A convex combination of unitary transforms converts any matrix to identityIs there always a complete, orthogonal set of unitary matrices?Why does $d_alpha$ divide $#G$ for $alphainhatG$?A general element of U(2)Derivative with Respect to a Variable Pauli MatrixUnitary matrix for matrix representationIs there any interesting relationship between a Hermitian matrix and its corresponding entrywise absolute?Constructing element of the Weyl GroupMeasure to compare a matrix with a given unitary matrixTrace of product of three Pauli matricesImplications of unitary matrix with sub-blocks of matricesFind the maximum of the set of square moduli of eigenvalues of a family of matrices
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A convex combination of unitary transforms converts any matrix to identity
Is there always a complete, orthogonal set of unitary matrices?Why does $d_alpha$ divide $#G$ for $alphainhatG$?A general element of U(2)Derivative with Respect to a Variable Pauli MatrixUnitary matrix for matrix representationIs there any interesting relationship between a Hermitian matrix and its corresponding entrywise absolute?Constructing element of the Weyl GroupMeasure to compare a matrix with a given unitary matrixTrace of product of three Pauli matricesImplications of unitary matrix with sub-blocks of matricesFind the maximum of the set of square moduli of eigenvalues of a family of matrices
$begingroup$
Question
Show that there exists a set of unitary matrices $U_i$, and probability $p_i$, such that for any $n times n$ matrix $A$
beginequation
tag1
sum_i p_i U_i A U^dagger_i = texttr(A) fracIn
endequation
Attempts
For $n=2$, it is easy to show
beginequation
frac14 ( sigma^x A sigma^x + sigma^y A sigma^y + sigma^z A sigma^z + I A I ) = texttr(A) I / 2
endequation
where $sigma^x,y,z$ are Pauli sigma matrices. The idea comes from kraus operator sum representation.
We can then generalize to dimension $n = 2^m$, where $U_i$ can be taken as the tensor products of these basis, but not arbitrary dimension.
In indices, Eq.(1) is equivalent to
beginequation
sum_i p_i (U_i)_ab (U_i^*)_dc = delta_bc delta_ad / n
endequation
This looks like the identity from the finite dimensional irreducible unitary representation of finite group, see Peter-Weyl theorem. But again this only works when group $G$ has irreducible representation at dimension $n$, and all $p_i$ in this case are equal.
I feel that "right proof" should not utilize these additional structures.
linear-algebra matrix-decomposition
asked Mar 20 at 17:49
anecdoteanecdote
508313
$endgroup$
add a comment |
$begingroup$
Question
Show that there exists a set of unitary matrices $U_i$, and probability $p_i$, such that for any $n times n$ matrix $A$
beginequation
tag1
sum_i p_i U_i A U^dagger_i = texttr(A) fracIn
endequation
Attempts
For $n=2$, it is easy to show
beginequation
frac14 ( sigma^x A sigma^x + sigma^y A sigma^y + sigma^z A sigma^z + I A I ) = texttr(A) I / 2
endequation
where $sigma^x,y,z$ are Pauli sigma matrices. The idea comes from kraus operator sum representation.
We can then generalize to dimension $n = 2^m$, where $U_i$ can be taken as the tensor products of these basis, but not arbitrary dimension.
In indices, Eq.(1) is equivalent to
beginequation
sum_i p_i (U_i)_ab (U_i^*)_dc = delta_bc delta_ad / n
endequation
This looks like the identity from the finite dimensional irreducible unitary representation of finite group, see Peter-Weyl theorem. But again this only works when group $G$ has irreducible representation at dimension $n$, and all $p_i$ in this case are equal.
