January 01, 1970
This paper introduces a new theoretical framework for the diagonalization of nonlinear functions defined in finite-dimensional real Euclidean spaces. Extending classical results from linear algebra, we provide a unified setting to determine when a nonlinear map can be represented in diagonal form via a change of basis. Due to the complexity of the equations involved, the final part of the paper focuses primarily on the two-dimensional case, for which clear characterizations can be obtained. We also illustrate how these theoretical findings can be applied to classical contexts, such as differential equations, dynamical systems, and the explicit computation of higher-order compositions and inverses of functions.
Although linear operators have universal applications in all areas of pure and applied science, the linear structure is not always natural in certain settings. In such cases, it may be more convenient to use less restrictive definitions that do not rely on the linear structure of the spaces. However, some of the properties of linear operators can be preserved when we do not use linearity, and this allows us to retain some of the advantages of linear maps, for example diagonalization.
Diagonalization is a fundamental procedure in linear algebra and has numerous useful applications. Recall that, in a real finite-dimensional setting, a linear function \(T\colon\mathbb{R}^n\to\mathbb{R}^n\) is said to be diagonalizable if the matrix representation of \(T\) with respect to some basis is diagonal. Diagonalizing a linear function can be understood as the process of finding a coordinate system in which the action of \(T\) becomes as simple as possible: each basis vector is merely scaled, and no mixing of directions occurs. This “simplification" is a significant advantage of linear diagonalizable functions, and it is the objective of this study to extend this to nonlinear functions.
While this is a general idea, it must be said that nonlinear diagonalization is a well-known and even common area of study. The question of how to define eigenvectors and eigenvalues for general maps has been addressed from many different perspectives, both theoretical [1], [2] and applied [3], [4], and the fundamental aspects are well established. However, in most of these cases, the focus is placed on fixing eigenvalues as numerical scalars and analyzing the corresponding eigenvectors. In the present paper, by contrast, we aim to determine eigenvalues as functions produced by the change of representation of the space induced by the (linear) transformation given by a change of basis. This allows us to study global properties of the operators instead of the usual local linealizations common in dynamical systems such as the classical Hartman–Grobman or Sternberg’s theorems (see, for example, [5]). In this way, we adopt a practical and concrete approach, offering explicit results that can be readily applied in a range of contexts. This will be illustrated through examples presented at the end of the paper across a variety of settings.
The foundations of the concepts studied in this work were originally presented in [6]. There, the lattice Lipschitz inequality \(|T(x) - T(y)| \leq_X K |x-y|\) for mappings \(T: X \to X\) in certain ordered spaces, is shown to be equivalent to a diagonal property. An eigendecomposition-type study is relevant in our context, similarly to the lattice Lipschitz case, although the functions that play the role of eigenvalues in our setting do not satisfy any Lipschitz condition. When the functions \(T: \mathbb{R}^n \to \mathbb{R}^n\) are of class \(\mathcal{C}^1\), the analysis of the linearization via the Jacobian matrix \(DT\) plays a central role. As in the nonlinear eigenvalue problem studied in [7], the eigendecomposition depends on the evaluation point of the matrix, a fact that is emphasized in the second part of the work. In the particular case of the plane, \(\mathbb{R}^2\), we provide explicit algebraic criteria to determine when such diagonalization is possible.
Parallel to theoretical advancements, the past decade has seen numerous practical applications for the diagonalization of nonlinear maps, particularly through the Koopman operator [8]. This approach offers a linearization of a finite dimensional dynamical system by embedding it onto an infinite dimensional space. Several models have been proposed to approximate this transformation using data-driven techniques, including kernel methods, dynamic mode decomposition, or deep neural networks [9]–[11]. While our work shares the goal of isolating one-dimensional dynamics, we focus on the explicit criteria that determines when the diagonalization is possible through a change of basis.
To do this, we structure the paper as follows. After this introduction, in Section 2 we present the context of the problem of diagonalization of non-linear functions. To this end, we will generalize known definitions and results from the linear case, showing illustrative examples of this new scenario. In addition, we will study in Section 3 our proposal in more depth for the case of the 2-dimensional Euclidean space, where we will obtain complete characterizations. Finally, Section 4 is devoted to further explore the potential applications of non-linear diagonalization, such as the study of ordinary differential equations.
We use standard notation from linear algebra and mathematical analysis [12], [13]. Concepts and notations that might be confusing will be explained when they are introduced. The finite dimensional space \(\mathbb{R}^n\) is assumed to be endowed with the Euclidean norm. A function \(\Phi: \Omega \subseteq \mathbb{R}^n \to \mathbb{R}^n\) is said to be Lipschitz if there exists a constant \(K>0\) such that for every \(x_1, x_2 \in \mathbb{R}^n,\) \[\big\| \Phi(x_1)- \Phi(x_2) \big\| \le K \| x_1-x_2\|.\] The infimum of such constants \(K\) is the Lipschitz constant of \(\Phi\), and it is denoted by \(\|\Phi\|_{Lip}\).
