FInding smallest eigenvalues using Lanczos algorithm
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I have some trouble understanding how Lanczos algorithm works for finding $K$ smallest eigenvalues of some large symmetric matrix $A$. For example if I want to calculate 50 smallest eigenvalues of $1000 times 1000$ matrix $A$ how shall I proceed? Right now, I think that first I need to set some number of iterations for Lanczos algorithm. For example $100$ iterations. At the end I will obtain $100times 100$ tridiagonal matrix $T$ and after diagonalizing this matrix I will get 100 eigenvalues that are close to eigenvalues of $A$. From these 100 eigenvalues I will choose $50$ smallest and I will hope that these $50$ smallest eigenvalues of $T$ approximate well enough $50$ smalees eigenvalues of $A$. I really doubt that this is correct, but after googling for some time I didnt find a better answer. Could you please explain me what shall I do?
numerical-linear-algebra
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
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add a comment |
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I have some trouble understanding how Lanczos algorithm works for finding $K$ smallest eigenvalues of some large symmetric matrix $A$. For example if I want to calculate 50 smallest eigenvalues of $1000 times 1000$ matrix $A$ how shall I proceed? Right now, I think that first I need to set some number of iterations for Lanczos algorithm. For example $100$ iterations. At the end I will obtain $100times 100$ tridiagonal matrix $T$ and after diagonalizing this matrix I will get 100 eigenvalues that are close to eigenvalues of $A$. From these 100 eigenvalues I will choose $50$ smallest and I will hope that these $50$ smallest eigenvalues of $T$ approximate well enough $50$ smalees eigenvalues of $A$. I really doubt that this is correct, but after googling for some time I didnt find a better answer. Could you please explain me what shall I do?
numerical-linear-algebra
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
$endgroup$
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Finding the smallest eigenvalues (in magnitude) using Lanczos is normally coupled with inverse iteration which turns the problem into finding the largest ones where Lanczos works much better.
$endgroup$
– Algebraic Pavel
Dec 20 '18 at 14:11
add a comment |
$begingroup$
I have some trouble understanding how Lanczos algorithm works for finding $K$ smallest eigenvalues of some large symmetric matrix $A$. For example if I want to calculate 50 smallest eigenvalues of $1000 times 1000$ matrix $A$ how shall I proceed? Right now, I think that first I need to set some number of iterations for Lanczos algorithm. For example $100$ iterations. At the end I will obtain $100times 100$ tridiagonal matrix $T$ and after diagonalizing this matrix I will get 100 eigenvalues that are close to eigenvalues of $A$. From these 100 eigenvalues I will choose $50$ smallest and I will hope that these $50$ smallest eigenvalues of $T$ approximate well enough $50$ smalees eigenvalues of $A$. I really doubt that this is correct, but after googling for some time I didnt find a better answer. Could you please explain me what shall I do?
numerical-linear-algebra
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
$endgroup$
I have some trouble understanding how Lanczos algorithm works for finding $K$ smallest eigenvalues of some large symmetric matrix $A$. For example if I want to calculate 50 smallest eigenvalues of $1000 times 1000$ matrix $A$ how shall I proceed? Right now, I think that first I need to set some number of iterations for Lanczos algorithm. For example $100$ iterations. At the end I will obtain $100times 100$ tridiagonal matrix $T$ and after diagonalizing this matrix I will get 100 eigenvalues that are close to eigenvalues of $A$. From these 100 eigenvalues I will choose $50$ smallest and I will hope that these $50$ smallest eigenvalues of $T$ approximate well enough $50$ smalees eigenvalues of $A$. I really doubt that this is correct, but after googling for some time I didnt find a better answer. Could you please explain me what shall I do?
numerical-linear-algebra
numerical-linear-algebra
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
asked Dec 19 '18 at 17:54
Studying OptimizationStudying Optimization
727
727
$begingroup$
Finding the smallest eigenvalues (in magnitude) using Lanczos is normally coupled with inverse iteration which turns the problem into finding the largest ones where Lanczos works much better.
$endgroup$
– Algebraic Pavel
Dec 20 '18 at 14:11
add a comment |
$begingroup$
Finding the smallest eigenvalues (in magnitude) using Lanczos is normally coupled with inverse iteration which turns the problem into finding the largest ones where Lanczos works much better.
$endgroup$
– Algebraic Pavel
Dec 20 '18 at 14:11
$begingroup$
Finding the smallest eigenvalues (in magnitude) using Lanczos is normally coupled with inverse iteration which turns the problem into finding the largest ones where Lanczos works much better.
$endgroup$
– Algebraic Pavel
Dec 20 '18 at 14:11
$begingroup$
Finding the smallest eigenvalues (in magnitude) using Lanczos is normally coupled with inverse iteration which turns the problem into finding the largest ones where Lanczos works much better.
$endgroup$
– Algebraic Pavel
Dec 20 '18 at 14:11
add a comment |
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$begingroup$
Finding the smallest eigenvalues (in magnitude) using Lanczos is normally coupled with inverse iteration which turns the problem into finding the largest ones where Lanczos works much better.
$endgroup$
– Algebraic Pavel
Dec 20 '18 at 14:11