I feel that "right proof" should not utilize these additional structures.
linear-algebra matrix-decomposition
asked Mar 20 at 17:49
anecdoteanecdote
508313
$endgroup$
$begingroup$
It may help to observe that the Choi matrix of your operator is given by $$ C_Phi = frac 1n I_n otimes I_n $$ since the desired decomposition is a Kraus decomposition, the matrices $U_i$ must satisfy $$ sum_ip_i operatornamevec(U_i)operatornamevec(U_i)^dagger = C_Phi = frac 1n I_n otimes I_n $$
$endgroup$
– Omnomnomnom
Mar 20 at 19:00
$begingroup$
An educated guess: I think that $U_i = W^i$ should work, where $W$ is the $n times n$ DFT matrix, with all $p_i = frac 1n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:07
$begingroup$
Never mind, my guess fails. The $U_i$ must span the set of all $n times n$ matrices, which my guess fails to do. Notably, the $U_i$ cannot be simultaneously diagonalizable.
$endgroup$
– Omnomnomnom
Mar 20 at 19:10
$begingroup$
Perhaps it helps though to know that you're looking for a set of unitary matrices $U_i$ whose span is $Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:31
$begingroup$
@Omnomnomnom, yes problem solved if you find a basis for matrix.
$endgroup$
– anecdote
Mar 21 at 20:56
add a comment |
$begingroup$
Question
Show that there exists a set of unitary matrices $U_i$, and probability $p_i$, such that for any $n times n$ matrix $A$
beginequation
tag1
sum_i p_i U_i A U^dagger_i = texttr(A) fracIn
endequation
Attempts
For $n=2$, it is easy to show
beginequation
frac14 ( sigma^x A sigma^x + sigma^y A sigma^y + sigma^z A sigma^z + I A I ) = texttr(A) I / 2
endequation
where $sigma^x,y,z$ are Pauli sigma matrices. The idea comes from kraus operator sum representation.
We can then generalize to dimension $n = 2^m$, where $U_i$ can be taken as the tensor products of these basis, but not arbitrary dimension.
In indices, Eq.(1) is equivalent to
beginequation
sum_i p_i (U_i)_ab (U_i^*)_dc = delta_bc delta_ad / n
endequation
This looks like the identity from the finite dimensional irreducible unitary representation of finite group, see Peter-Weyl theorem. But again this only works when group $G$ has irreducible representation at dimension $n$, and all $p_i$ in this case are equal.
I feel that "right proof" should not utilize these additional structures.
linear-algebra matrix-decomposition
asked Mar 20 at 17:49
anecdoteanecdote
508313
$endgroup$
Question
Show that there exists a set of unitary matrices $U_i$, and probability $p_i$, such that for any $n times n$ matrix $A$
beginequation
tag1
sum_i p_i U_i A U^dagger_i = texttr(A) fracIn
endequation
Attempts
For $n=2$, it is easy to show
beginequation
frac14 ( sigma^x A sigma^x + sigma^y A sigma^y + sigma^z A sigma^z + I A I ) = texttr(A) I / 2
endequation
where $sigma^x,y,z$ are Pauli sigma matrices. The idea comes from kraus operator sum representation.
We can then generalize to dimension $n = 2^m$, where $U_i$ can be taken as the tensor products of these basis, but not arbitrary dimension.
In indices, Eq.(1) is equivalent to
beginequation
sum_i p_i (U_i)_ab (U_i^*)_dc = delta_bc delta_ad / n
endequation
This looks like the identity from the finite dimensional irreducible unitary representation of finite group, see Peter-Weyl theorem. But again this only works when group $G$ has irreducible representation at dimension $n$, and all $p_i$ in this case are equal.
I feel that "right proof" should not utilize these additional structures.
linear-algebra matrix-decomposition
linear-algebra matrix-decomposition
asked Mar 20 at 17:49
anecdoteanecdote
508313
asked Mar 20 at 17:49
anecdoteanecdote
508313
asked Mar 20 at 17:49
anecdoteanecdote
508313
asked Mar 20 at 17:49
anecdoteanecdote
508313
asked Mar 20 at 17:49
anecdoteanecdote
508313
508313
$begingroup$
It may help to observe that the Choi matrix of your operator is given by $$ C_Phi = frac 1n I_n otimes I_n $$ since the desired decomposition is a Kraus decomposition, the matrices $U_i$ must satisfy $$ sum_ip_i operatornamevec(U_i)operatornamevec(U_i)^dagger = C_Phi = frac 1n I_n otimes I_n $$
$endgroup$
– Omnomnomnom
Mar 20 at 19:00
$begingroup$
An educated guess: I think that $U_i = W^i$ should work, where $W$ is the $n times n$ DFT matrix, with all $p_i = frac 1n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:07
$begingroup$
Never mind, my guess fails. The $U_i$ must span the set of all $n times n$ matrices, which my guess fails to do. Notably, the $U_i$ cannot be simultaneously diagonalizable.
