Distance from eigenspace of matrix
$begingroup$
In linear algebra, is there a separate name / concept for the notion of distance between linear vector subspaces?
I'm asking this because I'm considering a problem in numerical linear algebra where a Krylov subspace iterative method is used. Since for every subsequent $n$ a Krylov subspace method implicitly generates an additional basis vector in Krylov subspace, which approaches the eigenspace of the matrix for which the problem $$Ax=b$$ is being solved, it must be true that if $b$ is in the span of the eigenspace of $A$ then the convergence will happen faster.
But what if $b$ is very "far" from the eigenspace? I'm trying to think about what the notion of a distance between two vector subspaces could mean or how it could be defined. Would a vector $b$ contained in a subspace "far away" from the eigenspace of $A$ make iteration of a Krylov subspace method take longer than in a general case?
linear-algebra eigenvalues-eigenvectors algorithms terminology numerical-linear-algebra
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
$endgroup$
add a comment |
$begingroup$
In linear algebra, is there a separate name / concept for the notion of distance between linear vector subspaces?
I'm asking this because I'm considering a problem in numerical linear algebra where a Krylov subspace iterative method is used. Since for every subsequent $n$ a Krylov subspace method implicitly generates an additional basis vector in Krylov subspace, which approaches the eigenspace of the matrix for which the problem $$Ax=b$$ is being solved, it must be true that if $b$ is in the span of the eigenspace of $A$ then the convergence will happen faster.
But what if $b$ is very "far" from the eigenspace? I'm trying to think about what the notion of a distance between two vector subspaces could mean or how it could be defined. Would a vector $b$ contained in a subspace "far away" from the eigenspace of $A$ make iteration of a Krylov subspace method take longer than in a general case?
linear-algebra eigenvalues-eigenvectors algorithms terminology numerical-linear-algebra
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
$endgroup$
1
$begingroup$
A distance between two planes in $mathbb{R}^3$ can be defined by $|sin(theta)|$ where $theta$ is the angle between their normals (in particular, the triangular inequality is verified). I have the remembrance that such a result is generalizable, but I must look for references.
$endgroup$
– Jean Marie
Dec 9 '18 at 11:22
add a comment |
$begingroup$
In linear algebra, is there a separate name / concept for the notion of distance between linear vector subspaces?
I'm asking this because I'm considering a problem in numerical linear algebra where a Krylov subspace iterative method is used. Since for every subsequent $n$ a Krylov subspace method implicitly generates an additional basis vector in Krylov subspace, which approaches the eigenspace of the matrix for which the problem $$Ax=b$$ is being solved, it must be true that if $b$ is in the span of the eigenspace of $A$ then the convergence will happen faster.
But what if $b$ is very "far" from the eigenspace? I'm trying to think about what the notion of a distance between two vector subspaces could mean or how it could be defined. Would a vector $b$ contained in a subspace "far away" from the eigenspace of $A$ make iteration of a Krylov subspace method take longer than in a general case?
linear-algebra eigenvalues-eigenvectors algorithms terminology numerical-linear-algebra
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
$endgroup$
In linear algebra, is there a separate name / concept for the notion of distance between linear vector subspaces?
I'm asking this because I'm considering a problem in numerical linear algebra where a Krylov subspace iterative method is used. Since for every subsequent $n$ a Krylov subspace method implicitly generates an additional basis vector in Krylov subspace, which approaches the eigenspace of the matrix for which the problem $$Ax=b$$ is being solved, it must be true that if $b$ is in the span of the eigenspace of $A$ then the convergence will happen faster.
But what if $b$ is very "far" from the eigenspace? I'm trying to think about what the notion of a distance between two vector subspaces could mean or how it could be defined. Would a vector $b$ contained in a subspace "far away" from the eigenspace of $A$ make iteration of a Krylov subspace method take longer than in a general case?
linear-algebra eigenvalues-eigenvectors algorithms terminology numerical-linear-algebra
linear-algebra eigenvalues-eigenvectors algorithms terminology numerical-linear-algebra
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
edited Dec 9 '18 at 11:27
Omnomnomnom
129k792185
129k792185
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
asked Dec 9 '18 at 9:17
sequencesequence
4,27331437
4,27331437
1
$begingroup$
A distance between two planes in $mathbb{R}^3$ can be defined by $|sin(theta)|$ where $theta$ is the angle between their normals (in particular, the triangular inequality is verified). I have the remembrance that such a result is generalizable, but I must look for references.
