Is there always a complete, orthogonal set of unitary matrices?
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The set of size-$n$ unitary matrices span $Bbb C^{n times n}$ (this can be proven nicely using polar decomposition). If we select a maximal linear subset of unitary matrices, then we have a basis of $Bbb C^{n times n}$ consisting of $n^2$ unitary matrices. My question is whether we can find a basis that satisfies the additional constraint of orthogonality. That is:
Does there exist a basis $mathcal B$ of $Bbb C^{n times n}$ such that every $P in mathcal B$ is unitary (that is, $P^*P = I$) and for all distinct $P,Q in mathcal B$, we have $langle P,Q rangle = 0$?
Here, $langle cdot , cdot rangle$ refers to the Frobenius (AKA Hilbert-Schmidt) inner product, namely $langle P, Q rangle = operatorname{trace}(PQ^*)$.
When $n = 2$, the Pauli matrices provide a convenient solution. That is, we can take
$$
mathcal B = {I,sigma_1,sigma_2,sigma_3} subset Bbb C^{2 times 2}.
$$
We can use this to produce a solution whenever $n = 2^k$. In particular, if we define $sigma_0 = I$ for convenience, we can take
$$
mathcal B = {sigma_{m_1} otimes cdots otimes sigma_{m_k} : 0 leq m_j leq 3} subset Bbb C^{2^k times 2^k}.
$$
Could we come up with a basis for any other $n$? Could we do so for every $n$?
Some observations so far:
- Without loss of generality, we can assume that $mathcal B$ contains the $n times n$ identity matrix $I$. If $I$ is an element of the basis, it follows that the remaining matrices form a basis for the subspace of all trace-$0$ matrices.
- A commuting set of matrices spans at most an $n$-dimensional subset, so there must be elements of $mathcal B$ that fail to commute
fa.functional-analysis linear-algebra matrices
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
$endgroup$
add a comment |
$begingroup$
The set of size-$n$ unitary matrices span $Bbb C^{n times n}$ (this can be proven nicely using polar decomposition). If we select a maximal linear subset of unitary matrices, then we have a basis of $Bbb C^{n times n}$ consisting of $n^2$ unitary matrices. My question is whether we can find a basis that satisfies the additional constraint of orthogonality. That is:
Does there exist a basis $mathcal B$ of $Bbb C^{n times n}$ such that every $P in mathcal B$ is unitary (that is, $P^*P = I$) and for all distinct $P,Q in mathcal B$, we have $langle P,Q rangle = 0$?
Here, $langle cdot , cdot rangle$ refers to the Frobenius (AKA Hilbert-Schmidt) inner product, namely $langle P, Q rangle = operatorname{trace}(PQ^*)$.
When $n = 2$, the Pauli matrices provide a convenient solution. That is, we can take
$$
mathcal B = {I,sigma_1,sigma_2,sigma_3} subset Bbb C^{2 times 2}.
$$
We can use this to produce a solution whenever $n = 2^k$. In particular, if we define $sigma_0 = I$ for convenience, we can take
$$
mathcal B = {sigma_{m_1} otimes cdots otimes sigma_{m_k} : 0 leq m_j leq 3} subset Bbb C^{2^k times 2^k}.
$$
Could we come up with a basis for any other $n$? Could we do so for every $n$?
Some observations so far:
- Without loss of generality, we can assume that $mathcal B$ contains the $n times n$ identity matrix $I$. If $I$ is an element of the basis, it follows that the remaining matrices form a basis for the subspace of all trace-$0$ matrices.
- A commuting set of matrices spans at most an $n$-dimensional subset, so there must be elements of $mathcal B$ that fail to commute
fa.functional-analysis linear-algebra matrices
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
$endgroup$
$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
Mar 30 at 18:28
$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
Mar 30 at 18:40
$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
Mar 30 at 19:00
$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
Mar 30 at 20:22
add a comment |
$begingroup$
The set of size-$n$ unitary matrices span $Bbb C^{n times n}$ (this can be proven nicely using polar decomposition). If we select a maximal linear subset of unitary matrices, then we have a basis of $Bbb C^{n times n}$ consisting of $n^2$ unitary matrices. My question is whether we can find a basis that satisfies the additional constraint of orthogonality. That is:
Does there exist a basis $mathcal B$ of $Bbb C^{n times n}$ such that every $P in mathcal B$ is unitary (that is, $P^*P = I$) and for all distinct $P,Q in mathcal B$, we have $langle P,Q rangle = 0$?
Here, $langle cdot , cdot rangle$ refers to the Frobenius (AKA Hilbert-Schmidt) inner product, namely $langle P, Q rangle = operatorname{trace}(PQ^*)$.
When $n = 2$, the Pauli matrices provide a convenient solution. That is, we can take
$$
mathcal B = {I,sigma_1,sigma_2,sigma_3} subset Bbb C^{2 times 2}.
$$
We can use this to produce a solution whenever $n = 2^k$. In particular, if we define $sigma_0 = I$ for convenience, we can take
$$
mathcal B = {sigma_{m_1} otimes cdots otimes sigma_{m_k} : 0 leq m_j leq 3} subset Bbb C^{2^k times 2^k}.
$$
Could we come up with a basis for any other $n$? Could we do so for every $n$?
Some observations so far:
- Without loss of generality, we can assume that $mathcal B$ contains the $n times n$ identity matrix $I$. If $I$ is an element of the basis, it follows that the remaining matrices form a basis for the subspace of all trace-$0$ matrices.
