On an inequality involving operator norm of matrices and singular value












3












$begingroup$


Let $A, E in M_n(mathbb C)$ be as in this question On invertibility of $A+E$ where $||E||_2<$ smallest singular value of $A$ and $||A^{-1}E||_2<1$ .



How to prove that $dfrac {||A^{-1}b-(A+E)^{-1}b||_2}{||A^{-1}b||_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}$ ?






linear-algebra matrices norm svd singularvalues







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  • $begingroup$
    I don't have a whole lot of time to figure it out, but have you tried anything using the geometric series representation of the inverse when the spectral radius is less than one? The right hand side looks oddly reminiscent of the value obtained when summing up such a series.
    $endgroup$
    – OldGodzilla
    Jan 16 at 16:32
















3












$begingroup$


Let $A, E in M_n(mathbb C)$ be as in this question On invertibility of $A+E$ where $||E||_2<$ smallest singular value of $A$ and $||A^{-1}E||_2<1$ .



How to prove that $dfrac {||A^{-1}b-(A+E)^{-1}b||_2}{||A^{-1}b||_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}$ ?






linear-algebra matrices norm svd singularvalues







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't have a whole lot of time to figure it out, but have you tried anything using the geometric series representation of the inverse when the spectral radius is less than one? The right hand side looks oddly reminiscent of the value obtained when summing up such a series.
    $endgroup$
    – OldGodzilla
    Jan 16 at 16:32














3












3








3





$begingroup$


Let $A, E in M_n(mathbb C)$ be as in this question On invertibility of $A+E$ where $||E||_2<$ smallest singular value of $A$ and $||A^{-1}E||_2<1$ .



How to prove that $dfrac {||A^{-1}b-(A+E)^{-1}b||_2}{||A^{-1}b||_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}$ ?






linear-algebra matrices norm svd singularvalues







share|cite|improve this question











$endgroup$




Let $A, E in M_n(mathbb C)$ be as in this question On invertibility of $A+E$ where $||E||_2<$ smallest singular value of $A$ and $||A^{-1}E||_2<1$ .



How to prove that $dfrac {||A^{-1}b-(A+E)^{-1}b||_2}{||A^{-1}b||_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}$ ?






linear-algebra matrices norm svd singularvalues




linear-algebra matrices norm svd singularvalues






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share|cite|improve this question













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edited Jan 12 at 7:36







user521337

















asked Jan 12 at 7:00









user521337user521337

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1,2201417












  • $begingroup$
    I don't have a whole lot of time to figure it out, but have you tried anything using the geometric series representation of the inverse when the spectral radius is less than one? The right hand side looks oddly reminiscent of the value obtained when summing up such a series.
    $endgroup$
    – OldGodzilla
    Jan 16 at 16:32


















  • $begingroup$
    I don't have a whole lot of time to figure it out, but have you tried anything using the geometric series representation of the inverse when the spectral radius is less than one? The right hand side looks oddly reminiscent of the value obtained when summing up such a series.
    $endgroup$
    – OldGodzilla
    Jan 16 at 16:32
















$begingroup$
I don't have a whole lot of time to figure it out, but have you tried anything using the geometric series representation of the inverse when the spectral radius is less than one? The right hand side looks oddly reminiscent of the value obtained when summing up such a series.
$endgroup$
– OldGodzilla
Jan 16 at 16:32




$begingroup$
I don't have a whole lot of time to figure it out, but have you tried anything using the geometric series representation of the inverse when the spectral radius is less than one? The right hand side looks oddly reminiscent of the value obtained when summing up such a series.
$endgroup$
– OldGodzilla
Jan 16 at 16:32










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I assume that $bin Bbb C^n$ is an arbitrary non-zero vector and $$sigma_min(M)=inf {|Mx|_2: xinBbb C^nmbox{ and }|x|_2=1}$$ for each $MinBbb M_n(Bbb C)$. Put $c=(A+E)^{-1}b$. Then $A^{-1}b=A^{-1}(A+E)c=c+A^{-1}Ec$. Now we have to prove that



$$dfrac {|A^{-1}Ec |_2}{| c+A^{-1}Ec |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



Put $d=frac{c}{|c|_2}$. It suffices to prove that



$$dfrac {|A^{-1}Ed |_2}{| d+A^{-1}Ed |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