We start this section by introducing the main definitions and concepts related to the diagonalization problem in the nonlinear case. We then explore how some of the well-known results in the linear case can be extended to our nonlinear setting.
Throughout this paper \(\Phi\) denotes a function \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\). We use the usual notation \(\Phi(x_1,\ldots,x_n)\) to denote the image of the vector \(x\in\Omega\) whose coordinates in the canonical basis \(\mathcal{C}=\{e_1,\ldots,e_n\}\) are \((x_1,\ldots,x_n)\), that is, \[\Phi(x)=\Phi(x_1e_1+\ldots+x_ne_n)=\Phi(x_1,\ldots,x_n).\] We also write the coordinate \(i\in\{1,\ldots,n\}\) of the image of \(\Phi\) as \(\Phi_i\), so \[\Phi(x)=\Phi_1(x)e_1+\ldots\ldots+\Phi_n(x)e_n=(\Phi_1(x),\ldots,\Phi_n(x)).\] This notation is also used for any other bases when there is no confusion.
Definition 1. We say that a function \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\) is diagonal* if \(\Phi(x_1,\ldots,x_n)=(\Phi_1(x_1),\ldots,\Phi_n(x_n))\) for any \(x\in\Omega\). That is, the \(i\)-th component of the image of \(x\) only depends of the \(i\)-th coordinate of \(x\) in the canonical basis \(\mathcal{C}\). Obviously, if \(\Phi\) is a \(\mathcal{C}^1\) function, this is equivalent to the Jacobian matrix \(D\Phi(x)\) being diagonal.*
As in the linear case, a function \(\Phi\) is diagonalizable if there exist a basis \(\mathcal{B}\) of \(\mathbb{R}^n\) in which its expression is diagonal. In this case, the linear isomorphism given by the change of coordinates from the basis \(\mathcal{B}\) to the canonical basis \(\mathcal{C}\) is a key element.
Definition 2. A function \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\) is diagonalizable if there exists a basis \(\mathcal{B}=\{u_1,\ldots,u_n\}\) of \(\mathbb{R}^n\) such that the expression of \(\Phi\) in that basis is diagonal. That is, there exist scalar functions \(\lambda_1,\ldots,\lambda_n\colon\mathbb{R}\to\mathbb{R}\) such that for any \(x=a_1u_1+\ldots+a_nu_n\in\Omega\), we can write \(\Phi(x)=\lambda_1(a_1)u_1+\ldots+\lambda_n(a_n)u_n\).
We also propose the following definition, which is more practical and will be the one with which we will mainly work in this paper. Although equivalent to the previous definition, as we demonstrate in the following, this definition is consistent with the concept of conjugacy employed in dynamical systems.
Definition 3. For \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\), we say that \(\Phi\) is diagonalizable if there exist a linear isomorphism \(L\) and a diagonal function \(\Psi\) such that \(\Phi=L\circ\Psi\circ L^{-1}\). If there is no confusion, we will write the invertible matrix \(P\) which represents the linear isomorphism instead of \(L\), that is, \(\Phi=P\circ\Psi\circ P^{-1}\).
Proposition 1. Definitions 2 and 3 are consistent.
**Proof.* Consider any function \(\Phi\) diagonalizable in the sense of Definition 2. Denote by \(\mathcal{B}=\{u_1,\ldots,u_n\}\) the basis in which \(\Phi\) is diagonal in this sense, and \(\lambda_1,\ldots,\lambda_n\) the corresponding scalar functions. Let \(P\) be the invertible matrix which gives the change of coordinates from the canonical basis to the basis \(\mathcal{B}\). That is, if we have representations for \(x\in\Omega\) as \(x=x_1e_1+\ldots+x_ne_n\) and \(x=a_1u_1+\ldots+a_nu_n\), then \[P\begin{pmatrix} x_1\\\vdots\\x_n \end{pmatrix}=\begin{pmatrix} a_1\\\vdots\\a_n \end{pmatrix}.\] In particular, for each \(1 \leq i \leq n\) holds \(Pe_i=u_i\). Now, note that \[\begin{align} (P^{-1}\circ\Phi\circ P)\left(\sum_{i=1}^nx_ie_i\right)&=(P^{-1}\circ \Phi)\left(\sum_{i=1}^nx_iu_i\right)\\&=P^{-1}\sum_{i=1}^n\lambda_i(x_i)u_i\\&=\sum_{i=1}^n\lambda_i(x_i)e_i. \end{align}\] Therefore, naming \(\Psi:=P^{-1}\circ\Phi\circ P\) we have that \(\Psi\) is diagonal, and thus \(\Phi\) is diagonalizable in the sense of Definition 3.*
Conversely, let \(\Phi\) be diagonalizable according to Definition 3, and denote by \(P\) the matrix that diagonalizes \(\Phi\) and \(\Psi\) its diagonal function associated. Consider the basis \(\mathcal{B}=\{u_1,\ldots,u_n\}\) where \(u_i=Pe_i\). Then, by a similar argument to the one presented above, we can show that \(\Phi(x)=\lambda_1(a_1)u_1+\ldots+\lambda_n(a_n)u_n\), being \(x=a_1u_1+\ldots+a_nu_n\) and \(\lambda_i:=\Psi_i\). ◻
Similar to the linear case, we define the equivalent concepts of eigenvector and eigenvalues in the following way.
