1 Introduction

Local solar time is measured by a sundial. When the center of the sun is on an observer’s meridian, the observer’s local solar time is zero hours (noon). Because the earth moves with varying speed in its orbit at different times of the year and because the plane of the earth’s equator is inclined to its orbital plane, the length of the solar day is different depending on the time of year. It is more convenient to define time in terms of the average of local solar time. Such time, called mean solar time, may be thought of as being measured relative to an imaginary sun (the mean sun) that lies in the earth’s equatorial plane and about which the earth orbits with constant speed. Every mean solar day is of the same length.¹

In [1, 4] it is shown that the mean sun satisfies the functional equation

(1)

,

Then M(, )y(s) is the direction from the earth to the sun expressed in a local coordinate system on the surface of the earth in the point of longitude and latitude .

In the present paper we investigate generalizations of equation (1) for fixed . To be more precise, first we will solve the following functional equation

(2)

1. Determine all solutions (U, y) of (2).
2. For given U determine all y, such that (U, y) is a solution of (2).
3. For given y determine all U, such that (U, y) is a solution of (2).
4. Find relations between U and y for a solution (U, y) of (2).

We will mainly deal with problems of the second and third kind.

In Theorem 6 we describe in an appropriate system of coordinates the structure of the space S_U of all solutions of (2) for a given U: A GL(n, K). We also state in this theorem how such a mapping U necessarily looks if a nontrivial solution y (i.e. y0) exists. A similar description of U-invariant subspaces of S_U is given in Theorem 8. We emphasize that by our result (and similarly by the following theorems) the problem of solving (2) can be reduced, at least to some extent, to the problem of finding all exponential functions U₁₁: A GL(k, K) (cf. the representation of U in Theorem 6), i.e. non singular matrices U₁₁() satisfying the equation

Here we assume that these functions are known and we refer the reader to [3].

In Theorem 9 we construct to a given subspace S⁰ of Kⁿ the set of all mappings U and correspondingly the space S of all functions y, such that (U, y) satisfies (2) and S⁰ is exactly the set of all initial values y(0) for y S. It is clear that this yields together with Theorem 6 an implicit description of the set of all solutions (U, y) of (2) by varying the subspace S⁰ of Kⁿ. However, the space S and the mapping U obtained in this way from S⁰ may have the property that S is a proper subset of S_U . Therefore we also deal with the problem to characterize the situation when S = S_U .

From a mathematical point of view it seems also interesting to study the functional equation (2) for mappings U: A M_n(K). This situation is more complicated both with respect to the technical details and the construction (description) of the solutions U or y or (U, y). In Theorem 20 we start from a given mapping U: A M_n(K) and describe completely in appropriate coordinates the set of all functions y: A Kⁿ, such that (U, y) is a solution of (2). Again this theorem provides necessary conditions on U for the existence of nontrivial solutions y of (2). We also show in Theorem 21 how to construct all mappings U: A M_n(K) and corresponding spaces S of functions y: A Kⁿ, such that is a given subspace S⁰ of Kⁿ and (U, y) is a solution of (2), hence giving an implicit description of the general solution of (2) by varying S⁰. However, we were not able to contribute to the problem when S = S_U .

The main difficulties in this last part seem to arise from the fact that there can occur solutions y (to a given U) with y(0) = 0 but y0 (cf. Lemma 18).

2 Regular matrices U()

Here in this part we always assume that U is a mapping from the abelian group A to GL(n, K).

Proof.  The pair (U, y) is a solution of (2) if U(+)y() = U()y(+) for all , , A. Since B and C are regular matrices, this is equivalent to CU( + )B^-1By() = CU()B^-1By( + ) for all , , A.       

For C = U(0)^-1 we get CU(0) = I n, the identity matrix. Hence, without loss of generality we will always assume that U(0) = In.

Lemma 2. If (U, y) is a solution of (2), then

(3)

(4)

Proof.  Since U(0) = In, we get (3) from (2) for = = 0. And we get (4) from (2) and (3) for = 0.       

