Mixing
Non-determinantal positive-definiteness conditions for $4 times 4$ Hermitian matrices with certain null...
$begingroup$
I have classes of $4 times 4$ Hermitian matrices $D$, the two $2 times 2$ diagonal blocks of which are themselves diagonal. That is, the (1,2),(2,1),(3,4) and (4,3) entries of $D$ are zero. (The other off-diagonal entries of $D$ can be real, complex or quaternionic.) I want necessary and sufficient (maybe minimal) conditions that $D$ be positive definite, that do not involve the full determinant of $D$. (In the quaternionic case, this would be the "Moore determinant", but the real and complex cases are the ones of immediate interest to me.)
As some background, this pertains to a quantum-information-theoretic problem, in which I am also interested in the "partial transpose" $D^{PT}$ of $D$, obtained, in general, by transposing in place the four $2 times 2$ blocks of $D$. The determinants of both $D$ and $D^{PT}$ are quite cumbersome, while the difference of the two determinants simplifies greatly, which is the basic motivation for my question. (I want to enforce the positive-definite nature of $D^{PT}$, subject to $D$ itself being positive-definite. Hopefully, the $2 times 2$ and $3 times 3$ minors of these matrices are more computationally amenable than their determinants. This pertains to the problem of determining the probability that two quantum bits ["qubits"] are disentangled/separable.)
linear-algebra matrices complex-numbers determinant quaternions
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
$endgroup$
add a comment |
$begingroup$
I have classes of $4 times 4$ Hermitian matrices $D$, the two $2 times 2$ diagonal blocks of which are themselves diagonal. That is, the (1,2),(2,1),(3,4) and (4,3) entries of $D$ are zero. (The other off-diagonal entries of $D$ can be real, complex or quaternionic.) I want necessary and sufficient (maybe minimal) conditions that $D$ be positive definite, that do not involve the full determinant of $D$. (In the quaternionic case, this would be the "Moore determinant", but the real and complex cases are the ones of immediate interest to me.)
As some background, this pertains to a quantum-information-theoretic problem, in which I am also interested in the "partial transpose" $D^{PT}$ of $D$, obtained, in general, by transposing in place the four $2 times 2$ blocks of $D$. The determinants of both $D$ and $D^{PT}$ are quite cumbersome, while the difference of the two determinants simplifies greatly, which is the basic motivation for my question. (I want to enforce the positive-definite nature of $D^{PT}$, subject to $D$ itself being positive-definite. Hopefully, the $2 times 2$ and $3 times 3$ minors of these matrices are more computationally amenable than their determinants. This pertains to the problem of determining the probability that two quantum bits ["qubits"] are disentangled/separable.)
linear-algebra matrices complex-numbers determinant quaternions
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
$endgroup$
$begingroup$
One way to characterise positive matrices is by the condition that all eigenvalues are positive. So I do not understand your phrase "... conditions that $D$ be positive definite, that do not involve the full determinant of $D$.", because the determinant alone, which equals the product of all eigenvalues, is not sufficient to detect or disproof positivity.
$endgroup$
– Hanno
Jan 15 at 22:34
add a comment |
$begingroup$
I have classes of $4 times 4$ Hermitian matrices $D$, the two $2 times 2$ diagonal blocks of which are themselves diagonal. That is, the (1,2),(2,1),(3,4) and (4,3) entries of $D$ are zero. (The other off-diagonal entries of $D$ can be real, complex or quaternionic.) I want necessary and sufficient (maybe minimal) conditions that $D$ be positive definite, that do not involve the full determinant of $D$. (In the quaternionic case, this would be the "Moore determinant", but the real and complex cases are the ones of immediate interest to me.)
