Mixing
Is there any inequality between 2-norm condition number and Frobenius norm condition number for rectangular...
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What I have found in [1]
Condition number inequality between Frobenius norm and 2-norm for square matrix,
Consider a full rank matrix $X in mathbb{C}^{n times m}$, $m=n$, then we can have,
$$n - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X).$$
Question: Does this inequality works for the case when $m neq n$?
If not, have you seen other relationship between them?
Update:
The following paper [2] mentioned that it is a natural extension to non-rectangular case, I don't get why. read the two equations below equation (3.7) in [2].
Ref:
[1] Smith, Russell A. "The condition numbers of the matrix eigenvalue problem." Numerische Mathematik 10.3 (1967): 232-240.
[2] Bazán, F. S. V. (2000). Conditioning of Rectangular Vandermonde Matrices with Nodes in the Unit Disk. SIAM Journal on Matrix Analysis and Applications, 21(2), 679–693. https://doi.org/10.1137/S0895479898336021
linear-algebra matrices inequality condition-number matrix-analysis
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
$endgroup$
add a comment |
$begingroup$
What I have found in [1]
Condition number inequality between Frobenius norm and 2-norm for square matrix,
Consider a full rank matrix $X in mathbb{C}^{n times m}$, $m=n$, then we can have,
$$n - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X).$$
Question: Does this inequality works for the case when $m neq n$?
If not, have you seen other relationship between them?
Update:
The following paper [2] mentioned that it is a natural extension to non-rectangular case, I don't get why. read the two equations below equation (3.7) in [2].
Ref:
[1] Smith, Russell A. "The condition numbers of the matrix eigenvalue problem." Numerische Mathematik 10.3 (1967): 232-240.
[2] Bazán, F. S. V. (2000). Conditioning of Rectangular Vandermonde Matrices with Nodes in the Unit Disk. SIAM Journal on Matrix Analysis and Applications, 21(2), 679–693. https://doi.org/10.1137/S0895479898336021
linear-algebra matrices inequality condition-number matrix-analysis
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
$endgroup$
2
$begingroup$
How are you defining the condition numbers for non-square matrices?
$endgroup$
– tch
Jan 17 at 2:04
$begingroup$
@tch That's a good question. $kappa(A) = lVert A rVert cdot lVert A^{dagger} rVert $
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 4:26
$begingroup$
By reducing to the SVD, you can express the 2-norm condition number as the ratio of the largest and smallest nonzero singular values, and similarly the Frobenius condition number as the square root of the ratio of the sum of the squares of the singular values with the sum of the squares of their reciprocals. You might be able to manipulate these expressions to get some relationships.
$endgroup$
– Christopher A. Wong
Jan 17 at 5:35
add a comment |
$begingroup$
What I have found in [1]
Condition number inequality between Frobenius norm and 2-norm for square matrix,
Consider a full rank matrix $X in mathbb{C}^{n times m}$, $m=n$, then we can have,
$$n - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X).$$
Question: Does this inequality works for the case when $m neq n$?
If not, have you seen other relationship between them?
Update:
The following paper [2] mentioned that it is a natural extension to non-rectangular case, I don't get why. read the two equations below equation (3.7) in [2].
Ref:
[1] Smith, Russell A. "The condition numbers of the matrix eigenvalue problem." Numerische Mathematik 10.3 (1967): 232-240.
[2] Bazán, F. S. V. (2000). Conditioning of Rectangular Vandermonde Matrices with Nodes in the Unit Disk. SIAM Journal on Matrix Analysis and Applications, 21(2), 679–693. https://doi.org/10.1137/S0895479898336021
linear-algebra matrices inequality condition-number matrix-analysis
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
$endgroup$
What I have found in [1]
Condition number inequality between Frobenius norm and 2-norm for square matrix,
Consider a full rank matrix $X in mathbb{C}^{n times m}$, $m=n$, then we can have,
$$n - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X).$$
Question: Does this inequality works for the case when $m neq n$?
