Hamiltonian matrix

In mathematics, a Hamiltonian matrix is a $2 n$ -by- $2 n$ matrix $A$ such that $JA$ is symmetric, where $J$ is the skew-symmetric matrix

J={\begin{bmatrix}0_{n}&I_{n}\\-I_{n}&0_{n}\\\end{bmatrix}}

and $I n$ is the $n$ -by- $n$ identity matrix. In other words, $A$ is Hamiltonian if and only if $(JA) T = JA$ where $() T$ denotes the transpose.^[1]

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Properties

Suppose that the $2 n$ -by- $2 n$ matrix $A$ is written as the block matrix

A={\begin{bmatrix}a&b\\c&d\end{bmatrix}}

where $a$ , $b$ , $c$ , and $d$ are $n$ -by- $n$ matrices. Then the condition that $A$ be Hamiltonian is equivalent to requiring that the matrices $b$ and $c$ are symmetric, and that $a + d T = 0$ .^[1]^[2] Another equivalent condition is that $A$ is of the form $A = JS$ with $S$ symmetric.^[2]^: 34

It follows easily from the definition that the transpose of a Hamiltonian matrix is Hamiltonian. Furthermore, the sum (and any linear combination) of two Hamiltonian matrices is again Hamiltonian, as is their commutator. It follows that the space of all Hamiltonian matrices is a Lie algebra, denoted $sp(2 n)$ . The dimension of $sp(2 n)$ is $2 n 2 + n$ . The corresponding Lie group is the symplectic group $Sp(2 n)$ . This group consists of the symplectic matrices, those matrices $A$ which satisfy $A T JA = J$ . Thus, the matrix exponential of a Hamiltonian matrix is symplectic. However the logarithm of a symplectic matrix is not necessarily Hamiltonian because the exponential map from the Lie algebra to the group is not surjective.^[2]^: 34–36^[3]

The characteristic polynomial of a real Hamiltonian matrix is even. Thus, if a Hamiltonian matrix has $λ$ as an eigenvalue, then $-λ$ , $λ *$ and $-λ *$ are also eigenvalues.^[2]^: 45 It follows that the trace of a Hamiltonian matrix is zero.

The square of a Hamiltonian matrix is skew-Hamiltonian (a matrix $A$ is skew-Hamiltonian if $(JA) T = - JA$ ). Conversely, every skew-Hamiltonian matrix arises as the square of a Hamiltonian matrix.^[4]

Extension to complex matrices

As for symplectic matrices, the definition for Hamiltonian matrices can be extended to complex matrices in two ways. One possibility is to say that a matrix $A$ is Hamiltonian if $(JA) T = JA$ , as above.^[1]^[4] Another possibility is to use the condition $(JA) * = JA$ where the superscript asterisk ( $(\cdot) *$ ) denotes the conjugate transpose.^[5]

Hamiltonian operators

Let $V$ be a vector space, equipped with a symplectic form $Ω$ . A linear map $A:\;V\mapsto V$ is called a Hamiltonian operator with respect to $Ω$ if the form $x,y\mapsto \Omega (A(x),y)$ is symmetric. Equivalently, it should satisfy

\Omega (A(x),y)=-\Omega (x,A(y))

Choose a basis $e 1, \dots, e 2 n$ in $V$ , such that $Ω$ is written as ${\textstyle \sum _{i}e_{i}\wedge e_{n+i}}$ . A linear operator is Hamiltonian with respect to $Ω$ if and only if its matrix in this basis is Hamiltonian.^[4]

References

^ ^a ^b ^c Ikramov, Khakim D. (2001), "Hamiltonian square roots of skew-Hamiltonian matrices revisited", Linear Algebra and its Applications, 325: 101–107, doi:10.1016/S0024-3795(00)00304-9.
^ ^a ^b ^c ^d Meyer, K. R.; Hall, G. R. (1991), Introduction to Hamiltonian dynamical systems and the $N$ -body problem, Springer, ISBN 0-387-97637-X.
^ Dragt, Alex J. (2005), "The symplectic group and classical mechanics", Annals of the New York Academy of Sciences, 1045 (1): 291–307, doi:10.1196/annals.1350.025, PMID 15980319.
^ ^a ^b ^c Waterhouse, William C. (2005), "The structure of alternating-Hamiltonian matrices", Linear Algebra and its Applications, 396: 385–390, doi:10.1016/j.laa.2004.10.003.
^ Paige, Chris; Van Loan, Charles (1981), "A Schur decomposition for Hamiltonian matrices", Linear Algebra and its Applications, 41: 11–32, doi:10.1016/0024-3795(81)90086-0.

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