It is also possible to reverse the statement of Lemma 2.

For any mapping U: A GL(n, k) let

Some basic properties of these two sets are collected in the following

Proof.  It is clear that S_U and S_U⁰ are linear spaces. Assume that y⁰ S_U⁰, then there is some y S_U , such that y(0) = y⁰. Since (U, y) satisfies (3), the function is well defined. It is surjective, since for any y S_U we have (y(0)) = U(^.)y(0) = y(^.) according to (3). The mapping is also injective, since from (y₁⁰) = (y₂⁰) we derive U()y₁⁰ = U()y₂⁰, which implies for = 0 (and U(0) = In) that y₁⁰ = y₂⁰. Finally we have to prove that is a linear mapping. Let y₁⁰, y₂⁰ S_U⁰ and let ₁, ₂ K, then ₁y₁⁰ + ₂y₂⁰ S_U⁰ and (₁y₁⁰ + ₂y₂⁰) = U(^.)(₁y₁⁰ + ₂y₂⁰) = ₁U(^.)y₁⁰ + ₂U(^.)y₂⁰ = ₁(y₁⁰) + ₂(y₂⁰).       

In conclusion, both S_U and S_U⁰ are m-dimensional linear spaces for some 0 < m < n.

There are some more interesting properties of S_U and S_U⁰.

Proof. 

1. Assume that z() := U(₀)y(). Then (U, z) satisfies (2), since U( + )z() = U( + )U(₀)y() = U( + )U(₀)U()y(0) = U( + )U(₀ + )y(0) = U( + )y(₀ + ) = U()y( + ₀ + ) = U()U(0)y(₀ + + ) = U()U(₀)y( + ) = U()z( + ) by (3), (4), (3), (2), special form of U(0) and (2).
2. Let z() := y(+₀), then (U, z) satisfies (2), since U(+)z() = U(+)y( + ₀) = U()y( + + ₀) = U()z( + ) by (2).
3. If y⁰ S_U⁰, there exists some y S _U , such that y⁰ = y(0). From the first item of this lemma we know that U(₀)y S_U , hence U(₀)y(0) = U(₀)y⁰ S_U⁰.
4. According to the definition of S_U⁰ we know that S_U⁰ . Let y S_U and assume that ₀ A \ , then it follows from the second item of this lemma that z(^.) := y(^. + ₀) S_U and y(₀) = z(0) S_U⁰.       

If denotes a basis of SU0, then there exists a matrix B GL(n, k), such that Bbi = ei, the i-th unit vector in Kn. Applying this matrix B as a coordinate transformation on Kn as in Lemma 1 we get that SUB-1 - 1 = <e 1, ..., em>, the m-dimensional linear space generated by the first m unit vectors in Kn. Thus without loss of generality we may assume that SU = <e1, ..., em>.

Theorem 6. Let U: A GL(n, K) be a mapping, such that S_U⁰ = <e₁, ..., e_m> and U(0) = In. Then U() can be partitioned as a block matrix of the form

where U₁₁() GL(m, K), U₂₂() GL(n - m, K) and U₁₂() M_m,n-m(K). These matrices satisfy the boundary conditions U₁₁(0) = Im, U₂₂(0) = In-m and U₁₂(0) = (0)m,n-m, the zero matrix. Moreover, U₁₁ is an exponential function, i.e. U₁₁( + ) = U₁₁()U₁₁() for all , A.

Each y S_U can be expressed as

and (0) K^m.

Proof.  Let y S_U , then y() S_U⁰ = <e₁, ..., e_m>, so y() = and () K^m. We partition U() as a block matrix

(5)

such that U₁₁() is an m × m-matrix. From (3) we deduce that

Since (0) is an arbitrary element of K^m, it is clear that U ₂₁() = (0)n-m,m for all A. Because of the fact that U() is regular, both U₁₁ and U₂₂ are regular matrices as well. From U(0) = In the boundary conditions follow. Inserting into (4) the form of U and y just described we get

which means that U₁₁( + ) (0) = U₁₁()U₁₁() (0) for all (0) K^m and , A, so that U₁₁ is an exponential function.       