As some background, this pertains to a quantum-information-theoretic problem, in which I am also interested in the "partial transpose" $D^{PT}$ of $D$, obtained, in general, by transposing in place the four $2 times 2$ blocks of $D$. The determinants of both $D$ and $D^{PT}$ are quite cumbersome, while the difference of the two determinants simplifies greatly, which is the basic motivation for my question. (I want to enforce the positive-definite nature of $D^{PT}$, subject to $D$ itself being positive-definite. Hopefully, the $2 times 2$ and $3 times 3$ minors of these matrices are more computationally amenable than their determinants. This pertains to the problem of determining the probability that two quantum bits ["qubits"] are disentangled/separable.)
linear-algebra matrices complex-numbers determinant quaternions
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
$endgroup$
I have classes of $4 times 4$ Hermitian matrices $D$, the two $2 times 2$ diagonal blocks of which are themselves diagonal. That is, the (1,2),(2,1),(3,4) and (4,3) entries of $D$ are zero. (The other off-diagonal entries of $D$ can be real, complex or quaternionic.) I want necessary and sufficient (maybe minimal) conditions that $D$ be positive definite, that do not involve the full determinant of $D$. (In the quaternionic case, this would be the "Moore determinant", but the real and complex cases are the ones of immediate interest to me.)
As some background, this pertains to a quantum-information-theoretic problem, in which I am also interested in the "partial transpose" $D^{PT}$ of $D$, obtained, in general, by transposing in place the four $2 times 2$ blocks of $D$. The determinants of both $D$ and $D^{PT}$ are quite cumbersome, while the difference of the two determinants simplifies greatly, which is the basic motivation for my question. (I want to enforce the positive-definite nature of $D^{PT}$, subject to $D$ itself being positive-definite. Hopefully, the $2 times 2$ and $3 times 3$ minors of these matrices are more computationally amenable than their determinants. This pertains to the problem of determining the probability that two quantum bits ["qubits"] are disentangled/separable.)
linear-algebra matrices complex-numbers determinant quaternions
linear-algebra matrices complex-numbers determinant quaternions
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
edited May 27 '17 at 16:00
Paul B. Slater
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
asked May 27 '17 at 15:53
Paul B. SlaterPaul B. Slater
22319
22319
$begingroup$
One way to characterise positive matrices is by the condition that all eigenvalues are positive. So I do not understand your phrase "... conditions that $D$ be positive definite, that do not involve the full determinant of $D$.", because the determinant alone, which equals the product of all eigenvalues, is not sufficient to detect or disproof positivity.
$endgroup$
– Hanno
Jan 15 at 22:34
add a comment |
$begingroup$
One way to characterise positive matrices is by the condition that all eigenvalues are positive. So I do not understand your phrase "... conditions that $D$ be positive definite, that do not involve the full determinant of $D$.", because the determinant alone, which equals the product of all eigenvalues, is not sufficient to detect or disproof positivity.
$endgroup$
– Hanno
Jan 15 at 22:34
$begingroup$
One way to characterise positive matrices is by the condition that all eigenvalues are positive. So I do not understand your phrase "... conditions that $D$ be positive definite, that do not involve the full determinant of $D$.", because the determinant alone, which equals the product of all eigenvalues, is not sufficient to detect or disproof positivity.
$endgroup$
– Hanno
Jan 15 at 22:34
$begingroup$
One way to characterise positive matrices is by the condition that all eigenvalues are positive. So I do not understand your phrase "... conditions that $D$ be positive definite, that do not involve the full determinant of $D$.", because the determinant alone, which equals the product of all eigenvalues, is not sufficient to detect or disproof positivity.
$endgroup$
– Hanno
Jan 15 at 22:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's go and derive first an equivalent condition on a related $2times 2$ matrix which guarantees positive-definiteness of $,D$, valid in either the complex or the real case.
