Eigenvalues proof

Based on the above similarity transform, we ean now show that the traee of a matrix (i.e., the sum of its diagonal elements) is independent of the representation in whieh the matrix is formed, and, in partieular, the traee is equal to the sum of the eigenvalues of the matrix. The proof of this theorem proeeeds as follows ... 

We give a simple proof of this result The lemma says that T has a basis of common eigenvectors to T with the eigenvalues as in the above formula. By... 

The gap between the energy and bottom eigenvalue is nonnegative. If the fe-matrix P moves to further decrease the energy, and if the Pauli matrix S moves to further increase the bottom eigenvalue, the gap narrows and possibly shrinks to zero. It is important to note that there are semidefinite programs where this gap cannot shrink to zero we discuss such an example later. However, In our special case where we vary fc-matrices and Pauli matrices, as we have defined them, the gap shrinks to zero. This is an important result for both theoretical and practical reasons a proof is supplied below. 

By Eq. (6) the sum on the right-hand side of the above equation is equal to the energy E, and from Eq. (2) we realize that the sums on the left-hand side are just Hamiltonian operators in the second-quantized notation. Hence, when the 2-RDM corresponds to an A -particle wavefunction i//, Eq. (12) implies Eq. (13), and the proof of Nakatsuji s theorem is accomplished. Because the Hamiltonian is dehned in second quantization, the proof of Nakatsuji s theorem is also valid when the one-particle basis set is incomplete. Recall that the SE with a second-quantized Hamiltonian corresponds to a Hamiltonian eigenvalue equation with the given one-particle basis. Unlike the SE, the CSE only requires the 2- and 4-RDMs in the given one-particle basis rather than the full A -particle wavefunction. While Nakatsuji s theorem holds for the 2,4-CSE, it is not valid for the 1,3-CSE. This foreshadows the advantage of reconstructing from the 2-RDM instead of the 1-RDM, which we will discuss in the context of Rosina s theorem. 

The Teller proof (33) assumes that the eigenfunctions of the two states of concern and V/j say) may be written as a linear combination of two orthonormal basis functions and o, . The energies of the two states are then identical to the eigenvalues of the 2 x 2 Hamiltonian matrix... 

Longuet-Higgins has extended the Teller proof by considering the eigenvalues of the symmetric matrix obtained from a general orthonormal basis ofN states a, ... 

Proof. If A is diagonal, then an easy computation shows that AD — DA = 0. To prove the other implication, suppose that AD — DA = 0, Let e denote the fth standard basis vector of C . Then 0 = AD — DAid = DuAci — DAa. So Aci is an eigenvector of D with eigenvalue D. Because Da Djj unless i = j, it follows that Aa must be a scalar multiple of a for each L Hence A must be diagonal. ... 

Proof. Consider the characteristic polynomial of T, that is, det (A/ — T). Because of the A" term, this complex-coefficient polynomial has degree n > 0. Hence, by the Fundamental Theorem of Algebra, this polynomial has at least one complex root. In other words, there exists a A e C such that det(A/ — T) = 0. This implies that there is a nonzero r> e V such that (A/ — r)u = 0. Hence Xv = Tv and A is an eigenvalue of T. ... 

This proof does not give a method for finding real eigenvalues of real linear operators, because the Fundamental Theorem of Algebra does not guarantee real roots for polynomials with real coefficients. Proposition 2.11 does not hold for inhnite-dimensional complex vector spaces eiffier. See Fxercise 2.28. 

Proof. Suppose w is an eigenvector for T with eigenvalue A. We must show that for any g e G, the vector /)(g)w is an eigenvector for T with eigenvalue A. We have... 

Proof. Suppose that (G, V, p) is irreducible and the linear transformation T V —> V commutes with p. We must show that T is a scalar multiple of the identity. Because V is finite dimensional there must be at least one eigenvalue Z of T (by Proposition 2.11). By Proposition 5.2. the eigenspace corresponding to A must be an invariant space for p. This space is not trivial, so because p is irreducible it must be all of V. In other words, T =. 1. So T is a scalar multiple of the identity. ... 