Definition 4. Consider \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\). We say that a nonzero vector \(u\in\mathbb{R}^n\) is an eigenvector of \(\Phi\) if there exists a scalar function \(\lambda\colon\mathbb{R}\to\mathbb{R}\) such that \[\Phi(hu)=\lambda(h)u,\] for all \(h\in\mathbb{R}\) such that \(hu\in\Omega\). That is, \(\Phi\) is invariant under the direction of \(u\), and \(\lambda\) modules the scaling in such direction. We call \(\lambda\) an eigenfunction of \(\Phi\).
From this definition it follows that, for a diagonalizable function \(\Phi\), the elements of the basis in which \(\Phi\) is diagonal are eigenvectors, and the scalar functions associated (those appearing in Definition 2) are eigenfunctions. In what follows, we will present some results concerning the eigenvectors of general functions.
Proposition 2 (Composition results). Consider \(\Phi\colon\Omega\subset\mathbb{R}^n\to\Omega\) and let \(u\) be an eigenvector of \(\Phi\) with associated eigenfunction \(\lambda\).
For each \(k\in\mathbb{N}\), \(u\) is an eigenvector of the \(k\)-fold composition \(\Phi^{(k)}\) and its eigenfunction is \(\lambda^{(k)}\).
Let \(p\) be a polynomial \(p(t)=\sum_{j=0}^N\alpha_jt^j\) and define \(p(\Phi)(\cdot):= \sum_{j=0}^N\alpha_j\Phi^{(j)}(\cdot)\). Then, \(u\) is an eigenvector of \(p(\Phi)\) with eigenfunction \(p(\lambda)\).
**Proof.* To prove (i) note that \[\Phi^{(2)}(hu)=\Phi(\Phi(hu))=\Phi(\lambda(h)u)=\lambda(\lambda(h))u=\lambda^{(2)}(h)u.\] The desired result can be straightforwardly obtained from here by recursion. Finally, the result of (ii) is a consequence of (i) since \[p(\Phi)(hu)=\sum_{j=0}^N\alpha_j\Phi^{(j)}(hu)=\sum_{j=0}^N\alpha_j\lambda^{(j)}(h)u=p(\lambda)(h)u.\] ◻*
For a linear map, it is well known that eigenvectors associated to different eigenvalues are linearly independent. However, this situation may not hold for the linear case. For example, consider \(\Phi(x,y)=(x^2+y^2,2xy)\) and note that \[\begin{align} \Phi(h(1,0))&=(h^2,0)=h^2(1,0),\\ \Phi(h(2,0))&=(4h^2,0)= 2h^2(2,0). \end{align}\] That is, \((1,0)\) and \((2,0)\) are linearly dependent eigenvectors but their respective eigenfunctions are different. However, we can achieve a result similar to that in the linear case by imposing a condition that avoids the previously shown situation.
Proposition 3. Suppose that \(\Phi\colon\Omega\subset\mathbb{R}^n\to\mathbb{R}^n\) has eigenvectors \(u,v\in\mathbb{R}^n\) which eigenfunctions are, respectively, \(\mu\) and \(\lambda\). If for each nonzero \(\alpha\in\mathbb{R}\), we have \(\lambda(h)\alpha\neq\mu(\alpha h)\), for all \(h\in\mathbb{R}\) such that \(hu,hv\in\Omega\), then \(u\) and \(v\) are linearly independent.
**Proof.* Suppose that there exists \(\alpha\in\mathbb{R}\setminus\{0\}\) such that \(v=\alpha u\). Note that \[\label{eq:phiv} \Phi(hv)=\lambda(h)v=\lambda(h)\alpha u,\tag{1}\] and \[\label{eq:phiu} \Phi(hv)=\Phi(h\alpha u)=\mu(h\alpha)u.\tag{2}\] From 1 and 2 it follows that \(\lambda(h)\alpha=\mu(h\alpha)\), which is a contradiction. Therefore, we conclude that \(u\) and \(v\) are linearly independent. ◻*
Proposition 4. Suppose that \(\Phi\colon\Omega\subset\mathbb{R}^n\to\mathbb{R}^n\) is a diagonalizable function. Then, its eigenfunctions are of the form \(\lambda_j(h)=c_jh\), where \(c_j\in\mathbb{R}\), for \(j=1,\ldots,n\) if and only if \(\Phi\) is linear.
