Discrete time case in n dimension

Consider the system defined by the linear recursion xk+1 = Axk, where A is an n×n matrix. The C⁰ classification uses again the stable, unstable and center subspaces, that will be defined first, for discrete time systems. Let us denote the eigenvalues of the matrix with multiplicity by λ1, λ2, . . . , λn. Let u1, u2, . . . , un denote that basis in Rⁿ in which the matrix takes it Jordan canonical form. Using this basis the stable, unstable and center subspaces can be defined as follows.

Definition 2.15.. Let {u1, . . . , un} ⊂Rⁿ be the above basis and let λk be the eigenvalue

Figure 2.12: Center.

corresponding to uk (note that uk may not be an eigenvector). The subspaces E_s(A) =h{u_k :|λ_k|<1}i, E_u(A) =h{u_k :|λ_k|>1}i,

E_c(A) =h{u_k :|λ_k|= 1}i

are called the stable, unstable and center subspaces belonging to the matrix A∈GL(Rⁿ).

(The notation h·i denotes the subspace spanned by the vectors between the brackets.) The most important properties of these subspaces are summarised in the following theorem.

Theorem 2.16.. The subspaces E_s(A), E_u(A), E_c(A) have the following properties.

1. E_s(A)⊕E_u(A)⊕E_c(A) =Rⁿ

2. They are invariant under A (that is A(E_i(A))⊂E_i(A), i=s, u, c).

3. For any p∈Es(A) we have Aⁿp→0, if n →+∞.

4. For any p∈E_u(A) we have A⁻ⁿp→0, if n→+∞.

The dimensions of the stable, unstable and center subspaces will play an important role in the C⁰ classification of linear systems. First, we introduce notations for the dimensions of these invariant subspaces.

Definition 2.17.. Lets(A) = dim(E_s(A)), u(A) = dim(E_u(A))andc(A) = dim(E_c(A)) denote the dimensions of the stable, unstable and center subspaces of a matrix A, respec-tively.

The following set of matrices is important from the classification point of view. The elements of

HL(R) ={A∈GL(Rⁿ) :|λ| 6= 1 ∀λ∈σ(A)}, are called hyperbolic matrices in the discrete time case.

In the discrete time case only the hyperbolic systems will be classified according to C⁰-conjugacy. In order to carry out that we will need the Lemma below.

Lemma 2.18.. Let the hyperbolic matrices A, B ∈ HL(Rⁿ) be C⁰ conjugate, that is there exists a homeomorphism h : Rⁿ → Rⁿ, for which h(Ax) =Bh(x) for all x ∈ Rⁿ. E_u(B), that is h takes the unstable subspace to unstable subspace

3. s(A) =s(B), u(A) =u(B).

Proof. 1. Substituting x= 0 into equation h(Ax) =Bh(x) leads to h(0) =Bh(0). This implies that h(0) = 0, because the matrix B is hyperbolic, that is 1 is not an eigenvalue.

2. If x ∈ E_s(A), then Aⁿ → 0 as n → ∞, hence h(Aⁿx) = Bⁿh(x) implies that Bⁿh(x) tends also to zero. Therefore h(x) is in the stable subspace of B. Thus we showed that h E_s(A)

⊂ E_s(B). Using similar arguments for the function h⁻¹ we get that h⁻¹ E_s(A)

⊂ E_s(B) yielding E_s(B)⊂ h E_s(A)

. Since the two sets contain each other, they are equal h E_s(A)

=E_s(B).

3. Since there is a homeomorphism taking the subspaceE_s(A) to the subspaceE_s(B), their dimensions are equal, i.e. s(A) = s(B), implying also u(A) = u(B) since the dimensions of the center subspaces are zero.

In the case of continuous time linear systems we found that s(A) =s(B) is not only a necessary, but also a sufficient condition for the C⁰ conjugacy of two hyperbolic linear systems. Now we will investigate in the case of a one-dimensional and a two-dimensional example if this condition is sufficient or not for discrete time linear systems.

Example 2.2. Consider the one-dimensional linear equations given by the numbers (one-by-one matrices) A = ¹₂ and B = −¹₂. Both have one dimensional stable sub-space, that is s(A) = s(B) = 1, since the orbits of both systems are formed by geometric sequences converging to zero. However, as it was shown in Section 2.3.2, these two equations are not C⁰ conjugate. It was proved there that C⁰ conjugacy divides the space GL(R) into six classes.

This example shows that s(A) = s(B) is not sufficient for the C⁰ conjugacy of the two equations. Despite of this seemingly negative result, it is worth to investigate the following two dimensional example, in order to get intuition for the classification of hyperbolic linear systems.

Example 2.3. Consider the two-by-two matrices A= ¹₂I and B =−¹₂I, where I is the unit matrix. The stable subspace is two-dimensional for both systems, that is s(A) = s(B) = 2, since their orbits are given by sequences converging to zero. We will show that these matrices are C⁰ conjugate. We are looking for a homeomorphism h : R² → R² satisfying h(¹₂x) = −¹₂h(x) for all x ∈ R². It will be given in such a way that circles centered at the origin remain invariant, they will be rotated by different angles depending on the radius of the circle. Let us start from the unit circle with radius one and definehas the identity map on this circle, i.e. h(x) =x is |x|= 1. Then equation h(¹₂x) =−¹₂h(x) defines h along the circle of radius1/2, namely h rotates this circle byπ, i.e. h(x) =−x is |x|= 1/2. In the annulus between the two circles the homeomorphism can be defined arbitrarily. Then equation h(¹₂x) = −¹₂h(x) defines h again in the annulus between the circles of radius 1/2 and 1/4. Once the function is known in this annulus the equation defines again its values in the annulus between the circles of radius 1/4 and 1/8. In a similar way, the values of h in the annulus between the circles of radius 1 and 2 are defined by the equation h(¹₂x) = −¹₂h(x) based on the values in the annulus determined by the circles of radius 1/2 and 1. It can be easily seen that the angle of rotation on the circle of radius 2^k has to be −kπ. Thus let the angle of rotation on the circle of radius r be −πlog₂(r). This ensures that the angle of rotation is a continuous function of the radius and for r = 2^k it is −kπ. This way the function h can be given explicitly on the whole plane as follows

h(x) = R(−πlog₂(|x|))x, where R(α) =

cosα sinα

−sinα cosα

This function is obviously bijective, since the circles centered at the origin are invariant and the map is bijective on these circles. Its continuity is also obvious except at the origin. The continuity at the origin can be proved by using the above formula, we omit here the details of the proof.

We note that in 3-dimension the matrices A = ¹₂I and B = −¹₂I, where I is the unit matrix of size 3 ×3, are not C⁰ conjugate. Thus we found that s(A) = s(B) is not a sufficient condition of C⁰ conjugacy. The sufficient condition is formulated in the following lemma that we do not prove here.

Lemma 2.19.. Assume that s(A) = s(B) = n (or u(A) = u(B) = n). Then A and B are C⁰ conjugate, if and only if sgndetA=sgndetB.

This lemma enables us to formulate the following necessary and sufficient condition for the C⁰ conjugacy of matrices.

Theorem 2.20.. The hyperbolic linear maps A, B ∈ HL(Rⁿ) are C⁰ conjugate, if and the determinant of the matrix −I is also positive, while for odd values of n it is negative.

Thus, as we have already shown, the matrices A=

Example 2.5. Using the above theorem, it is easy to prove that the space HL(R²) of hyperbolic linear maps is divided into 8 classes according to C⁰ conjugacy. The proof of this is left to the Reader.