(1) Fixing notation:
$$D:=;begin{pmatrix} D_1& M\ M^*& D_2end{pmatrix}
:=;begin{pmatrix} begin{pmatrix}d_1& 0\ 0& d_2end{pmatrix}& begin{pmatrix}m_{13}& m_{14}\ m_{23}& m_{24}end{pmatrix}\[1ex]
M^*& begin{pmatrix}d_3& 0\ 0& d_4end{pmatrix}end{pmatrix}$$
There are no a priori restrictions on the entries of $M$, and their indices stick to the indexation as of $D$.
It is henceforth assumed that $,d_1,d_2,d_3,d_4>0$.
This feature is a necessary condition to $,D,$ being positive-definite, because
$,d_i =langle e_i| De_irangle>0$, with $e_i$ being a standard unit vector.
Using it right away one has the equivalence
$$0<D;iff;
0,<:begin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
Dbegin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
;=; begin{pmatrix}mathbb{1}& D_1^{-1/2}MD_2^{-1/2}\
D_2^{-1/2}M^*D_1^{-1/2}& mathbb{1}end{pmatrix}$$
since positive-definiteness is preserved under congruences
$Amapsto T^*AT$ with invertible $T$. Now use that
$$0 < begin{pmatrix}mathbb{1}& C\ C^*& mathbb{1}end{pmatrix}
;iff;|C|<1tag{1}$$
where $|cdot|$ denotes the operator norm. It equals the largest singular value.
The little sister of $(1)$ within $M_2(mathbb C)$ is the equivalence
$;0<left(begin{smallmatrix}1& z\ overline z& 1end{smallmatrix}right)$
iff $|z|<1$ (and she usually shows up in the proof(*) of the general case).
Thus the equivalence criterion for positive-definiteness of $,D,$ reads
$,left|D_1^{-1/2}MD_2^{-1/2}right|,stackrel{!}{<}, 1$
put another way: $,D_1^{-1/2}MD_2^{-1/2} = left(frac{m_{ij}}{sqrt{d_id_j}}right)_{i=1,2;j=3,4}$ is a strict contraction.
(2) An alternative, more hands-on approach would be based on the
Cholesky decomposition. It is the "$Rightarrow$" in
$$A>0:iff,exists! text{ lower-triangular matrix }Ltext{ with },LL^*=A,,$$
and many (to most) implementations of the Cholesky algorithm accept as input a matrix without definiteness properties, and return $L$, or some educated error message if the input was not positive (enough). Cholesky still works for positive-semidefinite input and yields a lower-triangular $L$ but its uniqueness is lost.
As $n=4$ this "positivity criterion" is checked within a blink of the eye $:ddotsmile$
* e.g. Proposition I.3.5 in R. Bhatia's book "Matrix Analysis"
(Springer, GTM 169, 1997)
answered Jan 16 at 15:45
HannoHanno
2,240528
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add a comment |
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$begingroup$
Let's go and derive first an equivalent condition on a related $2times 2$ matrix which guarantees positive-definiteness of $,D$, valid in either the complex or the real case.
(1) Fixing notation:
$$D:=;begin{pmatrix} D_1& M\ M^*& D_2end{pmatrix}
:=;begin{pmatrix} begin{pmatrix}d_1& 0\ 0& d_2end{pmatrix}& begin{pmatrix}m_{13}& m_{14}\ m_{23}& m_{24}end{pmatrix}\[1ex]
M^*& begin{pmatrix}d_3& 0\ 0& d_4end{pmatrix}end{pmatrix}$$
There are no a priori restrictions on the entries of $M$, and their indices stick to the indexation as of $D$.
It is henceforth assumed that $,d_1,d_2,d_3,d_4>0$.
This feature is a necessary condition to $,D,$ being positive-definite, because
$,d_i =langle e_i| De_irangle>0$, with $e_i$ being a standard unit vector.