The reader may wish to compare this Spectral Theorem to Proposition 4.4. Proof. To find the eigenvalues of A, we consider its characteristic polynomial. Then we use eigenvectors to construct the matrix M. 

Proof. Suppose v is an eigenvector of T with eigenvalue A. Then for every A e g we have... 

Proof. Since V is a finite-dimensional complex vector space, C must have at least one eigenvalue Z. Dehne... 

Proof. Choose any eigenvector of p(i). Let X denote the corresponding eigenvalue. For every k e N the vector is either trivial or an eigenvector for /)(i) with eigenvalue A + ik. Let A o denote the natural number such that... 

Proof. We are obliged to take somewhat different methods from those used in proving former theorems, for they all lead to the expansion of the eigenvalue up to even orders of k. Instead of applying the variational method directly to i x, we apply it to its inverse H l, which is bounded by (11. 2) and hence much more feasible than H% itself. 

As we can see, to get the correction of the n-th order to the eigenvalue (energy) one has to know the correction on the (n — l)-th order to the eigenvector (wave function). In fact a much stronger statement is valid, namely, that knowing the correction of the n-th order to the wave function allows us to know the correction of the (2n + l)-th order to the energy. Since we are interested here only in lower order corrections, we do not elaborate on this further. One can find proofs and detailed discussions in books by I. Mayer [18] and by L. Zulicke [27]. 

On analysis by Bell [30] the proof was shown to rely on the assumption that dispersion-free states have additive eigenvalues in the same way as quantum-mechanical eigenstates. Using the example of Stern-Gerlach measurements of spin states, the assumption is readily falsified. It is shown instead that the important effect, peculiar to quantum systems, is that eigenvalues of conjugate variables cannot be measured simultaneously and therefore are not additive. The uniqueness proof of the orthodox interpretation therefore falls away. 

Let us first mention the notion of ODLRO, a fundamental concept introduced by Yang [16] in connection with his celebrated proof of the largest bound for T(2). He demonstrated that the manifestation of a macroscopically large eigenvalue 2 in the second-order (fermion) density matrix may lead to a new physical order (cf. the theories of superconductivity and superfluidity). For additional information regarding these issues see Refs. [17-19]. 

One of the features of traditional eigenvalue analysis is that the disturbance held is assumed to grow either in space or in time. This distinction is only for ease of analysis and there are no general proofs or guidelines available that would tell an investigator which growth rate to investigate. Huerre Monkewitz (1985) have applied the so-called combined spatio-temporal... 

Proof. We begin by analyzing (5.2). The first step is to compute the stability of the rest points of system (5.2) by finding the eigenvalues of the Jacobian matrix evaluated at each of these rest points. At (0,0) this matrix takes the form... 

Proof. To see the sufficiency of the conditions, note that if the Jacobian of 2 has a positive eigenvalue then... 

If E exists (0 < Aq < 1), then A2 < Aq. Therefore the Jacobian at E has a positive eigenvalue (denoted by 112 in the previous discussion of the eigenvalues of this Jacobian). This completes the proof of the proposition. 

Proof. Assertion (1) is just Theorem C.4. The assertion concerning M (xf) follows from the Perron-Frobenius theory (Theorem A.5) and the monotonicity of the time-reversed system (6.3). If J is the Jacobian matrix of / at Xq, then (6.2) implies that —J satisfies the hypotheses of Theorem A.5. It follows that r = —s —J) < 0 is an eigenvalue of J corresponding to an eigenvector u > 0. Because M (Xo) is tangent at Xq to the line through Xq in the direction v, the local stable manifold of Xq is totally ordered. Since M X()) is the extension of the local stable manifold by the order-preserving backward (or time-reversed) system, it follows... 

Proof. If ai(0)a2(0) > 1 then both eigenvalues have the same sign, so the origin is an attractor or a repeller according to whether ai(0) + o 2(0) is positive or negative. 

Proof. A sufficient condition for J, evaluated at to have eigenvalues with negative real parts is that if the oflF-diagonal elements are replaced by their absolute values, then the determinants of the principal minors alternate in sign (Theorem A.11). 

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