Determinant of matrix formed from blocks of a $2 times 2$ block partitioned symplectic matrix.












5












$begingroup$


While working on a problem in quantum optics, I came across the following determinant of a complex matrix of size $n times n$ :



$$mathbb{G}=detleft[mathcal{U}_{11}^{}+mathcal{U}_{12}^{}mathcal{D}_{S}^{}right]$$



where $mathcal{U}_{xy}^{}$ are blocks of a $2 times 2$ block partitioned complex Symplectic matrix defined as,



$$begin{pmatrix}mathcal{U}_{11}^{} & mathcal{U}_{12}^{} \ mathcal{U}_{21}^{} & mathcal{U}_{22}^{}end{pmatrix}:=mathcal{U}_{2n times 2n}^{}=e_{}^{-mathcal{H}^{}Sigma_{}^{}}$$



with $mathcal{H}^{}Sigma_{}^{}$ being a complex Hamiltonian matrix expressed as a product of complex $2 times 2$ block partitioned symmetric matrix $$mathcal{H}^{}=begin{pmatrix}mathcal{A}_{n times n}^{}=mathcal{A}_{n times n}^{T} & mathcal{B}_{n times n}^{} \ mathcal{B}_{n times n}^{T} & mathcal{D}_{n times n}^{}=mathcal{D}_{n times n}^{T}end{pmatrix}$$
and the standard symplectic matrix
$$Sigma_{}^{}=begin{pmatrix}mathbb{O}_{n times n}^{} & mathbb{I}_{n times n}^{} \ -mathbb{I}_{n times n}^{} & mathbb{O}_{n times n}^{}end{pmatrix}.$$
Further $mathcal{D}_{S}^{}$ is a $n times n$ complex symmetric matrix and .
$mathbb{O}_{n times n}^{}$, $mathbb{I}_{n times n}^{}$ are respectively null and identity matrices of dimension $n times n$.



$mathbf{Question :}$ Is it possible to express $mathbb{G}$ in terms of expression involving traces or determinants of sums of products of $mathcal{A}_{n times n}^{}$, $mathcal{B}_{n times n}^{}$, $mathcal{D}_{n times n}^{}$, $mathcal{D}_{S}^{}$ matrices and eigenvalues of the symplectic matrix $mathcal{H}^{}Sigma_{}^{}$ without explicit computation of $mathcal{U}_{2n times 2n}^{}$, in an elegant general form? (I could do this for $n=2$ case using combination of Cayley–Hamilton theorem and Faddeev-Leverrier algorithm. For general case of $n$, I am not able to achieve this task. Without any rigorous reasoning, I suspect some elegant expression for $mathbb{G}$ can be given. Is this feasible?)






linear-algebra matrices symmetric-matrices symplectic-linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Shouldn't $mathbb{G}$ also depend on $mathcal{D}_{S}^{}$? Could you give your expression for $n=2$ so no one would waste time on rederiving it?
    $endgroup$
    – i9Fn
    Dec 27 '18 at 17:13










  • $begingroup$
    @i9Fn Thanks for pointing this out. I made appropriate edit regarding $mathcal{D}_{S}^{}$. Regarding $n=2$ case, I will add soon.
    $endgroup$
    – Sunyam
    Dec 27 '18 at 17:20
















5












$begingroup$


While working on a problem in quantum optics, I came across the following determinant of a complex matrix of size $n times n$ :



$$mathbb{G}=detleft[mathcal{U}_{11}^{}+mathcal{U}_{12}^{}mathcal{D}_{S}^{}right]$$



where $mathcal{U}_{xy}^{}$ are blocks of a $2 times 2$ block partitioned complex Symplectic matrix defined as,



$$begin{pmatrix}mathcal{U}_{11}^{} & mathcal{U}_{12}^{} \ mathcal{U}_{21}^{} & mathcal{U}_{22}^{}end{pmatrix}:=mathcal{U}_{2n times 2n}^{}=e_{}^{-mathcal{H}^{}Sigma_{}^{}}$$



with $mathcal{H}^{}Sigma_{}^{}$ being a complex Hamiltonian matrix expressed as a product of complex $2 times 2$ block partitioned symmetric matrix $$mathcal{H}^{}=begin{pmatrix}mathcal{A}_{n times n}^{}=mathcal{A}_{n times n}^{T} & mathcal{B}_{n times n}^{} \ mathcal{B}_{n times n}^{T} & mathcal{D}_{n times n}^{}=mathcal{D}_{n times n}^{T}end{pmatrix}$$
and the standard symplectic matrix
$$Sigma_{}^{}=begin{pmatrix}mathbb{O}_{n times n}^{} & mathbb{I}_{n times n}^{} \ -mathbb{I}_{n times n}^{} & mathbb{O}_{n times n}^{}end{pmatrix}.$$
Further $mathcal{D}_{S}^{}$ is a $n times n$ complex symmetric matrix and .
$mathbb{O}_{n times n}^{}$, $mathbb{I}_{n times n}^{}$ are respectively null and identity matrices of dimension $n times n$.



