Semi definition

We assume that A is a symmetric and positive semi-definite matrix. The case of interest is when the largest eigenvalue of A is significantly larger than the norm of the derivative of the nonlinear force f. A may be a constant matrix, or else A = A(y) is assumed to be slowly changing along solution trajectories, in which case A will be evaluated at the current averaged position in the numerical schemes below. In the standard Verlet scheme, which yields approximations y to y nAt) via... 

Here C = aa ) is the covariance of the basis functions used to model the turbulence. Covariance matrices are positive semi-definite by dehnition which implies a C a > 0, and thus a dehned maximum of Pr a exists. 

It can be shown that all symmetric matrices of the form X X and XX are positive semi-definite [2]. These cross-product matrices include the widely used dispersion matrices which can take the form of a variance-covariance or correlation matrix, among others (see Section 29.7). 

By way of example we construct a positive semi-definite matrix A of dimensions 2x2 from which we propose to determine the characteristic roots. The square matrix A is derived as the product of a rectangular matrix X with its transpose in order to ensure symmetry and positive semi-definitiveness ... 

If A is a symmetric positive definite matrix then we obtain that all eigenvalues are positive. As we have seen, this occurs when all columns (or rows) of the matrix A are linearly independent. Conversely, a linear dependence in the columns (or rows) of A will produce a zero eigenvalue. More generally, if A is symmetric and positive semi-definite of rank r[Pg.32]

Thus far we have considered the eigenvalue decomposition of a symmetric matrix which is of full rank, i.e. which is positive definite. In the more general case of a symmetric positive semi-definite pxp matrix A we will obtain r positive eigenvalues where r general case we obtain a pxr matrix of eigenvectors V such that ... 

Newton, the limit h —> 0 is singular. The symmetries underlying quantum and classical dynamics - unitarity and symplecticity, respectively - are fundamentally incompatible with the opposing theory s notion of a physical state quantum-mechanically, a positive semi-definite density matrix classically, a positive phase-space distribution function. 

The covariance matrix is positive semi-definite and symmetric. Thus, it can be written in terms of eigenvalues and eigenvectors as... 

Making use of the property of a semi-definite one-body operator, say, L ... 

To gain an understanding of this mechanism, consider the Hamiltonian operator (H — Egl) with only two-body interactions, where Eg is the lowest energy for an A -particle system with Hamiltonian H and the identity operator I. Because Eg is the lowest (or ground-state) energy, the Hamiltonian operator is positive semi-definite on the A -electron space that is, the expectation values of H with respect to all A -particle functions are nonnegative. Assume that the Hamiltonian may be expanded as a sum of operators G,G,... 

M. Nakata, H. Nakatsuji, M. Ehara, M. Fukuda, K. Nakata, and K. Fujisawa, Variational calculations of fermion second-order reduced density matrices by semi-definite programming algorithm. J. Chem. Phys. 114, 8282-8202 (2001). 

Because all of the components of J are Hermitian, and because the scalar product of any function with itself is positive semi-definite, the following identity holds ... 

Exercise. Take any r real numbers k1,k2,...,kr and consider the rxr matrix whose i, j element is G(/c, — kj). Prove that this matrix is positive definite or semi-definite for some special distributions. Functions G having this property for all sets k are called positive definite or of positive type . 

The expansion in eigenfunctions leads to expressions of the various quantities pertaining to the stochastic process - as in equations (7.13) through (7.16). It also simplifies some of the derivations, in particular the proof of the approach to equilibrium. In fact, according to (7.13) it is sufficient to prove that all X other than X = 0 are positive, i.e., that W is negative semi-definite. In the same notation as in V.5 one has for any vector pn = x pl in the Hilbert space... 

Exercise. Define jump moments for the case that Y has more components. Show that the matrix a (ye) must be negative definite, or at least semi-definite. 

Exercise. Consider the right-hand side of (3.5) as a linear operator W acting on the space of functions P(X) defined for 0 < X < oo and obeying (3.6). Verify that it has the symmetry property (V.7.5) and is negative semi-definite, the only eigenfunction with zero eigenvalue being (3.7). 

The coefficients and may be any real differentiable functions with the sole restriction that the matrix By is taken to be symmetric and must be positive definite. More precisely, at each point y of the r-dimensional space, Bij(y) must be nonnegative semi-definite ... 

Exercise. The matrix S is by definition symmetric and positive definite (or at least semi-definite). Show that S has the same properties, and that they are preserved by equation (6.9). 

It is a bivariate quasilinear Fokker-Planck equation of the form (6.1) with semi-definite fJ. 

Here x stands for the set x,, and summation over repeated indices is implied. The matrix By(x) is symmetric and positive semi-definite. Alternatively one may write... 

The Kramers equation (VIII.7.4) has the form (4.1) with two variables, nonlinear 4,(x), and constant Bijt which is semi-definite, while... 

Exercise. Prove the following lemma If H is a positive semi-definite Hermitian matrix, and F anti-Hermitian then the eigenvalues of A = H + F have nonnegative real parts. Moreover, if the real part is zero the corresponding eigenvector is an eigenvector of H and F separately. Use this lemma to show that (5.12) is the solution of (5.10). 

Note that since the matrix T is positive (semi)definite, the first K eigenvalues Tk, k - 1,. .., K, are all nonnegative. The other N - K eigenvectors of T all have zero as an eigenvalue. (Note that the nonvanishing eigenvalues of T must equal those of the K x K Hermitian matrix ft s V V.)... 

In theorem 12 of P it was shown that the first term is that of a positive definite Gramian matrix. The second is clearly the general term of a positive semi-definite matrix we cannot assert that it is definite since one or more of the A Hi might be zero. But the sum of a positive definite and a positive semi-definite matrix is definite and so the Jacobian nowhere vanishes. 

Hence, V0 is negative semi-definite this guarantees the boundedness of and 90. By invoking Barbalat s lemma [3], it can be recognized that V0 -> 0, which implies the global uniform convergence to 0 of as t -> oo, while 9a is only guaranteed to be uniformly bounded. ... 

The function V is guaranteed to be negative semi-definite if the arbitrary positive constant q is chosen so as to satisfy the inequality... 

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