Approximation of a matrix in the power method












2












$begingroup$


Here's the text of the problem ( here $lVertcdotrVert$ denotes any matrix induced norm):
Let be $Ain mathbb{R}^{ntimes n}$ a diagonalisable matrix $ntimes n$, with $lambda_{1}, lambda_{2},ldots,lambda_{n}$ real eigenvalues, such that $$lvertlambda_{1}rvert>lvertlambda_{2}rvertgeqldotsgeqlvertlambda_{n}rvert$$
with eigenvectors $v_1,ldots,v_n$. Let $x_0=sum_{i=1}^{n}alpha_i v_i$ ($alpha_1neq0)$ and define recursively
$$x_{k+1}=Ax_k $$ Moreover let $mu_k=(w^TAx_k)/(w^Tx_k)$, where $w$ is a vector in $mathbb{R}^n$ such that $w^Tv_1neq0$.
Prove that:



1) for all $varepsilon$, it exists a $bar{k}$ such that
$$ lVert Ax_k- mu_k x_k rVert<varepsilonlVert x_krVert$$
for every $k>bar{k}$;



2) Suppose that the previous inequality holds for a fixed $varepsilon$ and a fixed $k$. Show that it exists a matrix $tilde{A}$ such that $tilde{A}x_k=mu_k x_k$ and
$$ lVert A-tilde{A}rVert<varepsilon$$
End of exercise.



The point 1 is trivial, but on the other hand I don't know how to approach the second point. I noted that the matrix $tilde{A}=(x_k w^TA)/(w^T x_k)$ (where $x_k w^TA$ is the product of a column for a row) satisfies the condition $tilde{A}x_k=mu_k x_k$, but I don't know how to prove the inequality with the norm (assuming that this matrix works).
Someone could help me?






real-analysis eigenvalues-eigenvectors numerical-linear-algebra matrix-calculus







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Here's the text of the problem ( here $lVertcdotrVert$ denotes any matrix induced norm):
    Let be $Ain mathbb{R}^{ntimes n}$ a diagonalisable matrix $ntimes n$, with $lambda_{1}, lambda_{2},ldots,lambda_{n}$ real eigenvalues, such that $$lvertlambda_{1}rvert>lvertlambda_{2}rvertgeqldotsgeqlvertlambda_{n}rvert$$
    with eigenvectors $v_1,ldots,v_n$. Let $x_0=sum_{i=1}^{n}alpha_i v_i$ ($alpha_1neq0)$ and define recursively
    $$x_{k+1}=Ax_k $$ Moreover let $mu_k=(w^TAx_k)/(w^Tx_k)$, where $w$ is a vector in $mathbb{R}^n$ such that $w^Tv_1neq0$.
    Prove that:



    1) for all $varepsilon$, it exists a $bar{k}$ such that
    $$ lVert Ax_k- mu_k x_k rVert<varepsilonlVert x_krVert$$
    for every $k>bar{k}$;



    2) Suppose that the previous inequality holds for a fixed $varepsilon$ and a fixed $k$. Show that it exists a matrix $tilde{A}$ such that $tilde{A}x_k=mu_k x_k$ and
    $$ lVert A-tilde{A}rVert<varepsilon$$
    End of exercise.



    The point 1 is trivial, but on the other hand I don't know how to approach the second point. I noted that the matrix $tilde{A}=(x_k w^TA)/(w^T x_k)$ (where $x_k w^TA$ is the product of a column for a row) satisfies the condition $tilde{A}x_k=mu_k x_k$, but I don't know how to prove the inequality with the norm (assuming that this matrix works).
    Someone could help me?






    real-analysis eigenvalues-eigenvectors numerical-linear-algebra matrix-calculus







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Here's the text of the problem ( here $lVertcdotrVert$ denotes any matrix induced norm):
      Let be $Ain mathbb{R}^{ntimes n}$ a diagonalisable matrix $ntimes n$, with $lambda_{1}, lambda_{2},ldots,lambda_{n}$ real eigenvalues, such that $$lvertlambda_{1}rvert>lvertlambda_{2}rvertgeqldotsgeqlvertlambda_{n}rvert$$
      with eigenvectors $v_1,ldots,v_n$. Let $x_0=sum_{i=1}^{n}alpha_i v_i$ ($alpha_1neq0)$ and define recursively
      $$x_{k+1}=Ax_k $$ Moreover let $mu_k=(w^TAx_k)/(w^Tx_k)$, where $w$ is a vector in $mathbb{R}^n$ such that $w^Tv_1neq0$.
      Prove that:



      1) for all $varepsilon$, it exists a $bar{k}$ such that
      $$ lVert Ax_k- mu_k x_k rVert<varepsilonlVert x_krVert$$
      for every $k>bar{k}$;



      2) Suppose that the previous inequality holds for a fixed $varepsilon$ and a fixed $k$. Show that it exists a matrix $tilde{A}$ such that $tilde{A}x_k=mu_k x_k$ and
      $$ lVert A-tilde{A}rVert<varepsilon$$
      End of exercise.



