This outcome sounds fractional. Perhaps search fractional and linear dependence.

Systems with full rank generally can be linearised without changing the topology of the solutions. If the system has reduced rank then the dynamics gets much more interesting. Such systems are best studied using a Centre manifold which makes precise the effect of the higher order nonlinear terms. Such analysis is vital when studying the stability of such problems and they arise constantly in applications.

A point where the eigenvalues all have non zero real values is called hyperbolic. In this case, Lyapunov's first method allows one to conclude on local asymptotic stability (or instability).

Controllability is a different property (not to be confused with stability). It does not make sense if your system (as in your example) does not have any input.

Have you thought to use Jordan Canonical Form (aka Jordan Normal Form or just Jordan Form)? If you are seeking a solution x that is a function into a field (a commutative ring in which nonzero elements have inverses), then reformulating the problem in terms of the algebraic closure of the base field gives you existence of an invertible matrix P that defines a similarity transformation carrying the Jacobian to a Jordan form matrix that is unique up to the order of the elementary Jordan blocks. In this case, you do not uncouple the linearized system unless the Jacobian is diagonalizable, in which case, the columns of P are eigenvectors of the Jacobian. Since a Jordan canonical form is bi-diagonal, with all of its non-diagonal nonzero entries on the super-diagonal, it is as sparse as you can get in trying to uncouple the system. Moreover, the system is then treated as a system of smaller systems, each of which has an elementary Jordan matrix as its coefficient matrix. Each such system is ``minimally decoupled'', but is extremely easy to solve symbolically, using backward-substitution (but some books define Jordan forms so that the nonzero off-diagonal entries are on the main sub-diagonal, in which case, you would use forward substitution instead).

If you are seeking solutions x that are real-valued, reformulate the problem over the complex number field, which is a standard technique in differential equations and their applications.

http://en.wikipedia.org/wiki/Jordan_form

It is impossible that two eigenvectors of different eigenvalue are linear dependent because this means A maps one vector into two different vectors. This will never happen. When the real part of the eigenvalue is zero the system will definitely oscillates because the system have a pair of purely imaginary poles.

Yuzio Guo correctly states that ``It is impossible that two eigenvectors of different eigenvalue are linear dependent'', but I assumed that the original question is asked by someone who understands this simple fact. There are many systems however that have very few eigenvalues, and if the number of eigenvalues is small enough, it can happen that the Jacobian is not of full rank.

BTW, implicit in my above comments is the assumption that the Jacobian is a square matrix. If it is not square, you may want to try SVD or some other factorizations of non-square matrices.

http://en.wikipedia.org/wiki/Matrix_factorization

Rank of this matrix determines dimension of the space spanned by these eigenvectors and behavior of solutions in the neighborhood of the fixed point . For example, if rank of the corresponding matrix n x n is equal to n and all eigenvalues are purely imaginary numbers then solutions of the differential equations will be bounded in the neighborhood of the fixed point and this fixed point is said to be stable. If rank is less than n then solutions will be unbounded and the fixed point will be unstable. Some details can be found in any textbook on differential equations, for example, R. Grimshaw. Nonlinear Ordinary Differential Equations. CRC Press, 2000

To Matt Insall, from my point of view, the not full rank Jacobian have some zero eigenvalues. As C. Budd said, in this condition centre manifold method is a good choice.

A full rank linearization of an equilibrium point in a nonlinear system allows you the use of the Hartman–Grobman theorem and, therefore, the use on the Lyapunov's first method to conclues stability (or unstability) of that point, as Mr. Petit said.

When this linearization has no full rank, there at least an eigenvalue with real part equal to zero. Thus, it is no possible the use of the Hartman–Grobman theorem. And you should to use other results as the center manifold theorem (as mr. Budd said) or, maybe, the Lyapunov's second method to investigate the stability of that point, as Ms. Yucel mades mention.

To Yuzhu Guo, the original question did not mention a not full rank Jacobian. It mentioned a not full rank matrix of eigenvectors. This makes the question of whether the Jacobian has full rank irrelevant to the original question. (In my second response above, I mistakenly referred to the rank of the Jacobian. I meant the rank of the set of eigenvectors of the Jacobian. I apologize for the consequent confusion.) Again, C. Budd seems to be addressing the case where the Jacobian is not of full rank, and in this case, it is probably correct to use what you refer to as the ``centre manifold method''. But again, it's not relevant to the question asked. For one thing, the original system can be a linear system and have a full rank Jacobian that is horribly non-diagonizable. In particular, in this case, if the Jacobian is the (5x5) matrix J whose rows are [1 1 0 0 0], [0 1 1 0 0], [0 0 1 1 0], [0 0 0 1 1], and [0 0 0 0 1], respectively, then the only eigenvalue of the Jacobian is 1, and the corresponding eigenspace has dimension 1, since it is the null space of the (5x5) rank 4 matrix J-I whose rows are [0 1 0 0 0], [0 0 1 0 0], [0 0 0 1 0], [0 0 0 0 1], and [0 0 0 0 0]. But in this example, I have given you a matrix that is an elementary Jordan matrix, so it corresponds to a very simple system of differential equations. If x^T=[x_1 x_2 x_3 x_4 x_5], then the system x'=Jx can be written in the following form:

d(x_1)/dt=x_1+x_2

d(x_2)/dt=x_2+x_3

d(x_3)/dt=x_3+x_4

d(x_4)/dt=x_4+x_5

d(x_5)/dt=0

To solve this system, solve for x_5 first, to get a constant solution, which drives the x_4 equation (the x_4 equation is treated as a non-homogeneous equation), use x_4 to drive the x_3 equation, solve for x_3, use x_3 to drive the x_2 equation, solve for x_2, use x_2 to drive the x_1 equation, and solve for x_1, and you have the entire solution.