$endgroup$
– Omnomnomnom
Mar 20 at 19:10
$begingroup$
Perhaps it helps though to know that you're looking for a set of unitary matrices $U_i$ whose span is $Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:31
$begingroup$
@Omnomnomnom, yes problem solved if you find a basis for matrix.
$endgroup$
– anecdote
Mar 21 at 20:56
add a comment |
$begingroup$
It may help to observe that the Choi matrix of your operator is given by $$ C_Phi = frac 1n I_n otimes I_n $$ since the desired decomposition is a Kraus decomposition, the matrices $U_i$ must satisfy $$ sum_ip_i operatornamevec(U_i)operatornamevec(U_i)^dagger = C_Phi = frac 1n I_n otimes I_n $$
$endgroup$
– Omnomnomnom
Mar 20 at 19:00
$begingroup$
An educated guess: I think that $U_i = W^i$ should work, where $W$ is the $n times n$ DFT matrix, with all $p_i = frac 1n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:07
$begingroup$
Never mind, my guess fails. The $U_i$ must span the set of all $n times n$ matrices, which my guess fails to do. Notably, the $U_i$ cannot be simultaneously diagonalizable.
$endgroup$
– Omnomnomnom
Mar 20 at 19:10
$begingroup$
Perhaps it helps though to know that you're looking for a set of unitary matrices $U_i$ whose span is $Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:31
$begingroup$
@Omnomnomnom, yes problem solved if you find a basis for matrix.
$endgroup$
– anecdote
Mar 21 at 20:56
$begingroup$
It may help to observe that the Choi matrix of your operator is given by $$ C_Phi = frac 1n I_n otimes I_n $$ since the desired decomposition is a Kraus decomposition, the matrices $U_i$ must satisfy $$ sum_ip_i operatornamevec(U_i)operatornamevec(U_i)^dagger = C_Phi = frac 1n I_n otimes I_n $$
$endgroup$
– Omnomnomnom
Mar 20 at 19:00
$begingroup$
It may help to observe that the Choi matrix of your operator is given by $$ C_Phi = frac 1n I_n otimes I_n $$ since the desired decomposition is a Kraus decomposition, the matrices $U_i$ must satisfy $$ sum_ip_i operatornamevec(U_i)operatornamevec(U_i)^dagger = C_Phi = frac 1n I_n otimes I_n $$
$endgroup$
– Omnomnomnom
Mar 20 at 19:00
$begingroup$
An educated guess: I think that $U_i = W^i$ should work, where $W$ is the $n times n$ DFT matrix, with all $p_i = frac 1n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:07
$begingroup$
An educated guess: I think that $U_i = W^i$ should work, where $W$ is the $n times n$ DFT matrix, with all $p_i = frac 1n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:07
$begingroup$
Never mind, my guess fails. The $U_i$ must span the set of all $n times n$ matrices, which my guess fails to do. Notably, the $U_i$ cannot be simultaneously diagonalizable.
$endgroup$
– Omnomnomnom
Mar 20 at 19:10
$begingroup$
Never mind, my guess fails. The $U_i$ must span the set of all $n times n$ matrices, which my guess fails to do. Notably, the $U_i$ cannot be simultaneously diagonalizable.
$endgroup$
– Omnomnomnom
Mar 20 at 19:10
$begingroup$
Perhaps it helps though to know that you're looking for a set of unitary matrices $U_i$ whose span is $Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:31
$begingroup$
Perhaps it helps though to know that you're looking for a set of unitary matrices $U_i$ whose span is $Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 20 at 19:31
$begingroup$
@Omnomnomnom, yes problem solved if you find a basis for matrix.