      The point 1 is trivial, but on the other hand I don't know how to approach the second point. I noted that the matrix $tilde{A}=(x_k w^TA)/(w^T x_k)$ (where $x_k w^TA$ is the product of a column for a row) satisfies the condition $tilde{A}x_k=mu_k x_k$, but I don't know how to prove the inequality with the norm (assuming that this matrix works).
      Someone could help me?






      real-analysis eigenvalues-eigenvectors numerical-linear-algebra matrix-calculus







      share|cite|improve this question











      $endgroup$




      Here's the text of the problem ( here $lVertcdotrVert$ denotes any matrix induced norm):
      Let be $Ain mathbb{R}^{ntimes n}$ a diagonalisable matrix $ntimes n$, with $lambda_{1}, lambda_{2},ldots,lambda_{n}$ real eigenvalues, such that $$lvertlambda_{1}rvert>lvertlambda_{2}rvertgeqldotsgeqlvertlambda_{n}rvert$$
      with eigenvectors $v_1,ldots,v_n$. Let $x_0=sum_{i=1}^{n}alpha_i v_i$ ($alpha_1neq0)$ and define recursively
      $$x_{k+1}=Ax_k $$ Moreover let $mu_k=(w^TAx_k)/(w^Tx_k)$, where $w$ is a vector in $mathbb{R}^n$ such that $w^Tv_1neq0$.
      Prove that:



      1) for all $varepsilon$, it exists a $bar{k}$ such that
      $$ lVert Ax_k- mu_k x_k rVert<varepsilonlVert x_krVert$$
      for every $k>bar{k}$;



      2) Suppose that the previous inequality holds for a fixed $varepsilon$ and a fixed $k$. Show that it exists a matrix $tilde{A}$ such that $tilde{A}x_k=mu_k x_k$ and
      $$ lVert A-tilde{A}rVert<varepsilon$$
      End of exercise.



      The point 1 is trivial, but on the other hand I don't know how to approach the second point. I noted that the matrix $tilde{A}=(x_k w^TA)/(w^T x_k)$ (where $x_k w^TA$ is the product of a column for a row) satisfies the condition $tilde{A}x_k=mu_k x_k$, but I don't know how to prove the inequality with the norm (assuming that this matrix works).
      Someone could help me?






      real-analysis eigenvalues-eigenvectors numerical-linear-algebra matrix-calculus




      real-analysis eigenvalues-eigenvectors numerical-linear-algebra matrix-calculus






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      edited Dec 18 '18 at 13:37







      user627482

















      asked Dec 18 '18 at 10:34









      user627482user627482

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          $begingroup$

          But in the exercise I don't have $lVertcdotrVert_2$ but a generic induced norm, so $lVert rx^TrVert$ is equal to $lVert rlVertlVert xrVert^{ast}$, where $lVertcdotrVert^{ast}$ is the dual norm






          share|cite|improve this answer









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            $begingroup$

            But in the exercise I don't have $lVertcdotrVert_2$ but a generic induced norm, so $lVert rx^TrVert$ is equal to $lVert rlVertlVert xrVert^{ast}$, where $lVertcdotrVert^{ast}$ is the dual norm






            share|cite|improve this answer









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              0












              $begingroup$

              But in the exercise I don't have $lVertcdotrVert_2$ but a generic induced norm, so $lVert rx^TrVert$ is equal to $lVert rlVertlVert xrVert^{ast}$, where $lVertcdotrVert^{ast}$ is the dual norm






              share|cite|improve this answer









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                0





                $begingroup$

                But in the exercise I don't have $lVertcdotrVert_2$ but a generic induced norm, so $lVert rx^TrVert$ is equal to $lVert rlVertlVert xrVert^{ast}$, where $lVertcdotrVert^{ast}$ is the dual norm






                share|cite|improve this answer









                $endgroup$



                But in the exercise I don't have $lVertcdotrVert_2$ but a generic induced norm, so $lVert rx^TrVert$ is equal to $lVert rlVertlVert xrVert^{ast}$, where $lVertcdotrVert^{ast}$ is the dual norm







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 17:00









                user627482user627482

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