Using it right away one has the equivalence
$$0<D;iff;
0,<:begin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
Dbegin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
;=; begin{pmatrix}mathbb{1}& D_1^{-1/2}MD_2^{-1/2}\
D_2^{-1/2}M^*D_1^{-1/2}& mathbb{1}end{pmatrix}$$
since positive-definiteness is preserved under congruences
$Amapsto T^*AT$ with invertible $T$. Now use that
$$0 < begin{pmatrix}mathbb{1}& C\ C^*& mathbb{1}end{pmatrix}
;iff;|C|<1tag{1}$$
where $|cdot|$ denotes the operator norm. It equals the largest singular value.
The little sister of $(1)$ within $M_2(mathbb C)$ is the equivalence
$;0<left(begin{smallmatrix}1& z\ overline z& 1end{smallmatrix}right)$
iff $|z|<1$ (and she usually shows up in the proof(*) of the general case).
Thus the equivalence criterion for positive-definiteness of $,D,$ reads
$,left|D_1^{-1/2}MD_2^{-1/2}right|,stackrel{!}{<}, 1$
put another way: $,D_1^{-1/2}MD_2^{-1/2} = left(frac{m_{ij}}{sqrt{d_id_j}}right)_{i=1,2;j=3,4}$ is a strict contraction.
(2) An alternative, more hands-on approach would be based on the
Cholesky decomposition. It is the "$Rightarrow$" in
$$A>0:iff,exists! text{ lower-triangular matrix }Ltext{ with },LL^*=A,,$$
and many (to most) implementations of the Cholesky algorithm accept as input a matrix without definiteness properties, and return $L$, or some educated error message if the input was not positive (enough). Cholesky still works for positive-semidefinite input and yields a lower-triangular $L$ but its uniqueness is lost.
As $n=4$ this "positivity criterion" is checked within a blink of the eye $:ddotsmile$
* e.g. Proposition I.3.5 in R. Bhatia's book "Matrix Analysis"
(Springer, GTM 169, 1997)
answered Jan 16 at 15:45
HannoHanno
2,240528
$endgroup$
add a comment |
$begingroup$
Let's go and derive first an equivalent condition on a related $2times 2$ matrix which guarantees positive-definiteness of $,D$, valid in either the complex or the real case.
(1) Fixing notation:
$$D:=;begin{pmatrix} D_1& M\ M^*& D_2end{pmatrix}
:=;begin{pmatrix} begin{pmatrix}d_1& 0\ 0& d_2end{pmatrix}& begin{pmatrix}m_{13}& m_{14}\ m_{23}& m_{24}end{pmatrix}\[1ex]
M^*& begin{pmatrix}d_3& 0\ 0& d_4end{pmatrix}end{pmatrix}$$
There are no a priori restrictions on the entries of $M$, and their indices stick to the indexation as of $D$.
It is henceforth assumed that $,d_1,d_2,d_3,d_4>0$.
This feature is a necessary condition to $,D,$ being positive-definite, because
$,d_i =langle e_i| De_irangle>0$, with $e_i$ being a standard unit vector.
Using it right away one has the equivalence
$$0<D;iff;
0,<:begin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
Dbegin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
;=; begin{pmatrix}mathbb{1}& D_1^{-1/2}MD_2^{-1/2}\
D_2^{-1/2}M^*D_1^{-1/2}& mathbb{1}end{pmatrix}$$
since positive-definiteness is preserved under congruences
$Amapsto T^*AT$ with invertible $T$. Now use that
$$0 < begin{pmatrix}mathbb{1}& C\ C^*& mathbb{1}end{pmatrix}
;iff;|C|<1tag{1}$$
where $|cdot|$ denotes the operator norm. It equals the largest singular value.
The little sister of $(1)$ within $M_2(mathbb C)$ is the equivalence
$;0<left(begin{smallmatrix}1& z\ overline z& 1end{smallmatrix}right)$
iff $|z|<1$ (and she usually shows up in the proof(*) of the general case).
Thus the equivalence criterion for positive-definiteness of $,D,$ reads
$,left|D_1^{-1/2}MD_2^{-1/2}right|,stackrel{!}{<}, 1$
put another way: $,D_1^{-1/2}MD_2^{-1/2} = left(frac{m_{ij}}{sqrt{d_id_j}}right)_{i=1,2;j=3,4}$ is a strict contraction.
