Mixing
Examples of stochastic matrices that are also unitary?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
In general there is no relationship between stochastic matrices and unitary matrices because they are used in different fields.
The stochastic matrix presents all the non-negative elements with sum on each row (or column) equal to 1.
The unitary matrix, if real, presents inverse and transposed equal to each other.
Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time ?
matrices
asked Jul 20 at 20:18
user3520363
156
add a comment |Â
up vote
1
down vote
favorite
In general there is no relationship between stochastic matrices and unitary matrices because they are used in different fields.
The stochastic matrix presents all the non-negative elements with sum on each row (or column) equal to 1.
The unitary matrix, if real, presents inverse and transposed equal to each other.
Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time ?
matrices
asked Jul 20 at 20:18
user3520363
156
"Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time" - yes. It is both stochastic (actually even left and right stochastic!) and unitary. Same is true for an exchange matrix btw (all ones on its antidiagonal).
– Florian
Jul 20 at 20:39
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In general there is no relationship between stochastic matrices and unitary matrices because they are used in different fields.
The stochastic matrix presents all the non-negative elements with sum on each row (or column) equal to 1.
The unitary matrix, if real, presents inverse and transposed equal to each other.
Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time ?
matrices
asked Jul 20 at 20:18
user3520363
156
In general there is no relationship between stochastic matrices and unitary matrices because they are used in different fields.
The stochastic matrix presents all the non-negative elements with sum on each row (or column) equal to 1.
The unitary matrix, if real, presents inverse and transposed equal to each other.
Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time ?
matrices
asked Jul 20 at 20:18
user3520363
156
asked Jul 20 at 20:18
user3520363
156
asked Jul 20 at 20:18
user3520363
156
asked Jul 20 at 20:18
user3520363
156
156
"Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time" - yes. It is both stochastic (actually even left and right stochastic!) and unitary. Same is true for an exchange matrix btw (all ones on its antidiagonal).
– Florian
Jul 20 at 20:39
add a comment |Â
"Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time" - yes. It is both stochastic (actually even left and right stochastic!) and unitary. Same is true for an exchange matrix btw (all ones on its antidiagonal).
– Florian
Jul 20 at 20:39
"Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time" - yes. It is both stochastic (actually even left and right stochastic!) and unitary. Same is true for an exchange matrix btw (all ones on its antidiagonal).
– Florian
Jul 20 at 20:39
"Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time" - yes. It is both stochastic (actually even left and right stochastic!) and unitary. Same is true for an exchange matrix btw (all ones on its antidiagonal).
– Florian
Jul 20 at 20:39
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The only matrices that are both stochastic and unitary are permutation matrices (i.e. those with one $1$ in each row and in each column). Let the rows be vectors $a_i$ with entries $a_ij$. Firstly, note that to be unitary we need $a_i^daggera_i = 1$; to be stochastic, $1^daggera_i = 1$, where $1 = (1,1,dotsc,1)$, and We also need $a_ij>0$. It follows that $0 leq a_ij leq 1 $, since each summand in a sum of positive numbers is smaller than the sum. Of course this condition also means that $a_i^dagger=a_i^T$. By subtracting, we find
$$ (1-a_i)^Ta_i = 0. $$
But this is the same as
$$ sum_j (1-a_ij)a_ij = 0, $$
and since $0 leq a_ij leq 1 $, every term in this sum is nonnegative. Hence every term is zero, so $a_ij$ is either $0$ or $1$. There can then only be exactly nonzero entry in $a_i$ since $sum_j a_ij = 1$. Since $a_i^T a_j = 0$ for orthogonality, we also find that each column can only have one nonzero term in it (if there were more, the corresponding vectors would not be orthogonal). Hence the matrix is as stated, with one $1$ in each row and in each column. It is easy to check that every such matrix is both stochastic and unitary.
(We also see therefore that every stochastic unitary matrix is actually doubly stochastic.)
answered Jul 20 at 20:41
Chappers
55k74190
Thanks for answer, I read better now but I don't understand well why they tell also that "doubly stochastic matrices is a convex polytope known as the Birkhoff polytope"
– user3520363
Jul 20 at 21:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The only matrices that are both stochastic and unitary are permutation matrices (i.e. those with one $1$ in each row and in each column). Let the rows be vectors $a_i$ with entries $a_ij$. Firstly, note that to be unitary we need $a_i^daggera_i = 1$; to be stochastic, $1^daggera_i = 1$, where $1 = (1,1,dotsc,1)$, and We also need $a_ij>0$. It follows that $0 leq a_ij leq 1 $, since each summand in a sum of positive numbers is smaller than the sum. Of course this condition also means that $a_i^dagger=a_i^T$. By subtracting, we find
$$ (1-a_i)^Ta_i = 0. $$
But this is the same as
$$ sum_j (1-a_ij)a_ij = 0, $$
and since $0 leq a_ij leq 1 $, every term in this sum is nonnegative. Hence every term is zero, so $a_ij$ is either $0$ or $1$. There can then only be exactly nonzero entry in $a_i$ since $sum_j a_ij = 1$. Since $a_i^T a_j = 0$ for orthogonality, we also find that each column can only have one nonzero term in it (if there were more, the corresponding vectors would not be orthogonal). Hence the matrix is as stated, with one $1$ in each row and in each column. It is easy to check that every such matrix is both stochastic and unitary.
