Multiplications that preserve singular values
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What is the characterization of matrices $B$(not necessarily squared) such that $BA$ has the same largest singular value as $A$? How about when $BA$ preserves the same $k$ largest singular values of $A$?
What if $A$ is a positive definite matrix?
I know that $B$ where the columns of $B$ are orthonormal satisfies this, but is that all?
I would really appreciate it if someone can guide me in the right direction.
linear-algebra matrices numerical-linear-algebra matrix-decomposition
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
$endgroup$
add a comment |
$begingroup$
What is the characterization of matrices $B$(not necessarily squared) such that $BA$ has the same largest singular value as $A$? How about when $BA$ preserves the same $k$ largest singular values of $A$?
What if $A$ is a positive definite matrix?
I know that $B$ where the columns of $B$ are orthonormal satisfies this, but is that all?
I would really appreciate it if someone can guide me in the right direction.
linear-algebra matrices numerical-linear-algebra matrix-decomposition
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
$endgroup$
add a comment |
$begingroup$
What is the characterization of matrices $B$(not necessarily squared) such that $BA$ has the same largest singular value as $A$? How about when $BA$ preserves the same $k$ largest singular values of $A$?
What if $A$ is a positive definite matrix?
I know that $B$ where the columns of $B$ are orthonormal satisfies this, but is that all?
I would really appreciate it if someone can guide me in the right direction.
linear-algebra matrices numerical-linear-algebra matrix-decomposition
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
$endgroup$
What is the characterization of matrices $B$(not necessarily squared) such that $BA$ has the same largest singular value as $A$? How about when $BA$ preserves the same $k$ largest singular values of $A$?
What if $A$ is a positive definite matrix?
I know that $B$ where the columns of $B$ are orthonormal satisfies this, but is that all?
I would really appreciate it if someone can guide me in the right direction.
linear-algebra matrices numerical-linear-algebra matrix-decomposition
linear-algebra matrices numerical-linear-algebra matrix-decomposition
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
asked Dec 17 '18 at 1:31
Evan LiangEvan Liang
11
11
add a comment |
add a comment |
1 Answer
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I think it depends on the matching between left and right singular vectors. We can look at Singular Value Decomposition of both A and B:
begin{align}
A &= USigma V^T\
B &= YPsi Z^T \
BA &= YPsi Z^T USigma V^T
end{align}
Since we look for singular values, we can get rid of orthogonal matrices at the margins:
$$sigma(BA) equiv sigma( Psi Z^T USigma) $$
We do not know anything about the singular values (or eigenvalues in square case) of the matrix $D_1 C D_2$ where $D_1$ and $D_2$ are diagonal matrices and C is any matrix with known singular values. If $C$ is also diagonal matrix which is $Z^TU=I$ in your case, then of course, you can find the singular values. As a result of this analysis, positive definiteness of A is irrelevant.
If B is an orthogonal matrix then we have:
$$sigma(BA) equiv sigma( B USigma) $$
so, as a special case, if $B^TB = I$ i.e. column of $B$ constitutes an orthogonal set of vectors (no need B to be orthonormal) then:
$$sigma(BA) equiv sigma( B USigma) =sigma(USigma) = sigma(Sigma)$$
Moreover, if all singular values of $BU$ is 1, then again
$$sigma(BA) = sigma(Sigma).$$ These are the cases that we can say something about singular values of $D_1 C D_2$.
answered Dec 17 '18 at 16:49
KurbanKurban
406
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add a comment |
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$begingroup$
I think it depends on the matching between left and right singular vectors. We can look at Singular Value Decomposition of both A and B:
begin{align}
A &= USigma V^T\
B &= YPsi Z^T \
BA &= YPsi Z^T USigma V^T
end{align}
Since we look for singular values, we can get rid of orthogonal matrices at the margins:
$$sigma(BA) equiv sigma( Psi Z^T USigma) $$
We do not know anything about the singular values (or eigenvalues in square case) of the matrix $D_1 C D_2$ where $D_1$ and $D_2$ are diagonal matrices and C is any matrix with known singular values. If $C$ is also diagonal matrix which is $Z^TU=I$ in your case, then of course, you can find the singular values. As a result of this analysis, positive definiteness of A is irrelevant.
