Eigenvalues of two symmetric matrices












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For any two symmetric $ntimes n$ matrices $A$ and $B$ be their eigenvalues be ordered from largest to smallest . How to prove that for eigenvalues $|lambda _k^A-lambda _k^B|le ||A-B||$(operator Norm) for $1le kle n$. Where $lambda _k^A,lambda _k^B$ are respective eigenvalues of $A$ and $B$





I spend many time in this question but don't know how prove this !!!










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  • 3




    There are several matrix norms. Which one do you use ?
    – Jean Marie
    Dec 4 '18 at 20:53










  • I'm not sure but maybe you can use decomposition for symmetric matrices. And then use the same norm for the product. And finally use the property that in decomposition eigenvectors are orthogonal to each other.
    – kolobokish
    Dec 4 '18 at 21:22


















1














For any two symmetric $ntimes n$ matrices $A$ and $B$ be their eigenvalues be ordered from largest to smallest . How to prove that for eigenvalues $|lambda _k^A-lambda _k^B|le ||A-B||$(operator Norm) for $1le kle n$. Where $lambda _k^A,lambda _k^B$ are respective eigenvalues of $A$ and $B$





I spend many time in this question but don't know how prove this !!!










share|cite|improve this question




















  • 3




    There are several matrix norms. Which one do you use ?
    – Jean Marie
    Dec 4 '18 at 20:53










  • I'm not sure but maybe you can use decomposition for symmetric matrices. And then use the same norm for the product. And finally use the property that in decomposition eigenvectors are orthogonal to each other.
    – kolobokish
    Dec 4 '18 at 21:22
















1












1








1


1





For any two symmetric $ntimes n$ matrices $A$ and $B$ be their eigenvalues be ordered from largest to smallest . How to prove that for eigenvalues $|lambda _k^A-lambda _k^B|le ||A-B||$(operator Norm) for $1le kle n$. Where $lambda _k^A,lambda _k^B$ are respective eigenvalues of $A$ and $B$





I spend many time in this question but don't know how prove this !!!










share|cite|improve this question















For any two symmetric $ntimes n$ matrices $A$ and $B$ be their eigenvalues be ordered from largest to smallest . How to prove that for eigenvalues $|lambda _k^A-lambda _k^B|le ||A-B||$(operator Norm) for $1le kle n$. Where $lambda _k^A,lambda _k^B$ are respective eigenvalues of $A$ and $B$





I spend many time in this question but don't know how prove this !!!







linear-algebra






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share|cite|improve this question













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edited Dec 4 '18 at 20:55







G C R

















asked Dec 4 '18 at 20:49









G C RG C R

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215








  • 3




    There are several matrix norms. Which one do you use ?
    – Jean Marie
    Dec 4 '18 at 20:53










  • I'm not sure but maybe you can use decomposition for symmetric matrices. And then use the same norm for the product. And finally use the property that in decomposition eigenvectors are orthogonal to each other.
    – kolobokish
    Dec 4 '18 at 21:22
















  • 3




    There are several matrix norms. Which one do you use ?
    – Jean Marie
    Dec 4 '18 at 20:53










  • I'm not sure but maybe you can use decomposition for symmetric matrices. And then use the same norm for the product. And finally use the property that in decomposition eigenvectors are orthogonal to each other.
    – kolobokish
    Dec 4 '18 at 21:22










3




3




There are several matrix norms. Which one do you use ?
– Jean Marie
Dec 4 '18 at 20:53




There are several matrix norms. Which one do you use ?
– Jean Marie
Dec 4 '18 at 20:53












I'm not sure but maybe you can use decomposition for symmetric matrices. And then use the same norm for the product. And finally use the property that in decomposition eigenvectors are orthogonal to each other.
– kolobokish
Dec 4 '18 at 21:22






I'm not sure but maybe you can use decomposition for symmetric matrices. And then use the same norm for the product. And finally use the property that in decomposition eigenvectors are orthogonal to each other.
– kolobokish
Dec 4 '18 at 21:22












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