We are also interested in subspaces of S_U . First we present a generalization of Lemma 5.

Proof.  In order to prove that 1 implies 2, we set z() := y( + ₀) for arbitrary ₀ A. Since S is U(₀)-invariant, U(₀)y S. It is enough to prove that z = U(₀)y, since then z S. For A we get z() = y(₀ + ) = U(₀ + )y(0) = U(₀)U()y(0) = U(₀)y(), so z = U(₀)y by (3), (4) and (3).

To each y⁰ S⁰ there exists y S, such that y⁰ = y(0). Under the assumption 2, the function z() := y( + ₀) belongs to S for any ₀ A. So z(0) S⁰ and z(0) = y(₀) = U(₀)y(0) = U(₀)y⁰ by (3). Thus we proved that 2 implies 3.

In order to close the cycle of implications take y S. Then y(0) S⁰. For arbitrary ₀ A also U(₀)y(0) belongs to S⁰. Hence, there exists z S such that z(0) = U(₀)y(0). Taking into account that S is a subspace of S_U we can write z as z() = U()z(0) = U()U(₀)y(0) = U( + ₀)y(0) = U(₀)U()y(0) = U(₀)y() by (3), (4), (4) and (3). Thus U(₀)y S.       

A generalization of Theorem 6 is

Theorem 8. Let S be a k-dimensional U-invariant subspace of S_U . Then there exist coordinates in Kⁿ, such that S⁰ = <e ₁, ..., e_k>, and U() is a block matrix of the form

(6)

where U₁₁() GL(k, K), U₂₂() GL(n - k, K) and U₁₂() M_k,n-k(K), such that U₁₁(0) = Ik, U₂₂(0) = In-k, U₁₂(0) = (0)k,n-k. Moreover, U₁₁ is an exponential function, and y S if and only if

for K^k.

So far we described solutions (U, y) of (2) when the mapping U was given. Now we will assume that a linear subspace S⁰ of Kⁿ is given and we describe all solutions (U, y) of (2), such that S_U⁰ = S⁰. Let S⁰ be a k-dimensional U-invariant subspace of Kⁿ, then without loss of generality S⁰ = <e ₁, ..., e_k>.

Theorem 9. Let S⁰ = <e ₁, ..., e_k> be a subspace of Kⁿ, and let U ₁₁() GL(k, K), U₂₂() GL(n - k, K) and U₁₂() M_k,n-k(K), such that U₁₁(0) = Ik, U₂₂(0) = In-k, U₁₂(0) = (0)k,n-k. Moreover U₁₁ is assumed to be an exponential function. Then

is a U-invariant subspace of S_U , where U is given by (6).

Proof. 

      

When does S = S_U hold?

Lemma 10. The two spaces S and S_U coincide if and only if for all K^n-k \ there exists (₀, ₀) A², such that

(7)

Proof.  From Lemma 2 and Lemma 3 we know that S is a subspace of S_U different from S_U if and only if there exists y⁰ / S⁰, such that [U( + ) - U()U()]y⁰ = 0 for all , A. In other words, S = S_U if and only if for each y⁰ / S⁰ there exists ( ₀, ₀) A², such that [U(₀+₀)-U(₀)U(₀)]y⁰0. When writing y⁰ in the form for K^k and K^n-k, then y⁰ Kⁿ \ S⁰ if and only if 0. Together with (6) we get

which finishes the proof.       

Now we are going to present several examples for the situation S = S_U , i.e. by Lemma 10 examples, where condition (7) is satisfied. Here we always assume that A = K. First we will deal with the second line of condition (7). Secondly, if this condition is not satisfied by all , then let V denote the set

Thus V is an r-dimensional subspace of K^n-k for 0 < r < n - k. In order to satisfy the requirements of Lemma 10 in this situation as well, the first line in (7) must be satisfied for V .