If B is an orthogonal matrix then we have:
$$sigma(BA) equiv sigma( B USigma) $$
so, as a special case, if $B^TB = I$ i.e. column of $B$ constitutes an orthogonal set of vectors (no need B to be orthonormal) then:
$$sigma(BA) equiv sigma( B USigma) =sigma(USigma) = sigma(Sigma)$$
Moreover, if all singular values of $BU$ is 1, then again
$$sigma(BA) = sigma(Sigma).$$ These are the cases that we can say something about singular values of $D_1 C D_2$.
answered Dec 17 '18 at 16:49
KurbanKurban
406
$endgroup$
add a comment |
$begingroup$
I think it depends on the matching between left and right singular vectors. We can look at Singular Value Decomposition of both A and B:
begin{align}
A &= USigma V^T\
B &= YPsi Z^T \
BA &= YPsi Z^T USigma V^T
end{align}
Since we look for singular values, we can get rid of orthogonal matrices at the margins:
$$sigma(BA) equiv sigma( Psi Z^T USigma) $$
We do not know anything about the singular values (or eigenvalues in square case) of the matrix $D_1 C D_2$ where $D_1$ and $D_2$ are diagonal matrices and C is any matrix with known singular values. If $C$ is also diagonal matrix which is $Z^TU=I$ in your case, then of course, you can find the singular values. As a result of this analysis, positive definiteness of A is irrelevant.
If B is an orthogonal matrix then we have:
$$sigma(BA) equiv sigma( B USigma) $$
so, as a special case, if $B^TB = I$ i.e. column of $B$ constitutes an orthogonal set of vectors (no need B to be orthonormal) then:
$$sigma(BA) equiv sigma( B USigma) =sigma(USigma) = sigma(Sigma)$$
Moreover, if all singular values of $BU$ is 1, then again
$$sigma(BA) = sigma(Sigma).$$ These are the cases that we can say something about singular values of $D_1 C D_2$.
answered Dec 17 '18 at 16:49
KurbanKurban
406
$endgroup$
add a comment |
$begingroup$
I think it depends on the matching between left and right singular vectors. We can look at Singular Value Decomposition of both A and B:
begin{align}
A &= USigma V^T\
B &= YPsi Z^T \
BA &= YPsi Z^T USigma V^T
end{align}
Since we look for singular values, we can get rid of orthogonal matrices at the margins:
$$sigma(BA) equiv sigma( Psi Z^T USigma) $$
We do not know anything about the singular values (or eigenvalues in square case) of the matrix $D_1 C D_2$ where $D_1$ and $D_2$ are diagonal matrices and C is any matrix with known singular values. If $C$ is also diagonal matrix which is $Z^TU=I$ in your case, then of course, you can find the singular values. As a result of this analysis, positive definiteness of A is irrelevant.
If B is an orthogonal matrix then we have:
$$sigma(BA) equiv sigma( B USigma) $$
so, as a special case, if $B^TB = I$ i.e. column of $B$ constitutes an orthogonal set of vectors (no need B to be orthonormal) then:
$$sigma(BA) equiv sigma( B USigma) =sigma(USigma) = sigma(Sigma)$$
Moreover, if all singular values of $BU$ is 1, then again
$$sigma(BA) = sigma(Sigma).$$ These are the cases that we can say something about singular values of $D_1 C D_2$.
answered Dec 17 '18 at 16:49
KurbanKurban
406
$endgroup$
I think it depends on the matching between left and right singular vectors. We can look at Singular Value Decomposition of both A and B:
begin{align}
A &= USigma V^T\
B &= YPsi Z^T \
BA &= YPsi Z^T USigma V^T
end{align}
Since we look for singular values, we can get rid of orthogonal matrices at the margins:
$$sigma(BA) equiv sigma( Psi Z^T USigma) $$
We do not know anything about the singular values (or eigenvalues in square case) of the matrix $D_1 C D_2$ where $D_1$ and $D_2$ are diagonal matrices and C is any matrix with known singular values. If $C$ is also diagonal matrix which is $Z^TU=I$ in your case, then of course, you can find the singular values. As a result of this analysis, positive definiteness of A is irrelevant.
If B is an orthogonal matrix then we have:
$$sigma(BA) equiv sigma( B USigma) $$
so, as a special case, if $B^TB = I$ i.e. column of $B$ constitutes an orthogonal set of vectors (no need B to be orthonormal) then:
$$sigma(BA) equiv sigma( B USigma) =sigma(USigma) = sigma(Sigma)$$
Moreover, if all singular values of $BU$ is 1, then again
$$sigma(BA) = sigma(Sigma).$$ These are the cases that we can say something about singular values of $D_1 C D_2$.
answered Dec 17 '18 at 16:49
KurbanKurban
406
edited Dec 17 '18 at 16:54
answered Dec 17 '18 at 16:49
KurbanKurban
406
answered Dec 17 '18 at 16:49
KurbanKurban
406
answered Dec 17 '18 at 16:49
KurbanKurban
406
406
add a comment |
add a comment |
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