**Proof.* Since \(\Phi\) is diagonalizable, let \(u_1,\ldots,u_n\) be the eigenvectors of \(\Phi\) which form that basis in which \(\Phi\) is diagonal. Consider \(\lambda_j\) as the eigenfunction of \(u_j\), and consider \(a_1,\ldots,a_n\in\mathbb{R}\). Therefore, \[\Phi\left(\sum_{j=1}^n a_ju_j\right)=\sum_{j=1}^n\lambda_j(a_j)u_j=\sum_{j=1}^nc_ja_ju_j=\sum_{j=1}^na_j\lambda_j(1)u_j=\sum_{j=1}^na_j\Phi(u_j),\] and consequently \(\Phi\) is linear. To get the converse result, note that if \(\Phi\) is linear and diagonalizable in the sense of our definitions, then its eigenvalues are real numbers \(c_1,\ldots,c_n\). So, for its eigenvectors \(u_1,\ldots,u_n\) we get \[\Phi(hu_j)=h\Phi(u_j)=hc_ju_j=\lambda_j(h)u_j,\] for \(\lambda_j(h)=c_jh\). ◻*
Proposition 5 (Inversion results). For a function \(\Phi\colon\Omega\subset\mathbb{R}^n\to\mathbb{R}^n,\) the following statements hold:
If there exists an eigenvector \(u \in \mathbb{R}^n\) such that its eigenfunction \(\lambda\) is not injective on \(D := \{h \in \mathbb{R} : h u \in \Omega \}\), then \(\Phi\) is not injective either.
Suppose that \(\Phi\) is invertible on its image and denote its inverse by \(\Phi^{-1}\colon\Phi(\Omega)\to\Omega\). If \(u\in\mathbb{R}^n\) is an eigenvector for the eigenfunction \(\lambda\), then \(\lambda\colon D\to\lambda(D)\) is invertible.
Under the previous assumptions, \(u\) is also an eigenvector of \(\Phi^{-1}\) for the eigenfunction \(\lambda^{-1}\).
**Proof.* To get (i) let \(h_1,h_2\in D\) such that \(h_1\neq h_2\) and \(\lambda(h_1)=\lambda(h_2)\). Then, \[\Phi(h_1u)=\lambda(h_1)u=\lambda(h_2)u=\Phi(h_2u),\] and so \(\Phi\) is not injective. The result of (ii) is an immediate consequence of (i) and the fact that we have restricted the image of \(\Phi\) and \(\lambda\) to its ranges. Finally, for \(t\in\lambda(D)\) there exists \(h\in D\) such that \(\lambda(h)=t\). By (ii) we get \(h=\lambda^{-1}(t)\) and thus \[\Phi^{-1}(tu)=\Phi^{-1}(\lambda(h)u)=\Phi^{-1}(\Phi(hu))=hu=\lambda^{-1}(t)u.\] ◻*
The converse of these inversion results does not necessarily hold in general. Consider the function \(\Phi(x,y)=(e^x,0)\). Since \(\Phi(h(1,0))=(e^h,0)=\lambda(h)(1,0)\), then \((1,0)\) is an eigenvector of \(\Phi\) for the injective eigenfunction \(\lambda(h)=e^h\). However, \(\Phi\) is not injective since \(\Phi(a,b)=\Phi(a,c)\) for any \(a\) and \(b\neq c\). And more obviously, \(\Phi\) may be invertible and have no eigenvectors, such as \(\Phi(x,y)=(-y,x)\) which inverse is \(\Phi^{-1}(x,y)=(y,-x)\).
In general, the relationship between the diagonalization problem and the eigenvectors and eigenfunctions is weaker than in the linear case. In the linear case, if there exists a set of \(n\) linearly independent eigenvectors of a function \(T\), then it is known that \(T\) is diagonalizable in the basis defined by these vectors. However, this is not the case for nonlinear functions \(\Phi\), as demonstrated in the following example.
Example 1. Consider \(\Phi\colon\mathbb{R}^2\to\mathbb{R}^2\) defined by \(\Phi(x,y)=(y^2,xy)\). Note that \[\begin{align} \Phi(h(1,1))&=(h^2,h^2)=h^2(1,1),\\ \Phi(h(1,-1))&=(h^2,-h^2)=h^2(1,-1). \end{align}\] Now, for the matrix \(P=\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}\) which inverse is \(P^{-1}=\frac{1}{2}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}\), we observe that \(\Psi=P^{-1}\circ\Phi\circ P\) is not diagonal. Indeed, for \((s,t)\in\mathbb{R}^2\), we get \[\begin{align} \Psi(s,t)&=(P^{-1}\circ\Phi\circ P)(s,t)\\&=(P^{-1}\circ\Phi)(s+t,s-t)\\&=P^{-1}\Phi(s+t,s-t)\\&=P^{-1}\begin{pmatrix} s^2-2st+t^2\\s^2-t^2 \end{pmatrix}=\begin{pmatrix} s^2-st\\ t^2-st \end{pmatrix}. \end{align}\]
Nevertheless, it is possible to get a converse of this result, as we will show below. The key that relates the diagonality with the eigendecomposition is that the function must vanish at \(0\in\mathbb{R}^n\) (origin preserving). Note that this requirement can be satisfied by any function after a translation \(\Psi(x) = \Phi(x) - \Phi(0)\). This condition can be interpreted in different settings. For example, in dynamical systems it means that 0 is a fixed point of \(\Phi\). On the other hand, there are certain classes of mappings of high interest that are origin preserving, such as homogeneous functions with positive degree or, in particular, all linear functions.