Now we describe some examples how to construct U₂₂: K GL(s, K) for s < n - k, such that

(8)

Case charK2:

Set U₂₂() = cIs for all K \ with c K \ . Then c²c. For ₀ = ₀ = 1 we get [U₂₂(1)U₂₂(1)-U₂₂(1+1)] = [c²I s -cIs] = (c²-c)I s0 for all 0.

Case charK = 2 and > 2:

There exists c K\, such that c²1. Let U ₂₂() = cIs for all K \ , then for ₀ = ₀ = 1 we get [U₂₂(1)U₂₂(1) - U₂₂(1 + 1)] = [U₂₂(1)² - U ₂₂(0)] = [c²I s - Is] = (c² - 1)I s0 for all 0.

Case = 2:

If s = 1 each mapping U₂₂: K GL(1, K) = is a homomorphism, so (8) cannot be satisfied. If s > 1 there exist matrices M GL(s, K) of order 2^s-1. As a permutation of the vectors in K^s the cycle decomposition of M consists of one fixed point, the 0-vector, and a cycle of length 2^s-1. (Actually, cf. [2] 3.5 Theorem, there are (2^s - 1)/s irreducible polynomials of degree s over K = GF (2), such that the companion matrix of these polynomials is of order 2^s - 1.) If U ₂₂(1) = M, then for ₀ = ₀ = 1 we get [U₂₂(1)U₂₂(1) - U₂₂(1 + 1)] = [M² - I _s]0 for all 0, since 2 < 2^s - 1.

Now we describe examples how to construct U₁₂: K M_k,n-k(K), such that

(9)

Case charK2:

If k > r, then assume that U₁₂(1) = (0)k,r and

where the upper part is -Ir and the lower part is a 0-matrix of the dimension (k - r) × r. Then for ₀ = ₀ = 1 we get that U₁₁(1)U₁₂(1) + U₁₂(1)U₂₂(1) - U₁₂(1 + 1) = -U₁₂(2) and it is obvious that -U₁₂(2)0 for all K^r \ .

For k < r one possible way to proceed is indicated in

Lemma 11. If there are enough elements in K, to be more precise, if > 2 + 1, then it is always possible to find ₀ and ₀ satisfying (9).

Proof.  There exist uniquely determined integers q, s, such that r = kq + s and 0 < s < k. If q > 0 assume that ₁ K \ , then -₁ K \ . Let U₁₂(±₁) be given by

If q > 1 and is big enough, then there exists ₂ K \ and we assume that

Going on like this we can find elements ₁, ..., _q K and matrices U₁₂(±_i). If s > 0 and is big enough, then there exists _q+1 K \ and we assume that

When K^r \ , then there exists 1 < i < r, such that _i0. Hence, there exists j , such that (j - 1)k < i < jk. For ₀ = _j and ₀ = -_j we have U₁₁(_j)U₁₂(-_j) + U₁₂(_j)U₂₂(-_j) - U₁₂(0) = 2U₁₂(_j)U₂₂(-_j). According to the choice of i and j it is clear that (9) is satisfied.       

This is a very general result, but it is not the best result which is possible.

Example 12. Let K be the prime field of characteristic 3, k = 1 and r = 2. In this case < 2 + 1, but it is also possible to find U₁₂, such that (9) is satisfied. For instance U given by

satisfies (9).

Case charK = 2 and > 2:

Assume first k > r. There exists K \ and then + 1 / . Let furthermore U₁₂(1) = U₁₂() = (0)k,r and

then for ₀ = 1 and ₀ = we get U₁₁(1)U₁₂() + U₁₂(1)U₂₂() - U₁₂( + 1) = U₁₂( + 1) and (9) is satisfied. For k < r the following lemma holds:

Lemma 13. If there are enough elements in K, to be more precise, if > 2 + 2, then it is always possible to find ₀ and ₀ satisfying (9).

