Approximation of a matrix in the power method
$begingroup$
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
asked Dec 18 '18 at 10:34
user627482user627482
464
$endgroup$
add a comment |
$begingroup$
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
asked Dec 18 '18 at 10:34
user627482user627482
464
$endgroup$
add a comment |
$begingroup$
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
asked Dec 18 '18 at 10:34
user627482user627482
464
$endgroup$
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
asked Dec 18 '18 at 10:34
user627482user627482
464
asked Dec 18 '18 at 10:34
user627482user627482
464
edited Dec 18 '18 at 13:37
user627482
asked Dec 18 '18 at 10:34
user627482user627482
464
asked Dec 18 '18 at 10:34
user627482user627482
464
asked Dec 18 '18 at 10:34
user627482user627482
464
464
add a comment |
add a comment |
1 Answer
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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
answered Dec 18 '18 at 17:00
user627482user627482
464
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add a comment |
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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
answered Dec 18 '18 at 17:00
user627482user627482
464
$endgroup$
add a comment |
$begingroup$
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
answered Dec 18 '18 at 17:00
user627482user627482
464
$endgroup$
add a comment |
$begingroup$
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
answered Dec 18 '18 at 17:00
user627482user627482
464
$endgroup$
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
answered Dec 18 '18 at 17:00
user627482user627482
464
answered Dec 18 '18 at 17:00
user627482user627482
464
answered Dec 18 '18 at 17:00
user627482user627482
464
answered Dec 18 '18 at 17:00
user627482user627482
464
464
add a comment |
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