Proposition 6. Let \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\) be a function such that \(\Phi(0)=0\). If \(\Phi\) is diagonalizable, then there exist, at least, \(n\) linearly independent eigenvectors \(u_i\) and \(n\) eigenfunctions \(\lambda_i\colon\mathbb{R}\to\mathbb{R}\), not necessarily different.
**Proof.* Since \(\Phi\) is diagonalizable, there exists a basis \(\mathcal{B}=\{u_1,\ldots,u_n\}\) of \(\mathbb{R}^n\) such that \(\Phi(a_1u_1+\ldots+a_nu_n)=\lambda_1(a_1)u_1+\ldots+\lambda_n(a_n)u_n\). Note that the condition \(\Phi(0)=0\) leads to \(\lambda_i(0)=0\) for \(i=1,\ldots,n\) since \(\mathcal{B}\) is a basis of \(\mathbb{R}^n\). Therefore, fixing \(1\leq j\leq n\) and considering \(a_i=0\) if \(i\neq j\), then \[\Phi(a_ju_j)=\lambda_1(0)u_1+\ldots+\lambda_j(a_j)u_j+\ldots+\lambda_n(0)u_n=\lambda_j(a_j)u_j.\] ◻*
In the following example, the necessity of the condition \(\Phi(0)=0\) will be shown.
Example 2. Consider \(\Phi\colon\mathbb{R}^2\to\mathbb{R}^2\) defined as \(\Phi(x,y)=(e^y+x,e^y+y)\), where \(\Phi(0,0)=(1,1)\). This function is diagonalizable, for example, in the basis \(\{(1,1),(1,0)\}\). Indeed, if \(P=\begin{pmatrix} 1 &1\\ 1&0 \end{pmatrix}\) and so \(P^{-1}=\begin{pmatrix} 0 & 1\\1&-1 \end{pmatrix},\) then \[(P^{-1}\circ\Phi\circ P)(s,t)=P^{-1}\circ \Phi(s+t,s)=P^{-1}\begin{pmatrix} e^s+s+t\\ e^s+s \end{pmatrix}=\begin{pmatrix} e^s+s\\t \end{pmatrix}.\] The method for finding this basis will be explained later and is provided by Theorem 10, but this is not the objective of the current example. Besides, although \((1,1)\) is an eigenvector of \(\Phi\) since \[\Phi(h(1,1))=(e^h+h,e^h+h)=\lambda_1(h)(1,1), \quad \text{for} \;\lambda_1(h)=e^h+h,\] the vector \((1,0)\) is not an eigenvector, because \(\Phi(h(1,0))=(1+h,1)\). Nevertheless, although \(\Phi\) is not invariant under the direction of \((1,0)\), it is invariant under a parallel direction: \(\Phi(h,0)=\Phi(0)+\lambda_2(h)(1,0)\), for \(\lambda_2(h)=h\).
To establish the main characterization of diagonalizable functions, let us recall a classic result from linear algebra, see for example [13].
Lemma 1 (Theorem 1.3.21 of [13]). Consider an arbitrary family of diagonalizable matrices \(\{A_{\alpha}\}_{\alpha\in\Delta}\). Then, it is a commuting family if and only if it is a simultaneously diagonalizable family, that is, there exists a non singular \(P\) such that \(PA_{\alpha}P^{-1}\) is diagonal for all \(\alpha\in\Delta\).
Therefore, the characterization to the nonlinear case is as follows
Theorem 7. Given a function \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\) being \(\mathcal{C}^1\), it is diagonalizable if and only if \(\{D\Phi(x)\}_{x\in\Omega}\) is a commuting family of diagonalizable Jacobian matrices.
In the linear case it is known that the orthogonal diagonalization characterizes the symmetric matrices. The next results explore whether this holds for the nonlinear case.
Proposition 8. Let \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\) be a \(\mathcal{C}^1\) function. Suppose there exists an orthogonal matrix \(Q\) that diagonalizes \(\Phi\). Then, the Jacobian matrix \(D\Phi(x)\) is symmetric for all \(x\in\Omega\).