Proof.  There exist uniquely determined integers q, s, such that r = kq + s and 0 < s < k. Assume U₁₂(1) = (0)k,r. For 1 < i < q there exists _i K \ , such that _i + 1 / . Let U₁₂(_i) and U₁₂(_i + 1) be given by

If s > 0 and is big enough, then there exists _q+1 K, such that _q+1, _q+1 + 1 K \ and we assume that

Given K^r \ there exists 1 < i < r, such that _i0. Hence, there exists j , such that (j - 1)k < i < jk. For ₀ = _j and ₀ = 1 we have U₁₁(_j)U₁₂(1) + U₁₂(_j)U₂₂(1) - U₁₂(_j + 1) = -U₁₂(_j + 1). According to the choice of i and j it is clear that (9) is satisfied.       

Case = 2:

In the situation r > k we can only give partial results. If ₀ = 0 or ₀ = 0, then it is impossible to satisfy (9). Since U₁₁ and U₂₂ are exponential functions, the orders of U₁₁(1) and U₂₂(1) are divisors of 2. If both U₁₁(1) = Ik and U₂₂(1) = Ir, then it is also impossible to satisfy (9). If r = 1, then U₂₂(1) is the identity matrix I1. From the previous statements it is clear that necessarily k > 1 and U₁₁(1) must be a matrix of order 2. If U is defined by

then (9) is satisfied. For r = 2 assume that U₁₁(1) is given as above and

then again (9) is satisfied. For k = r = 3 and for any choice of U₁₁(1), U₂₂(1) GL(3, K) of order dividing 2 the computer did not find a matrix U₁₂(1) in M₃(K) such that (9) is satisfied. Other cases were not studied so far.

If k > r it is not possible to satisfy (9), since there is only one possible choice ₀ = ₀ = 1, which determines exactly one matrix U₁₁(1)U₁₂(1) + U₁₂(1)U₂₂(1). This matrix describes a homomorphism from K^k to K^r, which has a kernel of dimension > k - r > 0.

3 The general situation

In this part we generalize the functional equation (2) by assuming that U() is not necessarily a regular matrix, i.e. U: A M_n(K). Also in this situation Lemma 1 holds. When we define S_U and S_U⁰ as it was done earlier, then S_U and S_U⁰ are K-linear spaces (cf. Lemma 4). Again S_U⁰ is an m-dimensional subspace of Kⁿ for 0 < m < n, and S _U is invariant under translations, and S_U⁰ = (cf. Lemma 5). Without loss of generality we can assume (as in the earlier case) that there exists a basis of Kⁿ, such that S_U⁰ = <e₁, ..., e_m>.

Since U(0) need not be a regular matrix, we do not get the results of Lemma 2, and in general there is no isomorphism between S_U and S_U⁰.

For = = 0 or = 0 we derive from (2)

Lemma 14. Let (U, y) be a solution of (2), then

(10)

(11)

If U() is partitioned as in (5) and y() is written as for () K^m, then from (10) we get

which leads to the system of equations

(12)

Lemma 15. Let (U, y) be a solution of (2). Then there exists a system of coordinates of Kⁿ, such that

(13)

where the m × m-matrix U₁₁(0) is the block matrix of the form

(14)

for some k < m.

Proof.  According to Lemma 1 choose matrices C GL(n, K) and B' GL(m, K), such that

and

Without loss of generality assume that U = V . From the second line of (12) we deduce that 0 = 0 () = U₂₁() (0) for all A. Since (0) can arbitrarily be chosen in K^m, it is clear that U₂₁() = (0)n-m,m for all A.       

Since S_U⁰ = <e₁, ..., e_m>, there exist y₁, ..., y_m S_U , such that y_j(0) = e_j, the j-th unit vector in Kⁿ, for 1 < j < m. Let S _U ' := <y ₁, ..., y_m>, then S_U ' is an m-dimensional subspace of S_U . In order to prove this, it is only necessary to show that y₁, ..., y_m are linearly independent. Let ₁, ..., _m K, such that _{i = 1}^m_iy_i = 0, then also _{i = 1}^m_iy_i(0) = 0, which implies _{i = 1}^m_ie_i = 0, so that ₁ = ... = _m = 0.