- A commuting set of matrices spans at most an $n$-dimensional subset, so there must be elements of $mathcal B$ that fail to commute
fa.functional-analysis linear-algebra matrices
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
$endgroup$
The set of size-$n$ unitary matrices span $Bbb C^{n times n}$ (this can be proven nicely using polar decomposition). If we select a maximal linear subset of unitary matrices, then we have a basis of $Bbb C^{n times n}$ consisting of $n^2$ unitary matrices. My question is whether we can find a basis that satisfies the additional constraint of orthogonality. That is:
Does there exist a basis $mathcal B$ of $Bbb C^{n times n}$ such that every $P in mathcal B$ is unitary (that is, $P^*P = I$) and for all distinct $P,Q in mathcal B$, we have $langle P,Q rangle = 0$?
Here, $langle cdot , cdot rangle$ refers to the Frobenius (AKA Hilbert-Schmidt) inner product, namely $langle P, Q rangle = operatorname{trace}(PQ^*)$.
When $n = 2$, the Pauli matrices provide a convenient solution. That is, we can take
$$
mathcal B = {I,sigma_1,sigma_2,sigma_3} subset Bbb C^{2 times 2}.
$$
We can use this to produce a solution whenever $n = 2^k$. In particular, if we define $sigma_0 = I$ for convenience, we can take
$$
mathcal B = {sigma_{m_1} otimes cdots otimes sigma_{m_k} : 0 leq m_j leq 3} subset Bbb C^{2^k times 2^k}.
$$
Could we come up with a basis for any other $n$? Could we do so for every $n$?
Some observations so far:
- Without loss of generality, we can assume that $mathcal B$ contains the $n times n$ identity matrix $I$. If $I$ is an element of the basis, it follows that the remaining matrices form a basis for the subspace of all trace-$0$ matrices.
- A commuting set of matrices spans at most an $n$-dimensional subset, so there must be elements of $mathcal B$ that fail to commute
fa.functional-analysis linear-algebra matrices
fa.functional-analysis linear-algebra matrices
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
asked Mar 30 at 18:26
OmnomnomnomOmnomnomnom
1406
1406
$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
Mar 30 at 18:28
$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
Mar 30 at 18:40
$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
Mar 30 at 19:00
$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
Mar 30 at 20:22
add a comment |
$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
Mar 30 at 18:28
$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
Mar 30 at 18:40
$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
Mar 30 at 19:00
$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
Mar 30 at 20:22
$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
Mar 30 at 18:28
$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
Mar 30 at 18:28
$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
Mar 30 at 18:40
$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
Mar 30 at 18:40
$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
Mar 30 at 19:00
$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
Mar 30 at 19:00
$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
Mar 30 at 20:22
$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
Mar 30 at 20:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, consider the group of $n^2$ matrices generated by the shift $e_i mapsto e_{i+1}$ and the diagonal matrix with entries $(1,omega,omega^2,cdots,omega^{n-1})$ where $omega$ is a primitive $n$th root of unit.
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
$endgroup$
$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
Mar 30 at 20:26
1
$begingroup$
For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices.
$endgroup$
– Nathaniel Johnston
Mar 31 at 17:08
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
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$begingroup$
Yes, consider the group of $n^2$ matrices generated by the shift $e_i mapsto e_{i+1}$ and the diagonal matrix with entries $(1,omega,omega^2,cdots,omega^{n-1})$ where $omega$ is a primitive $n$th root of unit.
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
$endgroup$
$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
Mar 30 at 20:26
1
$begingroup$
For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices.
$endgroup$
– Nathaniel Johnston
Mar 31 at 17:08
add a comment |
$begingroup$
Yes, consider the group of $n^2$ matrices generated by the shift $e_i mapsto e_{i+1}$ and the diagonal matrix with entries $(1,omega,omega^2,cdots,omega^{n-1})$ where $omega$ is a primitive $n$th root of unit.
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
$endgroup$
$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
Mar 30 at 20:26
1
$begingroup$
For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices.
$endgroup$
– Nathaniel Johnston
Mar 31 at 17:08
add a comment |
$begingroup$
Yes, consider the group of $n^2$ matrices generated by the shift $e_i mapsto e_{i+1}$ and the diagonal matrix with entries $(1,omega,omega^2,cdots,omega^{n-1})$ where $omega$ is a primitive $n$th root of unit.
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
$endgroup$
Yes, consider the group of $n^2$ matrices generated by the shift $e_i mapsto e_{i+1}$ and the diagonal matrix with entries $(1,omega,omega^2,cdots,omega^{n-1})$ where $omega$ is a primitive $n$th root of unit.
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
answered Mar 30 at 18:37
Guillaume AubrunGuillaume Aubrun
2,1721625
2,1721625
$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
Mar 30 at 20:26
1
$begingroup$
For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices.
$endgroup$
– Nathaniel Johnston
Mar 31 at 17:08
add a comment |
$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
Mar 30 at 20:26
1
$begingroup$
For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices.
$endgroup$
– Nathaniel Johnston
Mar 31 at 17:08
$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
Mar 30 at 20:26
$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
Mar 30 at 20:26
1
1
$begingroup$
For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices.
$endgroup$
– Nathaniel Johnston
Mar 31 at 17:08
$begingroup$
For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices.
$endgroup$
– Nathaniel Johnston
Mar 31 at 17:08
add a comment |
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$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
Mar 30 at 18:28
$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
Mar 30 at 18:40
$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
Mar 30 at 19:00
$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
Mar 30 at 20:22