$endgroup$
– anecdote
Mar 21 at 20:56
$begingroup$
@Omnomnomnom, yes problem solved if you find a basis for matrix.
$endgroup$
– anecdote
Mar 21 at 20:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An attempted proof of existence that doesn't actually construct the spanning set and distribution.
First, we note that the set of unitary matrices spans $Bbb C^n times n$; we could prove this nicely using polar decomposition. From there, we note that there must exist a basis of $Bbb C^n times n$ $U_1,U_2,dots,U_n^2$ consisting of unitary matrices.
It follows that the vectors $operatornamevec(U_1),dots,operatornamevec(U_n^2)$ span $Bbb C^n^2$.
The argument below is incorrect
(Thus, there necessarily exist (positive) $p_k$ such that
$$
frac 1n I_n^2 = sum_i p_i operatornamevec(U_i)operatornamevec(U_i)^dagger
$$
We correspondingly find that these $U_i$ satisfy $sum_i p_i U_iA U_i^dagger = frac 1n operatornametr(A) I$, as desired.)
Some clarification:
First of all, the linear span bit. Let $langle cdot, cdot rangle$ denote the Frobenius (Hilbert-Schmidt) inner product. Suppose that $A$ lies in the orthogonal complement of the span of the unitary matrices. Let $A = UP$ be a polar decomposition. Then we have
$$
0 = langle U, A rangle = operatornametrace(U^dagger A) = operatornametrace(U^dagger UP) = operatornametrace(P)
$$
but $P$ is positive semidefinite, so $operatornametrace(P) = 0$ implies that $P = 0$. Thus, $A$ must be zero.
So, the span of the unitary matrices is all $Bbb C^n times n$.
Another obsrevation:
Let $mathcal C_U$ denote the convex cone generated by the set $uu^* : u = operatornamevec(U) text for some unitary U $. Showing that $sum_i p_i operatornamevec(U_i)operatornamevec(U_i) = I$ can be achieved with non-negative coefficients $p_i$ means that we're trying to show that $I in mathcal C_U$.
One orthogonal basis for $Bbb C^n times n$ consisting of unitary matrices is as follows: let
$$
X = pmatrix0&&&&1\1&0\&1&0\&&ddots\&&&1, Z = pmatrix1\ & omega \ && ddots \ &&& omega^n-1
$$
Then the matrices $Z^j X^k : 0 leq j,k leq n-1$ form our orthogonal basis.
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
I am a bit slow here... Can you show how to get the linear span from the polar decomposition?
$endgroup$
– anecdote
Mar 22 at 3:17
$begingroup$
About the resolution identity: let me write $textvec(U_i)$ as $v_i$. Now $ v_i $ is a set of basis, if $I = n sum_i p_i v_i v_i^dagger$, then $G_ij = n G_ik p_k G_kj$ where $G_ij$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal.
$endgroup$
– anecdote
Mar 22 at 3:21
$begingroup$
I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here.
$endgroup$
– Omnomnomnom
Mar 22 at 13:46
$begingroup$
Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ sum_i a_i operatornamevec(U_i)operatornamevec(U_i)^dagger = frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with.
$endgroup$
– Omnomnomnom
Mar 22 at 14:00
1
$begingroup$
@anecdote Let me rephrase the argument: define $W subseteq Bbb C^n times n$ to be the subspace spanned by the unitary matrices. What I show is that if $A in W^perp$, then $A = 0$. It follows that $W = Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 22 at 16:19
|
show 7 more comments
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1 Answer
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1 Answer
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oldest
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$begingroup$
An attempted proof of existence that doesn't actually construct the spanning set and distribution.
First, we note that the set of unitary matrices spans $Bbb C^n times n$; we could prove this nicely using polar decomposition. From there, we note that there must exist a basis of $Bbb C^n times n$ $U_1,U_2,dots,U_n^2$ consisting of unitary matrices.
It follows that the vectors $operatornamevec(U_1),dots,operatornamevec(U_n^2)$ span $Bbb C^n^2$.