For y S_U ' there exist uniquely defined ₁, ..., _m K such that y = _{i = 1}^m_iy_i. These _i can be read from y(0), since y(0) = _{i = 1}^m_ie_i.

Define the m × m-matrix Y () corresponding to the chosen y₁, ..., y_m by

(15)

(16)

These equations are collected to the matrix equation

(17)

Again these equations can be collected for j = 1, ..., m and we derive

(18)

(19)

such that Y ₁₁() is a k × k-matrix. We note that the “auxiliary” matrix function Y : A M_m(K), which will help us to describe the space S_U of solutions y (for given U), is in general not uniquely determined. However, from (17), from the decomposition of U₁₁(0) in Lemma 15 and the corresponding decomposition of Y () we see that Y ₁₁() and Y ₁₂() are uniquely determined by U₁₁(), namely

(20)

and we end up with the system of equations

(21)

(22)

(23)

A

Then it is possible to find a matrix B'' GL(m - k, K), such that the coordinate transformation on K^m induced by

satisfies

Let B be the corresponding coordinate transformation on Kⁿ

If U is decomposed as in (13) and (14), then also UB has this property.

Without loss of generality we assume that the basis of Kⁿ was chosen in such a way that Lemma 15 is satisfied and that and are a basis of V or W respectively. Then it is useful and important to partition Y () further as a 3 × 3 block matrix of the form

such that Z₁₁() = Y ₁₁() M_k(K), Z₂₂() M_m-k-r(K) and Z₃₃() M_r(K). Hence

Let x = denote a vector in K^m-k, where K^m-k-r and K^r. Then x belongs to W if and only if = 0. Moreover Y ₁₂()|_W = 0 for all A, which means Z₁₃() = (0)k,r for all A. From the definition of W it is clear that Z₁₂() = 0 for all A is equivalent to = 0.

The first line of (21) reads now as

From the second line of (21) we derive

and

Hence, each column of Z₂₃() is 0 K^m-k-r, so that Z ₂₃() = (0)m-k-r,m-k-r for all A.

From (22) we deduce

Let M denote the matrix between the two braces [ and ], then each column of M is 0 K^m-k-r and consequently M = (0) m-k-r,k. Hence, we proved that

The same way we deduce from (23) that

and correspondingly

This finishes the proof of

Theorem 16. There exists a coordinate system of K^m, such that Y () is a solution of (19) if and only if Y () can be written as

where

is an exponential function, Z₁₁() M_k(K), Z₂₂() M_m-k-r(K), Z₃₃() M_r(K), satisfying the conditions Z₁₁(0) = Ik, Z₂₂(0) = Im-k-r, Z₃₃(0) = Ir, Z₁₂(0) = (0)k,m-k-r, Z₂₁(0) = (0)m-k-r,k, Z₃₁(0) = (0)r,k and Z₃₂(0) = (0)r,m-k-r. For 0 the matrices Z₃₁(), Z₃₂(), Z₃₃() can be arbitrarily chosen.

Next we describe the structure of S_U in more details.

Proof.  Since y(0)0 also (0)0. Hence there exist z₂⁰, ..., z_m⁰ Kⁿ, such that is a basis of S_U⁰ = <e₁, ..., e_m>. Consequently, there exist z₂, ..., z_m S_U , such that z_j(0) = z_j⁰ for j = 2, ..., m. Then y, z₂, ..., z_m are linearly independent, which implies that <y, z₂, ..., z_m> is an m-dimensional subspace of S_U . Hence, there exist y₁, ..., y_m <y, z₂, ..., z_m>, such that y_j(0) = e_j for j = 1, ..., m and y <y, z₂, ..., z_m> = <y₁, ..., y_m> =: S_U '.       

