Nov 7 '11 at 3:22

Read Kato's peturbation theory for linear operators (chapter 2 is sufficient for your question), and all will be revealed.

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