(2) An alternative, more hands-on approach would be based on the
Cholesky decomposition. It is the "$Rightarrow$" in
$$A>0:iff,exists! text{ lower-triangular matrix }Ltext{ with },LL^*=A,,$$
and many (to most) implementations of the Cholesky algorithm accept as input a matrix without definiteness properties, and return $L$, or some educated error message if the input was not positive (enough). Cholesky still works for positive-semidefinite input and yields a lower-triangular $L$ but its uniqueness is lost.
As $n=4$ this "positivity criterion" is checked within a blink of the eye $:ddotsmile$
* e.g. Proposition I.3.5 in R. Bhatia's book "Matrix Analysis"
(Springer, GTM 169, 1997)
answered Jan 16 at 15:45
HannoHanno
2,240528
$endgroup$
add a comment |
$begingroup$
Let's go and derive first an equivalent condition on a related $2times 2$ matrix which guarantees positive-definiteness of $,D$, valid in either the complex or the real case.
(1) Fixing notation:
$$D:=;begin{pmatrix} D_1& M\ M^*& D_2end{pmatrix}
:=;begin{pmatrix} begin{pmatrix}d_1& 0\ 0& d_2end{pmatrix}& begin{pmatrix}m_{13}& m_{14}\ m_{23}& m_{24}end{pmatrix}\[1ex]
M^*& begin{pmatrix}d_3& 0\ 0& d_4end{pmatrix}end{pmatrix}$$
There are no a priori restrictions on the entries of $M$, and their indices stick to the indexation as of $D$.
It is henceforth assumed that $,d_1,d_2,d_3,d_4>0$.
This feature is a necessary condition to $,D,$ being positive-definite, because
$,d_i =langle e_i| De_irangle>0$, with $e_i$ being a standard unit vector.
Using it right away one has the equivalence
$$0<D;iff;
0,<:begin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
Dbegin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
;=; begin{pmatrix}mathbb{1}& D_1^{-1/2}MD_2^{-1/2}\
D_2^{-1/2}M^*D_1^{-1/2}& mathbb{1}end{pmatrix}$$
since positive-definiteness is preserved under congruences
$Amapsto T^*AT$ with invertible $T$. Now use that
$$0 < begin{pmatrix}mathbb{1}& C\ C^*& mathbb{1}end{pmatrix}
;iff;|C|<1tag{1}$$
where $|cdot|$ denotes the operator norm. It equals the largest singular value.
The little sister of $(1)$ within $M_2(mathbb C)$ is the equivalence
$;0<left(begin{smallmatrix}1& z\ overline z& 1end{smallmatrix}right)$
iff $|z|<1$ (and she usually shows up in the proof(*) of the general case).
Thus the equivalence criterion for positive-definiteness of $,D,$ reads
$,left|D_1^{-1/2}MD_2^{-1/2}right|,stackrel{!}{<}, 1$
put another way: $,D_1^{-1/2}MD_2^{-1/2} = left(frac{m_{ij}}{sqrt{d_id_j}}right)_{i=1,2;j=3,4}$ is a strict contraction.
(2) An alternative, more hands-on approach would be based on the
Cholesky decomposition. It is the "$Rightarrow$" in
$$A>0:iff,exists! text{ lower-triangular matrix }Ltext{ with },LL^*=A,,$$
and many (to most) implementations of the Cholesky algorithm accept as input a matrix without definiteness properties, and return $L$, or some educated error message if the input was not positive (enough). Cholesky still works for positive-semidefinite input and yields a lower-triangular $L$ but its uniqueness is lost.