Let N(S_U ) denote the set . Then N(S_U ) is a subspace of S_U . The appearance of this subspace N(S_U ) of S_U , which is in general not , is one of the main differences to the case of mappings U: A GL(n, K). We will see that N(S_U ) is closely related to the space W . This is described in

Lemma 18. A function z: A Kⁿ belongs to N(S _U ) if and only if

(24)

Proof.  The function z belongs to N(S_U ) if and only if (0) = 0 and U₁₁( + ) () = U₁₁() ( + ) for all , , A. Especially for = = 0 and because of the particular form of U₁₁ given in (20) we get

so that () = 0 for all A. For = 0 we derive

so that Y ₁₂() () = 0 for all , A. This however implies that () W for all A.

Assuming conversely that (0) = 0, () = 0 and () W for all A, then it is obvious that z N(S_U ).       

In conclusion we get the following result:

Proof.  It is clear that S_U ' N(S _U ) = and that S_U ' N(S _U ) is a subspace of S_U . Assume that x is an element of S_U , then there exists y S_U ', such that y(0) = x(0). Then z() := x() - y() belongs to N(S_U ). Hence x() = z() + y(), which finishes the proof.       

We notice that in the decomposition S_U = S_U ' N(S _U ) the space S_U ' is in general not uniquely determined, whereas N(S_U ) is unique by definition. However, S_U ' can be any m-dimensional subspace of S_U , such that the space of initial values y(0) (for y S_U ') is already S_U⁰.

As an immediate consequence we get

The following Theorem 20 yields together with Theorem 16 the structure of the space S_U , of all solutions of (2), for a given function U: A M_n(K). These theorems also contain necessary conditions on U in order to admit a nontrivial solution y.

We finish by Theorem 21, which provides a construction of all solutions (U, y) of (2) by starting from an arbitrary subspace of initial values y(0) of Kⁿ. This choice then leads via the block matrix Y () satisfying (19) to a matrix valued function U and a space S of solutions corresponding to U. In this general situation we do not discuss the problem when S = S_U .

Theorem 21. If Y () satisfies (19), U₁₁() is given by (20), is given by (16) for arbitrary (0) K^m and () given by (24), then (U, y) is a solution of (2) for

Proof.  From the special form of U() and y() it is clear that (2) is satisfied if and only if U₁₁( + )[ () + ()] = U₁₁()[ ( + ) + ( + )] for all , , A. This is equivalent to U₁₁( + )[Y () (0) + ()] = U₁₁()[Y ( + ) (0) + ( + )]. Due to the definition of () this is equivalent to U₁₁(0)[Y ( + )Y () - Y ()Y ( + )] (0) = 0 for all , , A. Since (0) is an arbitrary element of K^m we derive

We can rewrite U₁₁(0)[Y ( + )Y () - Y ()Y ( + )] as U₁₁(0)[Y ( + )Y () - Y ( + + ) + Y ( + + ) - Y ()Y ( + )] = U₁₁(0)[Y ( + )Y () - Y ( + + )] + U₁₁(0)[Y ( + + ) - Y ()Y ( + )], which is equal to (0)m,m since (19) holds.       

In order to determine all solutions (U, y) of (2) we start with an arbitrary m-dimensional subspace S⁰ of Kⁿ for some 0 < m < n. Let be a basis of S⁰, then there exists a matrix B GL(n, K), such that Bb_i = e_i for 1 < i < m. Hence BS⁰ = <e ₁, ..., e_m>. For each solution Y () of (19) described in Theorem 16 let U₁₁() be given by (20) and U() be given by (13) with arbitrary matrices U₁₂() and U₂₂(). Then each element y of

is together with U a solution of (2). Due to this construction T ⁰, the space of initial values y(0) for y T , is equal to <e₁, ..., e_m>. According to Lemma 1 each pair (UB, B^-1y) for y T is a solution of (2) and = S⁰. Hence, by varying S⁰ over all subspaces of Kⁿ we determine all solutions (U, y) of (2).

References