**Proof.* By hypothesis, the function \(\Psi:=Q^t\circ\Phi\circ Q\) satisfies that its Jacobian matrix \(D\Psi(s)\) is diagonal for each \(s\in Q^t\Omega\). In particular, it is symmetric. Moreover, by the chain rule we get \(D\Psi(s)=Q^t\Phi(x)Q\), being \(x=Qs\), and therefore \(D\Phi(x)= QD\Psi(s)Q^t\). So, we conclude that \[D\Phi(x)^t=Q D\Psi(s)^t Q^t=QD\Psi(s)Q^t= D\Phi(x),\] that is, \(D\Phi(x)\) is symmetric for each \(x\in\Omega\). ◻*
The converse of this result does not hold in general. There are functions with symmetric Jacobian matrices that are not diagonalizable, so the well-known characterization in the linear case can not be extended to our non-linear setting. Consider, for example, \(\Phi\colon\mathbb{R}^2\to\mathbb{R}^2\) defined as \(\Phi(x,y)=(x+y,y^2+x)\). Its Jacobian matrix is \[D\Phi(x,y)=\begin{pmatrix} 1 & 1\\ 1 & 2y \end{pmatrix},\] which is clearly symmetric. However, this family of Jacobian matrices do not commute, and thus \(\Phi\) is not diagonalizable by Theorem 7. The problem here is that, while \(D\Phi(x,y)\) is diagonalizable at any \((x,y)\), the eigenvectors differ between points. Nevertheless, if we assume that \(\Phi\) is diagonalizable and its Jacobian is symmetric, then the result holds.
Proposition 9. Let \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\) be a \(\mathcal{C}^1\) function. If \(\Phi\) is diagonalizable and \(D\Phi(x)\) is symmetric, then there exists an orthogonal matrix \(Q\) such that \(\Psi:=Q^t\circ\Phi\circ Q\) is diagonal.
The proof of this result is an immediate consequence of the linear case and Theorem 7.
The aim of this section is to provide an alternative and more practical characterization of diagonalizable functions, which also yields an explicit basis with respect to which the function is diagonal. Due to the technical complexity of this result, for the sake of clarity we restrict the results of this section to functions on \(\mathbb{R}^2\).
If \(\mathcal{B}\) is a basis of \(\mathbb{R}^2\), we denote by \(P_{\mathcal{B}, \mathcal{C}}\) the change of basis matrix from the coordinates in \(\mathcal{B}\) to the coordinates in the canonical basis \(\mathcal{C}\), and \(P_{\mathcal{C}, \mathcal{B}}\) for the inverse change. Recall that \(P_{\mathcal{C}, \mathcal{B}} = P_{\mathcal{B}, \mathcal{C}}^{-1}\). In particular, writing \(\mathcal{B} = \{ v_1, v_2 \} = \big\{ (\alpha_{11}, \alpha_{12}), (\alpha_{21}, \alpha_{22}) \big\}\), we have the change of basis matrix \[P_{\mathcal{B}, \mathcal{C}} = \begin{pmatrix} \alpha_{11} & \alpha_{21} \\ \alpha_{12} & \alpha_{22} \end{pmatrix} \quad \text{and} \quad P_{\mathcal{C}, \mathcal{B}} = \frac{1}{\det(P_{\mathcal{B}, \mathcal{C}})} \begin{pmatrix} \alpha_{22} & - \alpha_{21} \\ - \alpha_{12} & \alpha_{11} \end{pmatrix},\] where \(\det(P_{\mathcal{B}, \mathcal{C}}) = \alpha_{11} \alpha_{22} - \alpha_{12} \alpha_{21}\).
Consider now the mapping \(\Psi = (\Psi_1, \Psi_2) : \tilde{\Omega} \subseteq \mathbb{R}^2 \to \mathbb{R}^2\) to be \(\Phi\) in terms of the coordinates of the new basis \(\mathcal{B}\), that is, \[\Psi(s,t) = \Phi( s v_1 + t v_2 ) = \Psi_1(s,t) v_1 + \Psi_2(s,t) v_2.\] The relation between the definition sets is \(\tilde{\Omega} = P_{\mathcal{C}, \mathcal{B}}\Omega = \big\{ (s,t) \in \mathbb{R}^2 : s v_1 + t v_2 \in \Omega \big\}\). The relation between the mapping in its different coordinates is given by \(\Psi=P_{\mathcal{B}, \mathcal{C}}^{-1}\circ\Phi\circ P_{\mathcal{B}, \mathcal{C}}\), and so \[\label{eq:latlip95Psi95matrix} \begin{pmatrix} \Psi_1 \\ \Psi_2 \end{pmatrix} = \frac{1}{\det(P_{\mathcal{B}, \mathcal{C}})} \begin{pmatrix} \alpha_{22} & - \alpha_{21} \\ - \alpha_{12} & \alpha_{11} \end{pmatrix} \circ \begin{pmatrix} \Phi_1 \\ \Phi_2 \end{pmatrix} \circ \begin{pmatrix} \alpha_{11} & \alpha_{21} \\ \alpha_{12} & \alpha_{22} \end{pmatrix}.\tag{3}\] Expanding in terms of \(\alpha_{ij}\), the previous composition becomes \[\left\{ \begin{align} \det(P_{\mathcal{B},\mathcal{C}}) \Psi_1(s,t) = & \alpha_{22} \Phi_1( \alpha_{11} s + \alpha_{21} t, \alpha_{12} s + \alpha_{22} t ) \\ & - \alpha_{21} \Phi_2( \alpha_{11} s + \alpha_{21} t, \alpha_{12} s + \alpha_{22} t ) \quad \text{and} \\ \det(P_{\mathcal{B},\mathcal{C}}) \Psi_2(s,t) = & - \alpha_{12} \Phi_1( \alpha_{11} s + \alpha_{21} t, \alpha_{12} s + \alpha_{22} t ) \\ & + \alpha_{11} \Phi_2( \alpha_{11} s + \alpha_{21} t, \alpha_{12} s + \alpha_{22} t ), \end{align} \right.\] for any \((s,t) \in \tilde{\Omega}\) or, equivalently, \(( \alpha_{11} s + \alpha_{21} t, \alpha_{12} s + \alpha_{22} t ) \in \Omega\).
