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I jus want to know how to show that if a matrix X converge to Y ( with respect to any matrix norm) then the ith singular value of X converge to the ith singular value of Y.

Thank you

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    $\begingroup$ Is this a homework problem? What have you tried so far? $\endgroup$ – MCT Mar 23 '14 at 0:12
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    $\begingroup$ I'm reading an article about rank minimization where they use this fact, i think about caracteristic polynomial of transpose(X)*X, can we say that if a Polynom P converge to Q, then it's the same thing for the roots? $\endgroup$ – user47204 Mar 25 '14 at 9:40
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This is stated by Weyl's inequality: for any two matrices $A$ and $\tilde A$ which are related by $A - \tilde A = E$, their corresponding singular values satisfy $$ | \tilde\sigma_i - \sigma_i | \le \| E \|_2. $$

This is stated in http://www.math.msu.edu/~markiwen/Teaching/MTH995/Papers/SVD_Stewart.pdf, where also a reference is given. If you want to prove it yourself, it's rather easy when using the Courant-Fischer min-max principle as a starting point.

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