$mathbf{Question :}$ Is it possible to express $mathbb{G}$ in terms of expression involving traces or determinants of sums of products of $mathcal{A}_{n times n}^{}$, $mathcal{B}_{n times n}^{}$, $mathcal{D}_{n times n}^{}$, $mathcal{D}_{S}^{}$ matrices and eigenvalues of the symplectic matrix $mathcal{H}^{}Sigma_{}^{}$ without explicit computation of $mathcal{U}_{2n times 2n}^{}$, in an elegant general form? (I could do this for $n=2$ case using combination of Cayley–Hamilton theorem and Faddeev-Leverrier algorithm. For general case of $n$, I am not able to achieve this task. Without any rigorous reasoning, I suspect some elegant expression for $mathbb{G}$ can be given. Is this feasible?)






linear-algebra matrices symmetric-matrices symplectic-linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Shouldn't $mathbb{G}$ also depend on $mathcal{D}_{S}^{}$? Could you give your expression for $n=2$ so no one would waste time on rederiving it?
    $endgroup$
    – i9Fn
    Dec 27 '18 at 17:13










  • $begingroup$
    @i9Fn Thanks for pointing this out. I made appropriate edit regarding $mathcal{D}_{S}^{}$. Regarding $n=2$ case, I will add soon.
    $endgroup$
    – Sunyam
    Dec 27 '18 at 17:20














5












5








5





$begingroup$


While working on a problem in quantum optics, I came across the following determinant of a complex matrix of size $n times n$ :



$$mathbb{G}=detleft[mathcal{U}_{11}^{}+mathcal{U}_{12}^{}mathcal{D}_{S}^{}right]$$



where $mathcal{U}_{xy}^{}$ are blocks of a $2 times 2$ block partitioned complex Symplectic matrix defined as,



$$begin{pmatrix}mathcal{U}_{11}^{} & mathcal{U}_{12}^{} \ mathcal{U}_{21}^{} & mathcal{U}_{22}^{}end{pmatrix}:=mathcal{U}_{2n times 2n}^{}=e_{}^{-mathcal{H}^{}Sigma_{}^{}}$$



with $mathcal{H}^{}Sigma_{}^{}$ being a complex Hamiltonian matrix expressed as a product of complex $2 times 2$ block partitioned symmetric matrix $$mathcal{H}^{}=begin{pmatrix}mathcal{A}_{n times n}^{}=mathcal{A}_{n times n}^{T} & mathcal{B}_{n times n}^{} \ mathcal{B}_{n times n}^{T} & mathcal{D}_{n times n}^{}=mathcal{D}_{n times n}^{T}end{pmatrix}$$
and the standard symplectic matrix
$$Sigma_{}^{}=begin{pmatrix}mathbb{O}_{n times n}^{} & mathbb{I}_{n times n}^{} \ -mathbb{I}_{n times n}^{} & mathbb{O}_{n times n}^{}end{pmatrix}.$$
Further $mathcal{D}_{S}^{}$ is a $n times n$ complex symmetric matrix and .
$mathbb{O}_{n times n}^{}$, $mathbb{I}_{n times n}^{}$ are respectively null and identity matrices of dimension $n times n$.



$mathbf{Question :}$ Is it possible to express $mathbb{G}$ in terms of expression involving traces or determinants of sums of products of $mathcal{A}_{n times n}^{}$, $mathcal{B}_{n times n}^{}$, $mathcal{D}_{n times n}^{}$, $mathcal{D}_{S}^{}$ matrices and eigenvalues of the symplectic matrix $mathcal{H}^{}Sigma_{}^{}$ without explicit computation of $mathcal{U}_{2n times 2n}^{}$, in an elegant general form? (I could do this for $n=2$ case using combination of Cayley–Hamilton theorem and Faddeev-Leverrier algorithm. For general case of $n$, I am not able to achieve this task. Without any rigorous reasoning, I suspect some elegant expression for $mathbb{G}$ can be given. Is this feasible?)