$endgroup$
– Jean Marie
Dec 9 '18 at 11:22
add a comment |
1
$begingroup$
A distance between two planes in $mathbb{R}^3$ can be defined by $|sin(theta)|$ where $theta$ is the angle between their normals (in particular, the triangular inequality is verified). I have the remembrance that such a result is generalizable, but I must look for references.
$endgroup$
– Jean Marie
Dec 9 '18 at 11:22
1
1
$begingroup$
A distance between two planes in $mathbb{R}^3$ can be defined by $|sin(theta)|$ where $theta$ is the angle between their normals (in particular, the triangular inequality is verified). I have the remembrance that such a result is generalizable, but I must look for references.
$endgroup$
– Jean Marie
Dec 9 '18 at 11:22
$begingroup$
A distance between two planes in $mathbb{R}^3$ can be defined by $|sin(theta)|$ where $theta$ is the angle between their normals (in particular, the triangular inequality is verified). I have the remembrance that such a result is generalizable, but I must look for references.
$endgroup$
– Jean Marie
Dec 9 '18 at 11:22
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The common notion of distance is to consider an orthogonal projection $P$ onto the first linear subspace $V$, and an orthogonal projection $Q$ onto the other subspace $W$.
At this point we can define
$$d(V,W) = | P - Q |$$ as the distance between these subspaces, where the norm used is the operator norm. For properties and applications see Section 2.5.3 of Golub and Van Loan.
This distance metric is used throughout GVL’s exposition on unsymmetrical eigenvalue problems (which involve Krylov methods) — see Chapter 7.
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
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add a comment |
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$begingroup$
The common notion of distance is to consider an orthogonal projection $P$ onto the first linear subspace $V$, and an orthogonal projection $Q$ onto the other subspace $W$.
At this point we can define
$$d(V,W) = | P - Q |$$ as the distance between these subspaces, where the norm used is the operator norm. For properties and applications see Section 2.5.3 of Golub and Van Loan.
This distance metric is used throughout GVL’s exposition on unsymmetrical eigenvalue problems (which involve Krylov methods) — see Chapter 7.
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
$endgroup$
add a comment |
$begingroup$
The common notion of distance is to consider an orthogonal projection $P$ onto the first linear subspace $V$, and an orthogonal projection $Q$ onto the other subspace $W$.
At this point we can define
$$d(V,W) = | P - Q |$$ as the distance between these subspaces, where the norm used is the operator norm. For properties and applications see Section 2.5.3 of Golub and Van Loan.
This distance metric is used throughout GVL’s exposition on unsymmetrical eigenvalue problems (which involve Krylov methods) — see Chapter 7.
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
$endgroup$
add a comment |
$begingroup$
The common notion of distance is to consider an orthogonal projection $P$ onto the first linear subspace $V$, and an orthogonal projection $Q$ onto the other subspace $W$.
At this point we can define
$$d(V,W) = | P - Q |$$ as the distance between these subspaces, where the norm used is the operator norm. For properties and applications see Section 2.5.3 of Golub and Van Loan.
This distance metric is used throughout GVL’s exposition on unsymmetrical eigenvalue problems (which involve Krylov methods) — see Chapter 7.
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
$endgroup$
The common notion of distance is to consider an orthogonal projection $P$ onto the first linear subspace $V$, and an orthogonal projection $Q$ onto the other subspace $W$.
At this point we can define
$$d(V,W) = | P - Q |$$ as the distance between these subspaces, where the norm used is the operator norm. For properties and applications see Section 2.5.3 of Golub and Van Loan.
This distance metric is used throughout GVL’s exposition on unsymmetrical eigenvalue problems (which involve Krylov methods) — see Chapter 7.
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
answered Dec 9 '18 at 19:28
cdipaolocdipaolo
650313
650313
add a comment |
add a comment |
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$begingroup$
A distance between two planes in $mathbb{R}^3$ can be defined by $|sin(theta)|$ where $theta$ is the angle between their normals (in particular, the triangular inequality is verified). I have the remembrance that such a result is generalizable, but I must look for references.
$endgroup$
– Jean Marie
Dec 9 '18 at 11:22