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Is there always a complete, orthogonal set of unitary matrices?



The Next CEO of Stack OverflowWhen can a partial isometry $u$ in $mathcal B(H otimes K)$ be extended to a unitary in $1 otimes mathcal B(K)$?Cauchy-like inequality for Kronecker (tensor) productRecovering a linear map from a non-linear approximationHow to use Galerkin method to obtain existence with spaces $V subset H$ not compactly embeddedA linear combination problemAre Hilbert-Schmidt operators on separable Hilbert spaces “Hilbert Schmidt” on the space of Hilbert Schmidt Operators?Orthogonality of positive (semi-)definite matricesConditions on a $ntimes n$ Hermitian matrix such that its extremal eigenvectors have equal magnitude entriesIs the linear span of special orthogonal matrices equal to the whole space of $Ntimes N$ matrices?Existence of orthogonal basis of symmetric $ntimes n$ matrices, where each matrix is unitary?










2












$begingroup$


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 $langle P, Q rangle = operatornametrace(PQ^*)$.



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









share|cite|improve this question









$endgroup$











  • $begingroup$
    For what it's worth, I thought of this question while trying to answer this post on MSE
    $endgroup$
    – Omnomnomnom
    4 hours ago










  • $begingroup$
    This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
    $endgroup$
    – Michael Biro
    4 hours ago










  • $begingroup$
    google on "unitary error basis"
    $endgroup$
    – Chris Godsil
    4 hours ago










  • $begingroup$
    @ChrisGodsil Excellent, thanks for the tip!
    $endgroup$
    – Omnomnomnom
    2 hours ago















2












$begingroup$


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 $langle P, Q rangle = operatornametrace(PQ^*)$.



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









share|cite|improve this question









$endgroup$











  • $begingroup$
    For what it's worth, I thought of this question while trying to answer this post on MSE
    $endgroup$
    – Omnomnomnom
    4 hours ago










  • $begingroup$
    This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
    $endgroup$
    – Michael Biro
    4 hours ago










  • $begingroup$
    google on "unitary error basis"
    $endgroup$
    – Chris Godsil
    4 hours ago










  • $begingroup$
    @ChrisGodsil Excellent, thanks for the tip!
    $endgroup$
    – Omnomnomnom
    2 hours ago













2












2








2





$begingroup$


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 $langle P, Q rangle = operatornametrace(PQ^*)$.



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









share|cite|improve this question









$endgroup$




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 $langle P, Q rangle = operatornametrace(PQ^*)$.



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









OmnomnomnomOmnomnomnom

1356




1356











  • $begingroup$
    For what it's worth, I thought of this question while trying to answer this post on MSE
    $endgroup$
    – Omnomnomnom
    4 hours ago










  • $begingroup$
    This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
    $endgroup$
    – Michael Biro
    4 hours ago










  • $begingroup$
    google on "unitary error basis"
    $endgroup$
    – Chris Godsil
    4 hours ago










  • $begingroup$
    @ChrisGodsil Excellent, thanks for the tip!
    $endgroup$
    – Omnomnomnom
    2 hours ago
















  • $begingroup$
    For what it's worth, I thought of this question while trying to answer this post on MSE
    $endgroup$
    – Omnomnomnom
    4 hours ago










  • $begingroup$
    This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
    $endgroup$
    – Michael Biro
    4 hours ago










  • $begingroup$
    google on "unitary error basis"
    $endgroup$
    – Chris Godsil
    4 hours ago










  • $begingroup$
    @ChrisGodsil Excellent, thanks for the tip!
    $endgroup$
    – Omnomnomnom
    2 hours ago















$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
4 hours ago




$begingroup$
For what it's worth, I thought of this question while trying to answer this post on MSE
$endgroup$
– Omnomnomnom
4 hours ago












$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
4 hours ago




$begingroup$
This may be useful: reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/…
$endgroup$
– Michael Biro
4 hours ago












$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
4 hours ago




$begingroup$
google on "unitary error basis"
$endgroup$
– Chris Godsil
4 hours ago












$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
2 hours ago




$begingroup$
@ChrisGodsil Excellent, thanks for the tip!
$endgroup$
– Omnomnomnom
2 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Very elegant! Thank you
    $endgroup$
    – Omnomnomnom
    2 hours ago











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Very elegant! Thank you
    $endgroup$
    – Omnomnomnom
    2 hours ago















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Very elegant! Thank you
    $endgroup$
    – Omnomnomnom
    2 hours ago













5












5








5





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Guillaume AubrunGuillaume Aubrun

1,9771424




1,9771424











  • $begingroup$
    Very elegant! Thank you
    $endgroup$
    – Omnomnomnom
    2 hours ago
















  • $begingroup$
    Very elegant! Thank you
    $endgroup$
    – Omnomnomnom
    2 hours ago















$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
2 hours ago




$begingroup$
Very elegant! Thank you
$endgroup$
– Omnomnomnom
2 hours ago

















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