(We also see therefore that every stochastic unitary matrix is actually doubly stochastic.)
answered Jul 20 at 20:41
Chappers
55k74190
Thanks for answer, I read better now but I don't understand well why they tell also that "doubly stochastic matrices is a convex polytope known as the Birkhoff polytope"
– user3520363
Jul 20 at 21:06
add a comment |Â
up vote
1
down vote
accepted
The only matrices that are both stochastic and unitary are permutation matrices (i.e. those with one $1$ in each row and in each column). Let the rows be vectors $a_i$ with entries $a_ij$. Firstly, note that to be unitary we need $a_i^daggera_i = 1$; to be stochastic, $1^daggera_i = 1$, where $1 = (1,1,dotsc,1)$, and We also need $a_ij>0$. It follows that $0 leq a_ij leq 1 $, since each summand in a sum of positive numbers is smaller than the sum. Of course this condition also means that $a_i^dagger=a_i^T$. By subtracting, we find
$$ (1-a_i)^Ta_i = 0. $$
But this is the same as
$$ sum_j (1-a_ij)a_ij = 0, $$
and since $0 leq a_ij leq 1 $, every term in this sum is nonnegative. Hence every term is zero, so $a_ij$ is either $0$ or $1$. There can then only be exactly nonzero entry in $a_i$ since $sum_j a_ij = 1$. Since $a_i^T a_j = 0$ for orthogonality, we also find that each column can only have one nonzero term in it (if there were more, the corresponding vectors would not be orthogonal). Hence the matrix is as stated, with one $1$ in each row and in each column. It is easy to check that every such matrix is both stochastic and unitary.
(We also see therefore that every stochastic unitary matrix is actually doubly stochastic.)
answered Jul 20 at 20:41
Chappers
55k74190
Thanks for answer, I read better now but I don't understand well why they tell also that "doubly stochastic matrices is a convex polytope known as the Birkhoff polytope"
– user3520363
Jul 20 at 21:06
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The only matrices that are both stochastic and unitary are permutation matrices (i.e. those with one $1$ in each row and in each column). Let the rows be vectors $a_i$ with entries $a_ij$. Firstly, note that to be unitary we need $a_i^daggera_i = 1$; to be stochastic, $1^daggera_i = 1$, where $1 = (1,1,dotsc,1)$, and We also need $a_ij>0$. It follows that $0 leq a_ij leq 1 $, since each summand in a sum of positive numbers is smaller than the sum. Of course this condition also means that $a_i^dagger=a_i^T$. By subtracting, we find
$$ (1-a_i)^Ta_i = 0. $$
But this is the same as
$$ sum_j (1-a_ij)a_ij = 0, $$
and since $0 leq a_ij leq 1 $, every term in this sum is nonnegative. Hence every term is zero, so $a_ij$ is either $0$ or $1$. There can then only be exactly nonzero entry in $a_i$ since $sum_j a_ij = 1$. Since $a_i^T a_j = 0$ for orthogonality, we also find that each column can only have one nonzero term in it (if there were more, the corresponding vectors would not be orthogonal). Hence the matrix is as stated, with one $1$ in each row and in each column. It is easy to check that every such matrix is both stochastic and unitary.
(We also see therefore that every stochastic unitary matrix is actually doubly stochastic.)
answered Jul 20 at 20:41
Chappers
55k74190
The only matrices that are both stochastic and unitary are permutation matrices (i.e. those with one $1$ in each row and in each column). Let the rows be vectors $a_i$ with entries $a_ij$. Firstly, note that to be unitary we need $a_i^daggera_i = 1$; to be stochastic, $1^daggera_i = 1$, where $1 = (1,1,dotsc,1)$, and We also need $a_ij>0$. It follows that $0 leq a_ij leq 1 $, since each summand in a sum of positive numbers is smaller than the sum. Of course this condition also means that $a_i^dagger=a_i^T$. By subtracting, we find
$$ (1-a_i)^Ta_i = 0. $$
But this is the same as
$$ sum_j (1-a_ij)a_ij = 0, $$
and since $0 leq a_ij leq 1 $, every term in this sum is nonnegative. Hence every term is zero, so $a_ij$ is either $0$ or $1$. There can then only be exactly nonzero entry in $a_i$ since $sum_j a_ij = 1$. Since $a_i^T a_j = 0$ for orthogonality, we also find that each column can only have one nonzero term in it (if there were more, the corresponding vectors would not be orthogonal). Hence the matrix is as stated, with one $1$ in each row and in each column. It is easy to check that every such matrix is both stochastic and unitary.
(We also see therefore that every stochastic unitary matrix is actually doubly stochastic.)
answered Jul 20 at 20:41
Chappers
55k74190
answered Jul 20 at 20:41
Chappers
55k74190
answered Jul 20 at 20:41
Chappers
55k74190
answered Jul 20 at 20:41
Chappers
55k74190
55k74190
Thanks for answer, I read better now but I don't understand well why they tell also that "doubly stochastic matrices is a convex polytope known as the Birkhoff polytope"
– user3520363
Jul 20 at 21:06
add a comment |Â
Thanks for answer, I read better now but I don't understand well why they tell also that "doubly stochastic matrices is a convex polytope known as the Birkhoff polytope"
– user3520363
Jul 20 at 21:06
Thanks for answer, I read better now but I don't understand well why they tell also that "doubly stochastic matrices is a convex polytope known as the Birkhoff polytope"
– user3520363
Jul 20 at 21:06
Thanks for answer, I read better now but I don't understand well why they tell also that "doubly stochastic matrices is a convex polytope known as the Birkhoff polytope"
– user3520363
Jul 20 at 21:06
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857990%2fexamples-of-stochastic-matrices-that-are-also-unitary%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
"Is a unit matrix (or identity matrix) an example of stochastic & unitary matrix at same time" - yes. It is both stochastic (actually even left and right stochastic!) and unitary. Same is true for an exchange matrix btw (all ones on its antidiagonal).
– Florian
Jul 20 at 20:39