Negatives definition

The Usanovich theory is the most general of all acid-base theories. According to Usanovich (1939) any process leading to the formation of a salt is an acid-base reaction. The so-called positive-negative definition of Usanovich runs as follows. 

The status of H can be used to identify the character of extrema. A quadratic form <2(x) = xrHx is said to be positive-definite if Q(x) > 0 for all x = 0, and said to be positive-semidefinite if Q(x) > 0 for all x = 0. Negative-definite and negative-semidefinite are analogous except the inequality sign is reversed. If Q(x) is positive-definite (semidefinite), H(x) is said to be a positive-definite (semidefinite) matrix. These concepts can be summarized as follows ... 

It can be shown from a Taylor series expansion that if/(x) has continuous second partial derivatives, /(x) is concave if and only if its Hessian matrix is negative-semidefinite. For/(x) to be strictly concave, H must be negative-definite. For /(x) to be convex H(x) must be positive-semidefinite and for/(x) to be strictly convex, H(x) must be positive-definite. 

Positive-definite Negative-definite Unique ( isolated ) minimum Unique ( isolated ) maximum... 

Classify each of the following matrices as (a) positive-definite, (b) negative-definite,... 

Determine whether the following matrix is positive-definite, positive-semidefinite, negative-definite, negative-semidefinite, or none of the above. Show all calculations. 

This shows that the Hessian of / is positive definite (resp. negative definite) on (resp. N ). Therefore / is non-degenerate in the sense of Bott, i.e. the set of critical points is a disjoint union of submanifolds of X, and the Hessian is non-degenerate in the normal direction at any critical point. We put = dim N = 2 dime N which is the index of / at the critical manifold Cj,. Note that the index is always even in this case. 

Solution to the above Mth order equation always exists if the M + 1 by M + 1 Toeplitz matrix has a non-negative definite. In a typical experimental situation, lm f i-6-. the nonresonant contribution, dominates the CARS... 

Prove that the Hessian for the tobit model in (22-14) is negative definite after Olsen s transformation is applied to the parameters. 

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. 

If in atoms it is natural to split the Hamiltonian into an upper -turbed part with the desired characteristics and a non-negative definite perturbation J, in molecules, if one considers an unperturbed part for which the lower part of the spectrum is known, the perturbation is not positive. If one chooses the perturbation in the same way as in atoms i. e. the interelectronic repulsion, the unperturbed spectrum has no known spectrum. This dilemma was faced by Bazley and Fox, who suggested a method of truncations combined with a splitting of the unperturbed Hamiltonian into parts corresponding to the different nuclei. 

Sine e T is positive definite, even though Vi is not negative definite Then ... 

A term is positively defined if it belongs to a set that has some single characteristic or set of characteristics that can be used to test for inclusion in that set. Similarly a set is negatively-defined if it belongs to the complement of that set." The implications of such positive- or negative-definitions can be far-reaching, as described in [68]. 

A symmetric matrix A is said to be positive-definite if the quadratic form uTAu > 0 for all nonzero vectors u. Similarly, the symmetric matrix A is positive-semidefinite if uTAu 2 0 for all nonzero vectors u. Positive-definite matrices have strictly positive eigenvalues. We classify A as negative-definite if u Au < 0 for all nonzero vectors u. A is indefinite if uTAu is positive for some u and negative for others. 

Figure 5 illustrates more generally various cases that can occur for simple quadratic functions of form q x) — JxTHx, for n = 2, where H is a constant matrix. The contour plots display different characteristics when H is (a) positive-definite (elliptical contours with lowest function value at the center) and q is said to be a convex quadratic, (b) positive-semidefinite, (c) indefinite, or (d) negative-definite (elliptical contours with highest function value at the center), and q is a concave quadratic. For this figure, the following matrices are used for those different functions ... 

They have shown that the first term on the right is negative definite even for cases for which the linear phenomenological equations do not hold. By introducing the linear phenomenological equations J, = LikXk with constant coefficients we get... 

Equations (12.27) and (12.64) show the stability of the nonequilibrium stationary states in light of the fluctuations Sev The linear regime requires P > 0 and dP/dt < 0, which are Lyapunov conditions, as the matrix (dAJdej) is negative definite at near equilibrium. 

In accordance with the said criterion, the stationary state will only be stable at the negative definiteness of aU intrinsic values of matrix M p of eigenvalue equation ... 

This form is appealing because the first term in F.-,/2 can be interpreted as a gradient diffusion of turbulent kinetic energy, and the second is negative-definite (suggestive of dissipation of turbulence energy). However, the rate of entropy production is proportional to... 

Another concept that is important is that of positive (or negative) definite. For this, it is required that>l be a symmetric matrix, that is a,j = Cy,. An important theorem is the following. 

A symmetric matrix A is said to be positive definite if all of the eigenvalues are positive it is said to be negative definite if all of the eigenvalues are negative. Semidefinite is similarly defined. There is a simple test to determine if a symmetric matrix is positive or negative definite. 

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