In mathematics, a Hamiltonian matrix is a -by- matrix such that is symmetric, where is the skew-symmetric matrix
J= \begin{bmatrix} 0n&In\\ -In&0n\\ \end{bmatrix}
and is the -by- identity matrix. In other words, is Hamiltonian if and only if where denotes the transpose.[1]
Suppose that the -by- matrix is written as the block matrix
A=\begin{bmatrix}a&b\ c&d\end{bmatrix}
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 . The dimension of is . The corresponding Lie group is the symplectic group . This group consists of the symplectic matrices, those matrices which satisfy . 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.[3]
The characteristic polynomial of a real Hamiltonian matrix is even. Thus, if a Hamiltonian matrix has as an eigenvalue, then, and are also eigenvalues. It follows that the trace of a Hamiltonian matrix is zero.
The square of a Hamiltonian matrix is skew-Hamiltonian (a matrix is skew-Hamiltonian if). Conversely, every skew-Hamiltonian matrix arises as the square of a Hamiltonian matrix.[4]
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 is Hamiltonian if, as above. Another possibility is to use the condition where the superscript asterisk denotes the conjugate transpose.[5]
Let be a vector space, equipped with a symplectic form . A linear map
A: V\mapstoV
x,y\mapsto\Omega(A(x),y)
\Omega(A(x),y)=-\Omega(x,A(y))
Choose a basis in, such that is written as . A linear operator is Hamiltonian with respect to if and only if its matrix in this basis is Hamiltonian.