In a linear dynamical system, the variation of a state vector
(an
-dimensional vector denoted
) equals a constant matrix
(denoted
) multiplied by
. This variation can take two forms: either
as a flow, in which
varies
continuously with time


or as a mapping, in which
varies in discrete steps

These equations are linear in the following sense: if
and
are two valid solutions, then so is any linear combination
of the two solutions, e.g.,
where
and
are any two scalars. The matrix
need not be symmetric.
Linear dynamical systems can be solved exactly, in contrast to most nonlinear ones. Occasionally, a nonlinear system can be solved exactly by a change of variables to a linear system. Moreover, the solutions of (almost) any nonlinear system can be well-approximated by an equivalent linear system near its fixed points. Hence, understanding linear systems and their solutions is a crucial first step to understanding the more complex nonlinear systems.
Solution of linear dynamical systems
If the initial vector
is aligned with a right eigenvector
of
the matrix
, the dynamics are simple

where
is the corresponding eigenvalue;
the solution of this equation is

as may be confirmed by substitution.
If
is diagonalizable, then any vector in an
-dimensional space can be represented by a linear combination of the right and left eigenvectors (denoted
) of the matrix
.

Therefore, the general solution for
is
a linear combination of the individual solutions for the right
eigenvectors

Similar considerations apply to the discrete mappings.
Classification in two dimensions
The roots of the characteristic polynomial det(A - λI) are the eigenvalues of A. The sign and relation of these roots,
, to each other may be used to determine the stability of the dynamical system

For a 2-dimensional system, the characteristic polynomial is of the form
where
is the trace and
is the determinant of A. Thus the two roots are in the form:


Note also that
and
. Thus if
then the eigenvalues are of opposite sign, and the fixed point is a saddle. If
then the eigenvalues are of the same sign. Therefore if
both are positive and the point is unstable, and if
then both are negative and the point is stable. The discriminant will tell you if the point is nodal or spiral (i.e. if the eigenvalues are real or complex).
See also