If not, have you seen other relationship between them?
Update:
The following paper [2] mentioned that it is a natural extension to non-rectangular case, I don't get why. read the two equations below equation (3.7) in [2].
Ref:
[1] Smith, Russell A. "The condition numbers of the matrix eigenvalue problem." Numerische Mathematik 10.3 (1967): 232-240.
[2] Bazán, F. S. V. (2000). Conditioning of Rectangular Vandermonde Matrices with Nodes in the Unit Disk. SIAM Journal on Matrix Analysis and Applications, 21(2), 679–693. https://doi.org/10.1137/S0895479898336021
linear-algebra matrices inequality condition-number matrix-analysis
linear-algebra matrices inequality condition-number matrix-analysis
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
edited Jan 17 at 4:28
ArtificiallyIntelligence
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
asked Jan 17 at 1:45
ArtificiallyIntelligenceArtificiallyIntelligence
305111
305111
2
$begingroup$
How are you defining the condition numbers for non-square matrices?
$endgroup$
– tch
Jan 17 at 2:04
$begingroup$
@tch That's a good question. $kappa(A) = lVert A rVert cdot lVert A^{dagger} rVert $
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 4:26
$begingroup$
By reducing to the SVD, you can express the 2-norm condition number as the ratio of the largest and smallest nonzero singular values, and similarly the Frobenius condition number as the square root of the ratio of the sum of the squares of the singular values with the sum of the squares of their reciprocals. You might be able to manipulate these expressions to get some relationships.
$endgroup$
– Christopher A. Wong
Jan 17 at 5:35
add a comment |
2
$begingroup$
How are you defining the condition numbers for non-square matrices?
$endgroup$
– tch
Jan 17 at 2:04
$begingroup$
@tch That's a good question. $kappa(A) = lVert A rVert cdot lVert A^{dagger} rVert $
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 4:26
$begingroup$
By reducing to the SVD, you can express the 2-norm condition number as the ratio of the largest and smallest nonzero singular values, and similarly the Frobenius condition number as the square root of the ratio of the sum of the squares of the singular values with the sum of the squares of their reciprocals. You might be able to manipulate these expressions to get some relationships.
$endgroup$
– Christopher A. Wong
Jan 17 at 5:35
2
2
$begingroup$
How are you defining the condition numbers for non-square matrices?
$endgroup$
– tch
Jan 17 at 2:04
$begingroup$
How are you defining the condition numbers for non-square matrices?
$endgroup$
– tch
Jan 17 at 2:04
$begingroup$
@tch That's a good question. $kappa(A) = lVert A rVert cdot lVert A^{dagger} rVert $
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 4:26
$begingroup$
@tch That's a good question. $kappa(A) = lVert A rVert cdot lVert A^{dagger} rVert $
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 4:26
$begingroup$
By reducing to the SVD, you can express the 2-norm condition number as the ratio of the largest and smallest nonzero singular values, and similarly the Frobenius condition number as the square root of the ratio of the sum of the squares of the singular values with the sum of the squares of their reciprocals. You might be able to manipulate these expressions to get some relationships.
$endgroup$
– Christopher A. Wong
Jan 17 at 5:35
$begingroup$
By reducing to the SVD, you can express the 2-norm condition number as the ratio of the largest and smallest nonzero singular values, and similarly the Frobenius condition number as the square root of the ratio of the sum of the squares of the singular values with the sum of the squares of their reciprocals. You might be able to manipulate these expressions to get some relationships.
$endgroup$
– Christopher A. Wong
Jan 17 at 5:35
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Claim:
Define $kappa(A) = Vert A Vert cdot Vert A^dagger Vert$ and suppose $X$ is rank $k$. Then,
$$
k - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X)
$$
Moreover, this bound is tight since a diagonal matrix with $k$ identical values followed by zeros attains the bound.