To establish the main characterization of the diagonalizable functions, we will first present a key lemma on the existence of a basis in which \(\Phi\) is diagonal. The mappings are assumed to be \(\mathcal{C}^1\) and defined on subsets \(\Omega \subseteq \mathbb{R}^2;\) in particular, they can be defined on the whole \(\mathbb{R}^2\). We denote by \(\varphi_{ij}\) the partial derivative of \(\Phi_i\) over its \(j\)-component, i.e. the Jacobian of \(\Phi\) is given by \[D\Phi (x,y) = \begin{pmatrix} \nabla \Phi_1(x,y) \\ \nabla \Phi_2(x,y) \end{pmatrix} = \begin{pmatrix} \varphi_{11}(x,y) & \varphi_{12}(x,y) \\ \varphi_{21}(x,y) & \varphi_{22}(x,y) \end{pmatrix}.\]
Lemma 2. Let \(\Phi: \Omega \subseteq \mathbb{R}^2 \to \mathbb{R}^2\) be a \(\mathcal{C}^1\) function. Then, there exists a basis of \(\mathbb{R}^2\) such that \(\Phi\) is diagonal if and only if the quadratic form equation \[\label{eq:latlip95quadratic95b} 0 = - \beta_1^2 \varphi_{21}(x,y) + \beta_1 \beta_2 \big( \varphi_{11}(x,y) - \varphi_{22}(x,y) \big) + \beta_2^2 \varphi_{12}(x,y)\qquad{(1)}\] has two solutions \(( \hat{\beta}_1, \hat{\beta}_2 )\) and \(( \bar \beta_1, \bar \beta_2 )\) that are independent from \(x\) and \(y\) and linearly independent.
If this holds, \(\Phi\) is diagonal with respect to the basis given by the solutions.
Proof. Following the notation introduced before, the mapping \(\Phi\) is diagonal in the basis \(\mathcal{B} = \{ v_1, v_2 \} = \big\{ (\alpha_{11}, \alpha_{12}), (\alpha_{21}, \alpha_{22}) \big\}\) if and only if \(\Psi_1(s,t)\) only depends on \(s\) and \(\Psi_2(s,t)\) on \(t\). Equivalently, \[\label{eq:latlip95lemma95partials95095proof} \dfrac{ \partial \Psi_1(s,t) }{ \partial t } = 0 \quad \text{and} \quad \dfrac{ \partial \Psi_2(s,t) }{ \partial s } = 0, \quad \forall (s,t) \in \tilde{\Omega},\tag{4}\] which can be written in terms of \(\Phi\) as \[\left\{ \begin{align} 0 & = \alpha_{22}^2 \varphi_{12}(x,y) + \alpha_{21} \alpha_{22} \big( \varphi_{11}(x,y) - \varphi_{22}(x,y) \big) - \alpha_{21}^2 \varphi_{21}(x,y), \\ 0 & = - \alpha_{12}^2 \varphi_{12}(x,y) + \alpha_{11} \alpha_{12} \big( \varphi_{22}(x,y) - \varphi_{11}(x,y) \big) + \alpha_{11}^2 \varphi_{21}(x,y), \end{align} \right.\] for all \((x,y) = ( \alpha_{11} s + \alpha_{21} t, \alpha_{12} s + \alpha_{22} t ) \in \Omega\). Note that both equations represent the same quadratic equation ?? .
Therefore, if \(\Phi\) is diagonal in the basis \(\mathcal{B}\), then \(( \alpha_{11}, \alpha_{12} )\) and \(( \alpha_{21}, \alpha_{22} )\) are two solutions of ?? . As 4 is satisfied for any \((s,t) \in \tilde{\Omega}\), ?? holds for any \((x,y) \in \Omega\). The linear independence of the solutions is consequence of the fact that \(\mathcal{B}\) is a basis.