The argument below is incorrect
(Thus, there necessarily exist (positive) $p_k$ such that
$$
frac 1n I_n^2 = sum_i p_i operatornamevec(U_i)operatornamevec(U_i)^dagger
$$
We correspondingly find that these $U_i$ satisfy $sum_i p_i U_iA U_i^dagger = frac 1n operatornametr(A) I$, as desired.)
Some clarification:
First of all, the linear span bit. Let $langle cdot, cdot rangle$ denote the Frobenius (Hilbert-Schmidt) inner product. Suppose that $A$ lies in the orthogonal complement of the span of the unitary matrices. Let $A = UP$ be a polar decomposition. Then we have
$$
0 = langle U, A rangle = operatornametrace(U^dagger A) = operatornametrace(U^dagger UP) = operatornametrace(P)
$$
but $P$ is positive semidefinite, so $operatornametrace(P) = 0$ implies that $P = 0$. Thus, $A$ must be zero.
So, the span of the unitary matrices is all $Bbb C^n times n$.
Another obsrevation:
Let $mathcal C_U$ denote the convex cone generated by the set $uu^* : u = operatornamevec(U) text for some unitary U $. Showing that $sum_i p_i operatornamevec(U_i)operatornamevec(U_i) = I$ can be achieved with non-negative coefficients $p_i$ means that we're trying to show that $I in mathcal C_U$.
One orthogonal basis for $Bbb C^n times n$ consisting of unitary matrices is as follows: let
$$
X = pmatrix0&&&&1\1&0\&1&0\&&ddots\&&&1, Z = pmatrix1\ & omega \ && ddots \ &&& omega^n-1
$$
Then the matrices $Z^j X^k : 0 leq j,k leq n-1$ form our orthogonal basis.
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
I am a bit slow here... Can you show how to get the linear span from the polar decomposition?
$endgroup$
– anecdote
Mar 22 at 3:17
$begingroup$
About the resolution identity: let me write $textvec(U_i)$ as $v_i$. Now $ v_i $ is a set of basis, if $I = n sum_i p_i v_i v_i^dagger$, then $G_ij = n G_ik p_k G_kj$ where $G_ij$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal.
$endgroup$
– anecdote
Mar 22 at 3:21
$begingroup$
I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here.
$endgroup$
– Omnomnomnom
Mar 22 at 13:46
$begingroup$
Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ sum_i a_i operatornamevec(U_i)operatornamevec(U_i)^dagger = frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with.
$endgroup$
– Omnomnomnom
Mar 22 at 14:00
1
$begingroup$
@anecdote Let me rephrase the argument: define $W subseteq Bbb C^n times n$ to be the subspace spanned by the unitary matrices. What I show is that if $A in W^perp$, then $A = 0$. It follows that $W = Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 22 at 16:19
|
show 7 more comments
$begingroup$
An attempted proof of existence that doesn't actually construct the spanning set and distribution.
First, we note that the set of unitary matrices spans $Bbb C^n times n$; we could prove this nicely using polar decomposition. From there, we note that there must exist a basis of $Bbb C^n times n$ $U_1,U_2,dots,U_n^2$ consisting of unitary matrices.
It follows that the vectors $operatornamevec(U_1),dots,operatornamevec(U_n^2)$ span $Bbb C^n^2$.
The argument below is incorrect
(Thus, there necessarily exist (positive) $p_k$ such that
$$
frac 1n I_n^2 = sum_i p_i operatornamevec(U_i)operatornamevec(U_i)^dagger
$$
We correspondingly find that these $U_i$ satisfy $sum_i p_i U_iA U_i^dagger = frac 1n operatornametr(A) I$, as desired.)
Some clarification:
First of all, the linear span bit. Let $langle cdot, cdot rangle$ denote the Frobenius (Hilbert-Schmidt) inner product. Suppose that $A$ lies in the orthogonal complement of the span of the unitary matrices. Let $A = UP$ be a polar decomposition. Then we have
$$
0 = langle U, A rangle = operatornametrace(U^dagger A) = operatornametrace(U^dagger UP) = operatornametrace(P)
$$
but $P$ is positive semidefinite, so $operatornametrace(P) = 0$ implies that $P = 0$. Thus, $A$ must be zero.