Proof:
Suppose $X in mathbb{C}^{ntimes m}$ with rank $k$ and singular values
$$
sigma_1 geq cdots geq sigma_k geq 0 geq cdots geq 0
$$
Then $X^{dagger}$ has singular values,
$$
1/sigma_k geq cdots geq 1/sigma_1 geq 0 geq cdots geq 0
$$
Therefore,
$$
kappa_2(X) = Vert XVert_2 Vert X^dagger Vert_2 = sigma_1/sigma_k
$$
and
$$
kappa_F(X) = Vert XVert_F Vert X^dagger Vert_F = sqrt{left(sum_{i=1}^{k} sigma_i^2right)left(sum_{i=1}^{k} frac{1}{sigma_i^2}right)}
$$
Define $X'$ to be the $ktimes k$ matrix with $sigma_1,ldots, sigma_k$ on the diagonal. Then $X'$ is full rank and square so the result holds by observing that $kappa_2(X) = kappa_2(X')$ and $kappa_F(X) = kappa_F(X')$.
answered Jan 17 at 16:23
tchtch
803310
$endgroup$
$begingroup$
This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case.
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 18:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Claim:
Define $kappa(A) = Vert A Vert cdot Vert A^dagger Vert$ and suppose $X$ is rank $k$. Then,
$$
k - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X)
$$
Moreover, this bound is tight since a diagonal matrix with $k$ identical values followed by zeros attains the bound.
Proof:
Suppose $X in mathbb{C}^{ntimes m}$ with rank $k$ and singular values
$$
sigma_1 geq cdots geq sigma_k geq 0 geq cdots geq 0
$$
Then $X^{dagger}$ has singular values,
$$
1/sigma_k geq cdots geq 1/sigma_1 geq 0 geq cdots geq 0
$$
Therefore,
$$
kappa_2(X) = Vert XVert_2 Vert X^dagger Vert_2 = sigma_1/sigma_k
$$
and
$$
kappa_F(X) = Vert XVert_F Vert X^dagger Vert_F = sqrt{left(sum_{i=1}^{k} sigma_i^2right)left(sum_{i=1}^{k} frac{1}{sigma_i^2}right)}
$$
Define $X'$ to be the $ktimes k$ matrix with $sigma_1,ldots, sigma_k$ on the diagonal. Then $X'$ is full rank and square so the result holds by observing that $kappa_2(X) = kappa_2(X')$ and $kappa_F(X) = kappa_F(X')$.
answered Jan 17 at 16:23
tchtch
803310
$endgroup$
$begingroup$
This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case.
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 18:06
add a comment |
$begingroup$
Claim:
Define $kappa(A) = Vert A Vert cdot Vert A^dagger Vert$ and suppose $X$ is rank $k$. Then,
$$
k - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X)
$$
Moreover, this bound is tight since a diagonal matrix with $k$ identical values followed by zeros attains the bound.
Proof:
Suppose $X in mathbb{C}^{ntimes m}$ with rank $k$ and singular values
$$
sigma_1 geq cdots geq sigma_k geq 0 geq cdots geq 0
$$
Then $X^{dagger}$ has singular values,
$$
1/sigma_k geq cdots geq 1/sigma_1 geq 0 geq cdots geq 0
$$
Therefore,
$$
kappa_2(X) = Vert XVert_2 Vert X^dagger Vert_2 = sigma_1/sigma_k
$$
and
$$
kappa_F(X) = Vert XVert_F Vert X^dagger Vert_F = sqrt{left(sum_{i=1}^{k} sigma_i^2right)left(sum_{i=1}^{k} frac{1}{sigma_i^2}right)}
$$
Define $X'$ to be the $ktimes k$ matrix with $sigma_1,ldots, sigma_k$ on the diagonal. Then $X'$ is full rank and square so the result holds by observing that $kappa_2(X) = kappa_2(X')$ and $kappa_F(X) = kappa_F(X')$.
answered Jan 17 at 16:23
tchtch
803310
$endgroup$
$begingroup$
This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case.
