Estimate the number of iterations












0












$begingroup$


Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Any hint on how to solve this problem?
    $endgroup$
    – Jiexiong687691
    Nov 29 '18 at 9:08
















0












$begingroup$


Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Any hint on how to solve this problem?
    $endgroup$
    – Jiexiong687691
    Nov 29 '18 at 9:08














0












0








0





$begingroup$


Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra







share|cite|improve this question











$endgroup$




Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra




linear-algebra numerical-linear-algebra






share|cite|improve this question















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edited Dec 3 '18 at 8:11







Jiexiong687691

















asked Nov 29 '18 at 8:09









Jiexiong687691Jiexiong687691

825




825












  • $begingroup$
    Any hint on how to solve this problem?
    $endgroup$
    – Jiexiong687691
    Nov 29 '18 at 9:08


















  • $begingroup$
    Any hint on how to solve this problem?
    $endgroup$
    – Jiexiong687691
    Nov 29 '18 at 9:08
















$begingroup$
Any hint on how to solve this problem?
$endgroup$
– Jiexiong687691
Nov 29 '18 at 9:08




$begingroup$
Any hint on how to solve this problem?
$endgroup$
– Jiexiong687691
Nov 29 '18 at 9:08










1 Answer
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$begingroup$

Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






share|cite|improve this answer









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    $begingroup$

    Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



    Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



    If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



      Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



      If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



        Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



        If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






        share|cite|improve this answer









        $endgroup$



        Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



        Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



        If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 11:24









        loup blancloup blanc

        23.1k21850




        23.1k21850






























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