Since $|A^{-1}Ed |_2le ||E||_2||A^{-1}||_2$, it suffices to prove that



$$1-frac {||E||_2}{sigma_min}le | d+A^{-1}Ed |_2.$$



Since $| d+A^{-1}Ed |_2ge | d|_2- |A^{-1}Ed |_2=1-| A^{-1}Ed |_2$, it suffices to prove that



$${sigma_min}| A^{-1}Ed |_2le ||E||_2.$$



Since $$| A^{-1}Ed |_2le | A^{-1} |_2|Ed |_2le | A^{-1} |_2|E |_2|d|_2=| A^{-1} |_2|E |_2,$$



It suffices to check that $${sigma_min}| A^{-1}|_2le 1.$$



Indeed,



$$sigma_min| A^{-1}|_2=sigma_minsup {|A^{-1}x|_2: xinBbb C^nmbox{ and }|x|_2=1}.$$



Let $xinBbb C^n$ and $|x|_2=1$. We have $$1=|x|_2=|AA^{-1}x|_2ge sigma_{min} |A^{-1}x|_2,$$ so $sigma_min| A^{-1}|_2le 1$.






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    $begingroup$

    I assume that $bin Bbb C^n$ is an arbitrary non-zero vector and $$sigma_min(M)=inf {|Mx|_2: xinBbb C^nmbox{ and }|x|_2=1}$$ for each $MinBbb M_n(Bbb C)$. Put $c=(A+E)^{-1}b$. Then $A^{-1}b=A^{-1}(A+E)c=c+A^{-1}Ec$. Now we have to prove that



    $$dfrac {|A^{-1}Ec |_2}{| c+A^{-1}Ec |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



    Put $d=frac{c}{|c|_2}$. It suffices to prove that



    $$dfrac {|A^{-1}Ed |_2}{| d+A^{-1}Ed |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



    Since $|A^{-1}Ed |_2le ||E||_2||A^{-1}||_2$, it suffices to prove that



    $$1-frac {||E||_2}{sigma_min}le | d+A^{-1}Ed |_2.$$



    Since $| d+A^{-1}Ed |_2ge | d|_2- |A^{-1}Ed |_2=1-| A^{-1}Ed |_2$, it suffices to prove that



    $${sigma_min}| A^{-1}Ed |_2le ||E||_2.$$



    Since $$| A^{-1}Ed |_2le | A^{-1} |_2|Ed |_2le | A^{-1} |_2|E |_2|d|_2=| A^{-1} |_2|E |_2,$$



    It suffices to check that $${sigma_min}| A^{-1}|_2le 1.$$



    Indeed,



    $$sigma_min| A^{-1}|_2=sigma_minsup {|A^{-1}x|_2: xinBbb C^nmbox{ and }|x|_2=1}.$$



    Let $xinBbb C^n$ and $|x|_2=1$. We have $$1=|x|_2=|AA^{-1}x|_2ge sigma_{min} |A^{-1}x|_2,$$ so $sigma_min| A^{-1}|_2le 1$.






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      $begingroup$

      I assume that $bin Bbb C^n$ is an arbitrary non-zero vector and $$sigma_min(M)=inf {|Mx|_2: xinBbb C^nmbox{ and }|x|_2=1}$$ for each $MinBbb M_n(Bbb C)$. Put $c=(A+E)^{-1}b$. Then $A^{-1}b=A^{-1}(A+E)c=c+A^{-1}Ec$. Now we have to prove that



      $$dfrac {|A^{-1}Ec |_2}{| c+A^{-1}Ec |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



      Put $d=frac{c}{|c|_2}$. It suffices to prove that



      $$dfrac {|A^{-1}Ed |_2}{| d+A^{-1}Ed |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



      Since $|A^{-1}Ed |_2le ||E||_2||A^{-1}||_2$, it suffices to prove that



      $$1-frac {||E||_2}{sigma_min}le | d+A^{-1}Ed |_2.$$



      Since $| d+A^{-1}Ed |_2ge | d|_2- |A^{-1}Ed |_2=1-| A^{-1}Ed |_2$, it suffices to prove that



      $${sigma_min}| A^{-1}Ed |_2le ||E||_2.$$



      Since $$| A^{-1}Ed |_2le | A^{-1} |_2|Ed |_2le | A^{-1} |_2|E |_2|d|_2=| A^{-1} |_2|E |_2,$$



      It suffices to check that $${sigma_min}| A^{-1}|_2le 1.$$



      Indeed,



      $$sigma_min| A^{-1}|_2=sigma_minsup {|A^{-1}x|_2: xinBbb C^nmbox{ and }|x|_2=1}.$$



      Let $xinBbb C^n$ and $|x|_2=1$. We have $$1=|x|_2=|AA^{-1}x|_2ge sigma_{min} |A^{-1}x|_2,$$ so $sigma_min| A^{-1}|_2le 1$.