If this system is the linearized version of a given system, you now have an approximate solution to the original system, but its validity may be only in a small region around your initial condition. As C Budd indicates, one may need the higher-order terms of a Taylor approximation to get an approximation that is valid over a larger time interval (for partial differential equations, a region in space-time), but this does not depend upon the invertibility (or better, rank) of the Jacobian itself. It depends upon the amount of curvature of a level surface of the function that defines the original model, near the fixed-point that Alex Antrobus mentioned as a postulated use of the methods. Let p be such a fixed-point (Alex did not say whether it is a fixed-point of the right-hand side or a critical point of the equation, being then a fixed point of something else...). This, as I understand it, means that

dp/dt=p,

so that p is an ``exponential'' solution. But it also means that

f(Ap)=p,

so that if Df is the Jacobian of f, then the Jacobian Alex wants to consider is the product of (Df)(Ap), the real-valued matrix value of the Jacobian of f evaluated at the point Ap, with A (having applied the chain rule), in that order. Denoting by B the matrix (Df)(p), the matrix Alex is wanting to diagonalize is

BA

(not AB unless they happen to commute). If the level surface of the function g defined by g(x)=f(Ax) has near p very high curvature, then the linearization, L, given by

L(x)=g(p)+BA(x-p), i.e. L(x)=p+BA(x-p), or L(x)=(BA-I)p+BAx

will not be a very good approximation, and the Hessian (second-order derivative matrix, see http://en.wikipedia.org/wiki/Hessian_matrix) should be used, or a ``manifold'' method, which accommodates this high curvature, should be used.

Some references include

http://www.amazon.com/Course-Applied-Functional-Analysis-Mathematics/dp/0471962384

http://www.amazon.com/books/dp/0898741432

The rank is NOT important provided the eigenvalue is not neutral (this means that the real part of it is not equal to 0). For neutral eigenvalues the rank is very importantt/ The central manifold technique should be employed as it has been mentioned by C. Budd. The appearence of the neutral eigenvalues is in some sence typical feature of the families of ODE's wiich depends on a parameter . See V. I. Arnold "V. I. Arnold, Geometrical methods in the theory of ordinary differential equations.

Dear Alex,

Really, the linearization of a non-linear system of equations is important, to find a solution x(t) of this system. In this case, we encounter the difficulty that, on different intervals of time [t_i , t_{i+1}], while a coordinate x_m(t) is slowly varying, another coordinate x_n(t) is rapidly varying in time. That means that, in a certain interval of time, you have to choose independent variations for the most rapidly varying coordinate x_n(t), the other coordinates x_m(t) being obtained by calculations. As a literature recommendation, I can mention only my article

http://www.sciencedirect.com/science/article/pii/S0079672710000261 ,

where, nice solutions are found for such a system of equations.

Concerning the rank of the matrices, which was your first question, I think that an explicit dependence of something on this rank is difficult to be found, and not very important for the solution of the problem. However, we could suppose that higher ranks will lead to more and more complicated problems, while the solution itself essentially depends on the structure of the nonlinear system, and not especially on rank.

Kind regards,

Eliade

http://www.sciencedirect.com/science/article/pii/S0079672710000261

The rank of the eignevector matrix of the Jacobian can be very important. The sensitivity of the eigenvalues is related to the condition number of the eigenvector matrix (I forget the exact raltionship - see JH Wilkinson's book for details). So if the eigenvectors are aligned, the eigenvalues are very sensitive. Try this MATLAB code to show it:

format long

A1=[-2*sqrt(2),-2;1 0]

A2=[-sqrt(3)-sqrt(2),-sqrt(6);1 0]

[V,L]=eig(A1)

[V,L]=eig(A2)

sqrt(2)

sqrt(3)

The eigenvectors of A1 are aligned - the resulting eigenvalue calculation is accurate to only 7 dp. The eigenvalues of A2 are accurate to 14 dp.

A second consequence is that the Jacobian matrix is very non-normal. This means that there is potential for an initial small perturbation from a stable equilibrium to be amplified enormously before converging to the equilibrium point. This can drive the system unstable even though it is locally stable. For more about this, see the paper:

Hydrodynamic stability without eigenvalues L. N. Trefethen, A. E. Trefethen, S. C. Reddy, and T. A. Driscoll, Science, 261 (1993), 578-584.

A few things could be going on here. The linearization may be ill-conditioned (check the condition number) and the rank issue may stem from numerical issues. Another alternative is that the NL system is, locally, of lower dimension (not entirely unlikely as the linearization is a fixed point that, a priori, places a constraint on the flow which, quite plausibly, can manifest as a lower dimensional solution manifold). One thing that you might want to look at is linearizing about a trajectory and examine the linearization for a nominal trajectory passing through the fixed point. Center manifold theory will be able to tell you something for those parameter values where the eigenvalues lie on the imaginary axis (otherwise, they're not centers and the theory doesn't hold) but in the general case, won't be applicable.

check this thesis Section 4.1 and 4.2. http://ecommons.usask.ca/handle/10388/etd-12082008-154626

or for simplicity, Supplementary Information for this paper http://www.plosone.org/article/info:doi/10.1371/journal.pone.0006886