So, the span of the unitary matrices is all $Bbb C^n times n$.
Another obsrevation:
Let $mathcal C_U$ denote the convex cone generated by the set $uu^* : u = operatornamevec(U) text for some unitary U $. Showing that $sum_i p_i operatornamevec(U_i)operatornamevec(U_i) = I$ can be achieved with non-negative coefficients $p_i$ means that we're trying to show that $I in mathcal C_U$.
One orthogonal basis for $Bbb C^n times n$ consisting of unitary matrices is as follows: let
$$
X = pmatrix0&&&&1\1&0\&1&0\&&ddots\&&&1, Z = pmatrix1\ & omega \ && ddots \ &&& omega^n-1
$$
Then the matrices $Z^j X^k : 0 leq j,k leq n-1$ form our orthogonal basis.
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
I am a bit slow here... Can you show how to get the linear span from the polar decomposition?
$endgroup$
– anecdote
Mar 22 at 3:17
$begingroup$
About the resolution identity: let me write $textvec(U_i)$ as $v_i$. Now $ v_i $ is a set of basis, if $I = n sum_i p_i v_i v_i^dagger$, then $G_ij = n G_ik p_k G_kj$ where $G_ij$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal.
$endgroup$
– anecdote
Mar 22 at 3:21
$begingroup$
I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here.
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– Omnomnomnom
Mar 22 at 13:46
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Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ sum_i a_i operatornamevec(U_i)operatornamevec(U_i)^dagger = frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with.
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– Omnomnomnom
Mar 22 at 14:00
1
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@anecdote Let me rephrase the argument: define $W subseteq Bbb C^n times n$ to be the subspace spanned by the unitary matrices. What I show is that if $A in W^perp$, then $A = 0$. It follows that $W = Bbb C^n times n$
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– Omnomnomnom
Mar 22 at 16:19
|
show 7 more comments
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An attempted proof of existence that doesn't actually construct the spanning set and distribution.
First, we note that the set of unitary matrices spans $Bbb C^n times n$; we could prove this nicely using polar decomposition. From there, we note that there must exist a basis of $Bbb C^n times n$ $U_1,U_2,dots,U_n^2$ consisting of unitary matrices.
It follows that the vectors $operatornamevec(U_1),dots,operatornamevec(U_n^2)$ span $Bbb C^n^2$.
The argument below is incorrect
(Thus, there necessarily exist (positive) $p_k$ such that
$$
frac 1n I_n^2 = sum_i p_i operatornamevec(U_i)operatornamevec(U_i)^dagger
$$
We correspondingly find that these $U_i$ satisfy $sum_i p_i U_iA U_i^dagger = frac 1n operatornametr(A) I$, as desired.)
Some clarification:
First of all, the linear span bit. Let $langle cdot, cdot rangle$ denote the Frobenius (Hilbert-Schmidt) inner product. Suppose that $A$ lies in the orthogonal complement of the span of the unitary matrices. Let $A = UP$ be a polar decomposition. Then we have
$$
0 = langle U, A rangle = operatornametrace(U^dagger A) = operatornametrace(U^dagger UP) = operatornametrace(P)
$$
but $P$ is positive semidefinite, so $operatornametrace(P) = 0$ implies that $P = 0$. Thus, $A$ must be zero.
So, the span of the unitary matrices is all $Bbb C^n times n$.
Another obsrevation:
Let $mathcal C_U$ denote the convex cone generated by the set $uu^* : u = operatornamevec(U) text for some unitary U $. Showing that $sum_i p_i operatornamevec(U_i)operatornamevec(U_i) = I$ can be achieved with non-negative coefficients $p_i$ means that we're trying to show that $I in mathcal C_U$.