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 18:06
add a comment |
$begingroup$
Claim:
Define $kappa(A) = Vert A Vert cdot Vert A^dagger Vert$ and suppose $X$ is rank $k$. Then,
$$
k - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X)
$$
Moreover, this bound is tight since a diagonal matrix with $k$ identical values followed by zeros attains the bound.
Proof:
Suppose $X in mathbb{C}^{ntimes m}$ with rank $k$ and singular values
$$
sigma_1 geq cdots geq sigma_k geq 0 geq cdots geq 0
$$
Then $X^{dagger}$ has singular values,
$$
1/sigma_k geq cdots geq 1/sigma_1 geq 0 geq cdots geq 0
$$
Therefore,
$$
kappa_2(X) = Vert XVert_2 Vert X^dagger Vert_2 = sigma_1/sigma_k
$$
and
$$
kappa_F(X) = Vert XVert_F Vert X^dagger Vert_F = sqrt{left(sum_{i=1}^{k} sigma_i^2right)left(sum_{i=1}^{k} frac{1}{sigma_i^2}right)}
$$
Define $X'$ to be the $ktimes k$ matrix with $sigma_1,ldots, sigma_k$ on the diagonal. Then $X'$ is full rank and square so the result holds by observing that $kappa_2(X) = kappa_2(X')$ and $kappa_F(X) = kappa_F(X')$.
answered Jan 17 at 16:23
tchtch
803310
$endgroup$
Claim:
Define $kappa(A) = Vert A Vert cdot Vert A^dagger Vert$ and suppose $X$ is rank $k$. Then,
$$
k - 2 + frac{1}{kappa_2(X)} + kappa_2(X) le kappa_F(X)
$$
Moreover, this bound is tight since a diagonal matrix with $k$ identical values followed by zeros attains the bound.
Proof:
Suppose $X in mathbb{C}^{ntimes m}$ with rank $k$ and singular values
$$
sigma_1 geq cdots geq sigma_k geq 0 geq cdots geq 0
$$
Then $X^{dagger}$ has singular values,
$$
1/sigma_k geq cdots geq 1/sigma_1 geq 0 geq cdots geq 0
$$
Therefore,
$$
kappa_2(X) = Vert XVert_2 Vert X^dagger Vert_2 = sigma_1/sigma_k
$$
and
$$
kappa_F(X) = Vert XVert_F Vert X^dagger Vert_F = sqrt{left(sum_{i=1}^{k} sigma_i^2right)left(sum_{i=1}^{k} frac{1}{sigma_i^2}right)}
$$
Define $X'$ to be the $ktimes k$ matrix with $sigma_1,ldots, sigma_k$ on the diagonal. Then $X'$ is full rank and square so the result holds by observing that $kappa_2(X) = kappa_2(X')$ and $kappa_F(X) = kappa_F(X')$.
answered Jan 17 at 16:23
tchtch
803310
answered Jan 17 at 16:23
tchtch
803310
answered Jan 17 at 16:23
tchtch
803310
answered Jan 17 at 16:23
tchtch
803310
803310
$begingroup$
This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case.
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 18:06
add a comment |
$begingroup$
This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case.
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 18:06
$begingroup$
This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case.
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 18:06
$begingroup$
This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case.
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 18:06
add a comment |
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2
$begingroup$
How are you defining the condition numbers for non-square matrices?
$endgroup$
– tch
Jan 17 at 2:04
$begingroup$
@tch That's a good question. $kappa(A) = lVert A rVert cdot lVert A^{dagger} rVert $
$endgroup$
– ArtificiallyIntelligence
Jan 17 at 4:26
$begingroup$
By reducing to the SVD, you can express the 2-norm condition number as the ratio of the largest and smallest nonzero singular values, and similarly the Frobenius condition number as the square root of the ratio of the sum of the squares of the singular values with the sum of the squares of their reciprocals. You might be able to manipulate these expressions to get some relationships.
$endgroup$
– Christopher A. Wong
Jan 17 at 5:35