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        $begingroup$

        I assume that $bin Bbb C^n$ is an arbitrary non-zero vector and $$sigma_min(M)=inf {|Mx|_2: xinBbb C^nmbox{ and }|x|_2=1}$$ for each $MinBbb M_n(Bbb C)$. Put $c=(A+E)^{-1}b$. Then $A^{-1}b=A^{-1}(A+E)c=c+A^{-1}Ec$. Now we have to prove that



        $$dfrac {|A^{-1}Ec |_2}{| c+A^{-1}Ec |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



        Put $d=frac{c}{|c|_2}$. It suffices to prove that



        $$dfrac {|A^{-1}Ed |_2}{| d+A^{-1}Ed |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



        Since $|A^{-1}Ed |_2le ||E||_2||A^{-1}||_2$, it suffices to prove that



        $$1-frac {||E||_2}{sigma_min}le | d+A^{-1}Ed |_2.$$



        Since $| d+A^{-1}Ed |_2ge | d|_2- |A^{-1}Ed |_2=1-| A^{-1}Ed |_2$, it suffices to prove that



        $${sigma_min}| A^{-1}Ed |_2le ||E||_2.$$



        Since $$| A^{-1}Ed |_2le | A^{-1} |_2|Ed |_2le | A^{-1} |_2|E |_2|d|_2=| A^{-1} |_2|E |_2,$$



        It suffices to check that $${sigma_min}| A^{-1}|_2le 1.$$



        Indeed,



        $$sigma_min| A^{-1}|_2=sigma_minsup {|A^{-1}x|_2: xinBbb C^nmbox{ and }|x|_2=1}.$$



        Let $xinBbb C^n$ and $|x|_2=1$. We have $$1=|x|_2=|AA^{-1}x|_2ge sigma_{min} |A^{-1}x|_2,$$ so $sigma_min| A^{-1}|_2le 1$.






        share|cite|improve this answer









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        I assume that $bin Bbb C^n$ is an arbitrary non-zero vector and $$sigma_min(M)=inf {|Mx|_2: xinBbb C^nmbox{ and }|x|_2=1}$$ for each $MinBbb M_n(Bbb C)$. Put $c=(A+E)^{-1}b$. Then $A^{-1}b=A^{-1}(A+E)c=c+A^{-1}Ec$. Now we have to prove that



        $$dfrac {|A^{-1}Ec |_2}{| c+A^{-1}Ec |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



        Put $d=frac{c}{|c|_2}$. It suffices to prove that



        $$dfrac {|A^{-1}Ed |_2}{| d+A^{-1}Ed |_2}le dfrac {||E||_2||A^{-1}||_2}{1-frac {||E||_2}{sigma_min}}.$$



        Since $|A^{-1}Ed |_2le ||E||_2||A^{-1}||_2$, it suffices to prove that



        $$1-frac {||E||_2}{sigma_min}le | d+A^{-1}Ed |_2.$$



        Since $| d+A^{-1}Ed |_2ge | d|_2- |A^{-1}Ed |_2=1-| A^{-1}Ed |_2$, it suffices to prove that



        $${sigma_min}| A^{-1}Ed |_2le ||E||_2.$$



        Since $$| A^{-1}Ed |_2le | A^{-1} |_2|Ed |_2le | A^{-1} |_2|E |_2|d|_2=| A^{-1} |_2|E |_2,$$



        It suffices to check that $${sigma_min}| A^{-1}|_2le 1.$$



        Indeed,



        $$sigma_min| A^{-1}|_2=sigma_minsup {|A^{-1}x|_2: xinBbb C^nmbox{ and }|x|_2=1}.$$



        Let $xinBbb C^n$ and $|x|_2=1$. We have $$1=|x|_2=|AA^{-1}x|_2ge sigma_{min} |A^{-1}x|_2,$$ so $sigma_min| A^{-1}|_2le 1$.







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        answered Jan 17 at 11:29









        Alex RavskyAlex Ravsky

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