One orthogonal basis for $Bbb C^n times n$ consisting of unitary matrices is as follows: let
$$
X = pmatrix0&&&&1\1&0\&1&0\&&ddots\&&&1, Z = pmatrix1\ & omega \ && ddots \ &&& omega^n-1
$$
Then the matrices $Z^j X^k : 0 leq j,k leq n-1$ form our orthogonal basis.
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
An attempted proof of existence that doesn't actually construct the spanning set and distribution.
First, we note that the set of unitary matrices spans $Bbb C^n times n$; we could prove this nicely using polar decomposition. From there, we note that there must exist a basis of $Bbb C^n times n$ $U_1,U_2,dots,U_n^2$ consisting of unitary matrices.
It follows that the vectors $operatornamevec(U_1),dots,operatornamevec(U_n^2)$ span $Bbb C^n^2$.
The argument below is incorrect
(Thus, there necessarily exist (positive) $p_k$ such that
$$
frac 1n I_n^2 = sum_i p_i operatornamevec(U_i)operatornamevec(U_i)^dagger
$$
We correspondingly find that these $U_i$ satisfy $sum_i p_i U_iA U_i^dagger = frac 1n operatornametr(A) I$, as desired.)
Some clarification:
First of all, the linear span bit. Let $langle cdot, cdot rangle$ denote the Frobenius (Hilbert-Schmidt) inner product. Suppose that $A$ lies in the orthogonal complement of the span of the unitary matrices. Let $A = UP$ be a polar decomposition. Then we have
$$
0 = langle U, A rangle = operatornametrace(U^dagger A) = operatornametrace(U^dagger UP) = operatornametrace(P)
$$
but $P$ is positive semidefinite, so $operatornametrace(P) = 0$ implies that $P = 0$. Thus, $A$ must be zero.
So, the span of the unitary matrices is all $Bbb C^n times n$.
Another obsrevation:
Let $mathcal C_U$ denote the convex cone generated by the set $uu^* : u = operatornamevec(U) text for some unitary U $. Showing that $sum_i p_i operatornamevec(U_i)operatornamevec(U_i) = I$ can be achieved with non-negative coefficients $p_i$ means that we're trying to show that $I in mathcal C_U$.
One orthogonal basis for $Bbb C^n times n$ consisting of unitary matrices is as follows: let
$$
X = pmatrix0&&&&1\1&0\&1&0\&&ddots\&&&1, Z = pmatrix1\ & omega \ && ddots \ &&& omega^n-1
$$
Then the matrices $Z^j X^k : 0 leq j,k leq n-1$ form our orthogonal basis.
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
edited 2 days ago
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
answered Mar 21 at 22:33
OmnomnomnomOmnomnomnom
129k793187
129k793187
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I am a bit slow here... Can you show how to get the linear span from the polar decomposition?
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– anecdote
Mar 22 at 3:17
$begingroup$
About the resolution identity: let me write $textvec(U_i)$ as $v_i$. Now $ v_i $ is a set of basis, if $I = n sum_i p_i v_i v_i^dagger$, then $G_ij = n G_ik p_k G_kj$ where $G_ij$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal.
$endgroup$
– anecdote
Mar 22 at 3:21
$begingroup$
I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here.
$endgroup$
– Omnomnomnom
Mar 22 at 13:46
$begingroup$
Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ sum_i a_i operatornamevec(U_i)operatornamevec(U_i)^dagger = frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with.
$endgroup$
– Omnomnomnom
Mar 22 at 14:00
1
$begingroup$
@anecdote Let me rephrase the argument: define $W subseteq Bbb C^n times n$ to be the subspace spanned by the unitary matrices. What I show is that if $A in W^perp$, then $A = 0$. It follows that $W = Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 22 at 16:19
|
show 7 more comments
$begingroup$
I am a bit slow here... Can you show how to get the linear span from the polar decomposition?
$endgroup$
– anecdote
Mar 22 at 3:17
$begingroup$
About the resolution identity: let me write $textvec(U_i)$ as $v_i$. Now $ v_i $ is a set of basis, if $I = n sum_i p_i v_i v_i^dagger$, then $G_ij = n G_ik p_k G_kj$ where $G_ij$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal.
