Diagonalizability

Fig. 2. The BO model is the adiabatic limit of full QD if energy level crossings do not appear. QCMD is connected to QD by the semiclassical approach if no caustics are present. Its adiabatic limit is again the BO solution, this time if the Hamiltonian H is smoothly diagonalizable. Thus, QCMD may be justified indirectly by the adiabatic limit excluding energy level crossings and other discontinuities of the spectral decomposition.

Since has been already constrained to be hermitian, it is legitimate to assume, withoutany loss of generality that is always diagonalizable into, say, , by a unitary transformation of the basis elements [10], The diagonal elements of , then called its eigenvalues, are real. The rank constraint on P (which is basis independent) further reduces the number of non-zero eigenvalues toN. Let % (i = l,. .., N), be these non-zero eigenvalues. 

It is easy to see a matrix has a closed orbit if and only if it is diagonalizable. Hence the set of closed orbits can be identified with the set of eigenvalues. On the other hand, a matrix B with [B, 5I] = 0 (i.e. a normal matrix) can be diagonalizable by a unitary matrix. Hence the quotient space is also identified with the set of eigenvalues. The identification can be seen directly in this example. 

Let us briefly summarize the derivation of At (see Ref. 14). To construct it, we use the known fact that for the nonintegrable case, Friedrichs Hamiltonian is diagonalizable in the complex energy plane as [10, 15, 16]... 

Exercise 3.16 Suppose that V is a complex scalar product space and Tl V V is an orthogonal projection. Show that the only possible eigenvalues for n are 0 and 1. Show that H is diagonalizable, i.e., show that there is a basis ofV composed of eigenvectors C / Fl. 

If the result of a similarity transformation is to produce a diagonal matrix (see Appendix A.4-l(b)), then the process is called diagonaliz-ation. If the matrices A and B can be diagonalized by the same matrix, then A and B commute. 

Remark. The detailed balance relation (4.2) or (6.1) asserts that the matrix W is virtually symmetric and will be seen in the next section to guarantee that W can be diagonalized. The relation (6.14) is also a property of W but does not by itself guarantee diagonalizability, and wil be referred to as extended detailed balance . The relations (6.12) and (6.13) are not properties of W but relate the transition probabilities in one system to those in another system. They will therefore not be honored with the name detailed balance. The extended detailed balance property will be important in XI.4. 

As mentioned before, the properties (V.2.5), which characterize a W-matrix, are not sufficient to guarantee that there exists a matrix S such that S 1WS is diagonal. The additional detailed balance property (4.2) or (6.1), however, makes W symmetric in a certain sense and thereby diagonalizable, see (6.15). In this section we develop the consequences. 

If W is a finite matrix, linear algebra tells us that this can in fact be asserted when W is symmetric. If W is an operator in an infinite-dimensional space the mathematical conditions are considerably more complicated, but as a rule of thumb one may also regard any symmetrical operator as diagonaliz-able - as is customary in quantum mechanics. It will now be shown that the detailed balance property guarantees that the operator W is symmetrical. We adopt the notation of a continuous range. 

S. R. Jain When Prof. Rice talks about optimal control schemes, his Lagrange function follows a time-reversed Schrodinger equation. Is it assumed in the variational deduction that the Hamiltonian is time reversal invariant that is, is it always diagonalizable by orthogonal transformations ... 

Some matrices are diagonalizable over C in the sense that they have a basis of - possibly complex - eigenvectors and these matrices can therefore be diagonalized by matrix similarity. For example, this is so for all n by n matrices with n distinct eigenvalues since eigenvectors for different eigenvalues are linearly independent. Other matrices such... 

Specifically for normal matrices, defined by the matrix equation A A = AA, this implies orthogonal diagonalizability for all normal matrices, such as symmetric (with AT = A e R" "), hermitian A = A Cn,n), orthogonal (ATA = /), unitary (A A = /), and skew-symmetric (AT = —A) matrices. 

The situation is more complex if A is not diagonalizable. But in either case the location of the eigenvalues A G C of A determines the behavior of the solution y(t) g1 as t grows. In particular, if all eigenvalues A have negative real parts, i.e., if they lie in the left halfplane of C, then the solution y(t) will decay to zero over time from any initial... 

A matrix is diagonalizable if it is equivalent to a diagonal matrix D. The characteristic equation of A is invariant under a similarity transformation, for... 

In the general case, if A is not a normal matrix, then it is not necessarily diagonalizable. However, it is diagonalizable if the characteristic equation has n distinct roots. 

The presence of the continuum, coupled with the SVCA, results in a complex-I mmetiTC g matrix. Such matrices are diagonalizable using complex-orthogonal... 

Normality is a convenient test for diagonalizability every normal matrix can be converted to a diagonal matrix by a unitary transform, and every matrix which can be made diagonal by a unitary transform is also normal (but finding the desired transform requires much more work than simply testing to see whether the matrix is normal). d Changing the signs simultaneously of certain trios (e.g. g, g )) of elements of g will leave G untouched. Our... 

Theorem. Let G represented by A be diagonalizable, and suppose A is a finitely generated k-algebra. Then G is a finite product of copies of Gm and various. ... 

The name diagonalizable will be justified in (4.6). But we can already distinguish these groups Hopf-algebraically over fields. We first need the following result, which in group language states the independence of characters. 

Looking back to the previous section, we find the duals of the finite constant groups are precisely the finite diagonalizable groups the dual algebra of kr is kpT]. In general this would not be one of our Hopf algebras, since it is not commutative. But when T is commutative we can write it as a product of various Z/nZ and compute that the dual of Z/nZ is ft,. 

Show that H is diagonalizable, that all characters of H extend to G, and that H is definable as the common kernel of a set of characters of G. 

Theorem. Let M be a subgroup of GL fc). The elements of M can be simuhan-eously diagonalized iff the group scheme G corresponding to M is diagonalizable. 

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