linear-algebra matrices symmetric-matrices symplectic-linear-algebra







share|cite|improve this question











$endgroup$




While working on a problem in quantum optics, I came across the following determinant of a complex matrix of size $n times n$ :



$$mathbb{G}=detleft[mathcal{U}_{11}^{}+mathcal{U}_{12}^{}mathcal{D}_{S}^{}right]$$



where $mathcal{U}_{xy}^{}$ are blocks of a $2 times 2$ block partitioned complex Symplectic matrix defined as,



$$begin{pmatrix}mathcal{U}_{11}^{} & mathcal{U}_{12}^{} \ mathcal{U}_{21}^{} & mathcal{U}_{22}^{}end{pmatrix}:=mathcal{U}_{2n times 2n}^{}=e_{}^{-mathcal{H}^{}Sigma_{}^{}}$$



with $mathcal{H}^{}Sigma_{}^{}$ being a complex Hamiltonian matrix expressed as a product of complex $2 times 2$ block partitioned symmetric matrix $$mathcal{H}^{}=begin{pmatrix}mathcal{A}_{n times n}^{}=mathcal{A}_{n times n}^{T} & mathcal{B}_{n times n}^{} \ mathcal{B}_{n times n}^{T} & mathcal{D}_{n times n}^{}=mathcal{D}_{n times n}^{T}end{pmatrix}$$
and the standard symplectic matrix
$$Sigma_{}^{}=begin{pmatrix}mathbb{O}_{n times n}^{} & mathbb{I}_{n times n}^{} \ -mathbb{I}_{n times n}^{} & mathbb{O}_{n times n}^{}end{pmatrix}.$$
Further $mathcal{D}_{S}^{}$ is a $n times n$ complex symmetric matrix and .
$mathbb{O}_{n times n}^{}$, $mathbb{I}_{n times n}^{}$ are respectively null and identity matrices of dimension $n times n$.



$mathbf{Question :}$ Is it possible to express $mathbb{G}$ in terms of expression involving traces or determinants of sums of products of $mathcal{A}_{n times n}^{}$, $mathcal{B}_{n times n}^{}$, $mathcal{D}_{n times n}^{}$, $mathcal{D}_{S}^{}$ matrices and eigenvalues of the symplectic matrix $mathcal{H}^{}Sigma_{}^{}$ without explicit computation of $mathcal{U}_{2n times 2n}^{}$, in an elegant general form? (I could do this for $n=2$ case using combination of Cayley–Hamilton theorem and Faddeev-Leverrier algorithm. For general case of $n$, I am not able to achieve this task. Without any rigorous reasoning, I suspect some elegant expression for $mathbb{G}$ can be given. Is this feasible?)






linear-algebra matrices symmetric-matrices symplectic-linear-algebra




linear-algebra matrices symmetric-matrices symplectic-linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 19:59







Sunyam

















asked Sep 27 '18 at 18:42









SunyamSunyam

166214




166214












  • $begingroup$
    Shouldn't $mathbb{G}$ also depend on $mathcal{D}_{S}^{}$? Could you give your expression for $n=2$ so no one would waste time on rederiving it?
    $endgroup$
    – i9Fn
    Dec 27 '18 at 17:13










  • $begingroup$
    @i9Fn Thanks for pointing this out. I made appropriate edit regarding $mathcal{D}_{S}^{}$. Regarding $n=2$ case, I will add soon.
    $endgroup$
    – Sunyam
    Dec 27 '18 at 17:20


















  • $begingroup$
    Shouldn't $mathbb{G}$ also depend on $mathcal{D}_{S}^{}$? Could you give your expression for $n=2$ so no one would waste time on rederiving it?
    $endgroup$
    – i9Fn
    Dec 27 '18 at 17:13










  • $begingroup$
    @i9Fn Thanks for pointing this out. I made appropriate edit regarding $mathcal{D}_{S}^{}$. Regarding $n=2$ case, I will add soon.
    $endgroup$
    – Sunyam
    Dec 27 '18 at 17:20
















$begingroup$
Shouldn't $mathbb{G}$ also depend on $mathcal{D}_{S}^{}$? Could you give your expression for $n=2$ so no one would waste time on rederiving it?
$endgroup$
– i9Fn
Dec 27 '18 at 17:13




$begingroup$
Shouldn't $mathbb{G}$ also depend on $mathcal{D}_{S}^{}$? Could you give your expression for $n=2$ so no one would waste time on rederiving it?
$endgroup$
– i9Fn
Dec 27 '18 at 17:13












$begingroup$
@i9Fn Thanks for pointing this out. I made appropriate edit regarding $mathcal{D}_{S}^{}$. Regarding $n=2$ case, I will add soon.
$endgroup$
– Sunyam
Dec 27 '18 at 17:20




$begingroup$
@i9Fn Thanks for pointing this out. I made appropriate edit regarding $mathcal{D}_{S}^{}$. Regarding $n=2$ case, I will add soon.
$endgroup$
– Sunyam
Dec 27 '18 at 17:20










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