As $n=4$ this "positivity criterion" is checked within a blink of the eye $:ddotsmile$
* e.g. Proposition I.3.5 in R. Bhatia's book "Matrix Analysis"
(Springer, GTM 169, 1997)
answered Jan 16 at 15:45
HannoHanno
2,240528
$endgroup$
Let's go and derive first an equivalent condition on a related $2times 2$ matrix which guarantees positive-definiteness of $,D$, valid in either the complex or the real case.
(1) Fixing notation:
$$D:=;begin{pmatrix} D_1& M\ M^*& D_2end{pmatrix}
:=;begin{pmatrix} begin{pmatrix}d_1& 0\ 0& d_2end{pmatrix}& begin{pmatrix}m_{13}& m_{14}\ m_{23}& m_{24}end{pmatrix}\[1ex]
M^*& begin{pmatrix}d_3& 0\ 0& d_4end{pmatrix}end{pmatrix}$$
There are no a priori restrictions on the entries of $M$, and their indices stick to the indexation as of $D$.
It is henceforth assumed that $,d_1,d_2,d_3,d_4>0$.
This feature is a necessary condition to $,D,$ being positive-definite, because
$,d_i =langle e_i| De_irangle>0$, with $e_i$ being a standard unit vector.
Using it right away one has the equivalence
$$0<D;iff;
0,<:begin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
Dbegin{pmatrix} D_1^{-1/2}& mathbb{0}\ mathbb{0}& D_2^{-1/2}end{pmatrix}
;=; begin{pmatrix}mathbb{1}& D_1^{-1/2}MD_2^{-1/2}\
D_2^{-1/2}M^*D_1^{-1/2}& mathbb{1}end{pmatrix}$$
since positive-definiteness is preserved under congruences
$Amapsto T^*AT$ with invertible $T$. Now use that
$$0 < begin{pmatrix}mathbb{1}& C\ C^*& mathbb{1}end{pmatrix}
;iff;|C|<1tag{1}$$
where $|cdot|$ denotes the operator norm. It equals the largest singular value.
The little sister of $(1)$ within $M_2(mathbb C)$ is the equivalence
$;0<left(begin{smallmatrix}1& z\ overline z& 1end{smallmatrix}right)$
iff $|z|<1$ (and she usually shows up in the proof(*) of the general case).
Thus the equivalence criterion for positive-definiteness of $,D,$ reads
$,left|D_1^{-1/2}MD_2^{-1/2}right|,stackrel{!}{<}, 1$
put another way: $,D_1^{-1/2}MD_2^{-1/2} = left(frac{m_{ij}}{sqrt{d_id_j}}right)_{i=1,2;j=3,4}$ is a strict contraction.
(2) An alternative, more hands-on approach would be based on the
Cholesky decomposition. It is the "$Rightarrow$" in
$$A>0:iff,exists! text{ lower-triangular matrix }Ltext{ with },LL^*=A,,$$
and many (to most) implementations of the Cholesky algorithm accept as input a matrix without definiteness properties, and return $L$, or some educated error message if the input was not positive (enough). Cholesky still works for positive-semidefinite input and yields a lower-triangular $L$ but its uniqueness is lost.
As $n=4$ this "positivity criterion" is checked within a blink of the eye $:ddotsmile$
* e.g. Proposition I.3.5 in R. Bhatia's book "Matrix Analysis"
(Springer, GTM 169, 1997)
answered Jan 16 at 15:45
HannoHanno
2,240528
answered Jan 16 at 15:45
HannoHanno
2,240528
answered Jan 16 at 15:45
HannoHanno
2,240528
answered Jan 16 at 15:45
HannoHanno
2,240528
2,240528
add a comment |
add a comment |
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One way to characterise positive matrices is by the condition that all eigenvalues are positive. So I do not understand your phrase "... conditions that $D$ be positive definite, that do not involve the full determinant of $D$.", because the determinant alone, which equals the product of all eigenvalues, is not sufficient to detect or disproof positivity.
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– Hanno
Jan 15 at 22:34