$endgroup$
– anecdote
Mar 22 at 3:21
$begingroup$
I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here.
$endgroup$
– Omnomnomnom
Mar 22 at 13:46
$begingroup$
Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ sum_i a_i operatornamevec(U_i)operatornamevec(U_i)^dagger = frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with.
$endgroup$
– Omnomnomnom
Mar 22 at 14:00
1
$begingroup$
@anecdote Let me rephrase the argument: define $W subseteq Bbb C^n times n$ to be the subspace spanned by the unitary matrices. What I show is that if $A in W^perp$, then $A = 0$. It follows that $W = Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 22 at 16:19
$begingroup$
I am a bit slow here... Can you show how to get the linear span from the polar decomposition?
$endgroup$
– anecdote
Mar 22 at 3:17
$begingroup$
I am a bit slow here... Can you show how to get the linear span from the polar decomposition?
$endgroup$
– anecdote
Mar 22 at 3:17
$begingroup$
About the resolution identity: let me write $textvec(U_i)$ as $v_i$. Now $ v_i $ is a set of basis, if $I = n sum_i p_i v_i v_i^dagger$, then $G_ij = n G_ik p_k G_kj$ where $G_ij$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal.
$endgroup$
– anecdote
Mar 22 at 3:21
$begingroup$
About the resolution identity: let me write $textvec(U_i)$ as $v_i$. Now $ v_i $ is a set of basis, if $I = n sum_i p_i v_i v_i^dagger$, then $G_ij = n G_ik p_k G_kj$ where $G_ij$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal.
$endgroup$
– anecdote
Mar 22 at 3:21
$begingroup$
I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here.
$endgroup$
– Omnomnomnom
Mar 22 at 13:46
$begingroup$
I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here.
$endgroup$
– Omnomnomnom
Mar 22 at 13:46
$begingroup$
Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ sum_i a_i operatornamevec(U_i)operatornamevec(U_i)^dagger = frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with.
$endgroup$
– Omnomnomnom
Mar 22 at 14:00
$begingroup$
Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ sum_i a_i operatornamevec(U_i)operatornamevec(U_i)^dagger = frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with.
$endgroup$
– Omnomnomnom
Mar 22 at 14:00
1
1
$begingroup$
@anecdote Let me rephrase the argument: define $W subseteq Bbb C^n times n$ to be the subspace spanned by the unitary matrices. What I show is that if $A in W^perp$, then $A = 0$. It follows that $W = Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 22 at 16:19
$begingroup$
@anecdote Let me rephrase the argument: define $W subseteq Bbb C^n times n$ to be the subspace spanned by the unitary matrices. What I show is that if $A in W^perp$, then $A = 0$. It follows that $W = Bbb C^n times n$
$endgroup$
– Omnomnomnom
Mar 22 at 16:19
|
show 7 more comments
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$begingroup$
It may help to observe that the Choi matrix of your operator is given by $$ C_Phi = frac 1n I_n otimes I_n $$ since the desired decomposition is a Kraus decomposition, the matrices $U_i$ must satisfy $$ sum_ip_i operatornamevec(U_i)operatornamevec(U_i)^dagger = C_Phi = frac 1n I_n otimes I_n $$
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– Omnomnomnom
Mar 20 at 19:00
$begingroup$
An educated guess: I think that $U_i = W^i$ should work, where $W$ is the $n times n$ DFT matrix, with all $p_i = frac 1n$
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– Omnomnomnom
Mar 20 at 19:07
$begingroup$
Never mind, my guess fails. The $U_i$ must span the set of all $n times n$ matrices, which my guess fails to do. Notably, the $U_i$ cannot be simultaneously diagonalizable.
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– Omnomnomnom
Mar 20 at 19:10
$begingroup$
Perhaps it helps though to know that you're looking for a set of unitary matrices $U_i$ whose span is $Bbb C^n times n$
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– Omnomnomnom
Mar 20 at 19:31
$begingroup$
@Omnomnomnom, yes problem solved if you find a basis for matrix.
$endgroup$
– anecdote
Mar 21 at 20:56