Conversely, if two linearly independent solutions \(( \hat{\beta}_1, \hat{\beta}_2 )\) and \(( \bar \beta_1, \bar \beta_2 )\) exist, one can consider the basis \(\mathcal{B} = \big\{ ( \hat{\beta}_1, \hat{\beta}_2 ), ( \bar \beta_1, \bar \beta_2 ) \big\}\), which clearly satisfies 4 , and so \(\Phi\) has a diagonal representation in \(\mathcal{B}\). Then, the corresponding \(\Psi\) is diagonal. ◻
In other words, previous lemma states that there must exists two directions in the plane that are invariant under the Jacobian matrix \(D\Phi(x,y)\) at every point.
Note that if \(\Phi\) is diagonal with respect to a basis \(\mathcal{B} = \{ v_1, v_2 \}\), any other basis formed by multiples of it such as \(\mathcal{D} = \{ \lambda v_1, \mu v_2 \}\) with \(\lambda, \mu \neq 0\) also satisfies the condition. Denoting by \(\hat{\Psi} : \tilde \Omega \subseteq \mathbb{R}^2 \to \mathbb{R}^2\) the functions 3 with respect to \(\mathcal{D}\), we have that \[\hat{\Psi}_1 (s) = \lambda \Psi_1 \left( \frac{s}{\lambda} \right), \quad \text{and} \quad \hat{\Psi}_2 (t) = \mu \Psi_2 \left( \frac{t}{\mu} \right), \quad \forall (s,t) \in \tilde\Omega.\] Therefore, we will assume in some of the following cases that the vectors have norm one. In what follows, we will omit the variables \((x,y)\) when referring to \(\varphi_{i,j}\), since the dependence is clear.
Let us present the main characterization of diagonalizable functions in \(\mathbb{R}^2\).
Theorem 10. Let \(\Phi : \Omega \subseteq \mathbb{R}^2 \to \mathbb{R}^2\) be a \(\mathcal{C}^1\) function. It is diagonalizable in some basis of \(\mathbb{R}^2\) if and only if \[\label{eq:P95discriminant} ( \varphi_{11} - \varphi_{22} )^2 + 4 \varphi_{21} \varphi_{12} \geq 0\qquad{(2)}\] and \[\label{eq:abconstants} \begin{align} A & = \frac{ ( \varphi_{11} - \varphi_{22} )^2 + 2 \varphi_{21}^2 + 2 \varphi_{21} \varphi_{12} }{ ( \varphi_{11} - \varphi_{22} )^2 + ( \varphi_{12} + \varphi_{21} )^2 }, \\ B & = \frac{ ( \varphi_{11} - \varphi_{22} )^4 + 4 ( \varphi_{11} - \varphi_{22} )^2 \varphi_{21} \varphi_{12} }{ \big( ( \varphi_{11} - \varphi_{22} )^2 + ( \varphi_{12} + \varphi_{21} )^2 \big)^2 } \end{align}\qquad{(3)}\] are constant numbers (independent from \(x\) and \(y\)), except when the denominator is \(0\) (\(\varphi_{1,1} = \varphi_{2,2}\) and \(\varphi_{1,2} = \varphi_{2,1} = 0\)).
Proof. By the previous comment, we assume without loss of generality that the basis \(\mathcal{B} = \{ v_1, v_2 \}\) is formed by the norm one vectors \(v_1 = ( \cos \theta, \sin \theta )\) and \(v_2 = ( \cos \xi, \sin \xi )\), with \(\theta, \xi \in [ 0, \pi [\) and \(\theta \neq \xi\). Therefore, by Lemma 2 and rewriting ?? in terms of \(\mathcal{B}\), \(\Phi\) is diagonal in some basis (that would be \(\mathcal{B}\)) if and only if \[\label{eq:teo95quadratic95zeta} \sin \zeta \cdot \cos \zeta \cdot (\varphi_{11} - \varphi_{22}) = (\cos^2 \zeta - \sin^2 \zeta) \cdot \varphi_{21}\tag{5}\] has two different solutions in \([ 0, \pi [\), which we call \(\theta\) and \(\xi\).
By squaring both sides of 5 and denoting by \(t = \sin^2 \zeta\), the equation becomes \(P(t) = 0\), where \(P(t)\) is the second degree polynomial \[\label{eq:P95t} \begin{align} P(t) & = \big( ( \varphi_{11} - \varphi_{22} )^2 + ( \varphi_{12} + \varphi_{21} )^2 \big) \cdot t^2 \\ & + \big( - ( \varphi_{11} - \varphi_{22} )^2 - 2 \varphi_{21}^2 - 2 \varphi_{21} \varphi_{12} \big) \cdot t + \varphi_{21}^2. \end{align}\tag{6}\] Then, the solutions for \(t\) (assuming a non-negative discriminant and that the denominator is not \(0\)) are given by \[\label{eq:P95roots} t=\frac{ ( \varphi_{11} - \varphi_{22} )^2 + 2 \varphi_{21}^2 + 2 \varphi_{21} \varphi_{12} \pm \sqrt{ ( \varphi_{11} - \varphi_{22} )^4 + 4 \varphi_{21} \varphi_{12} ( \varphi_{11} - \varphi_{22} )^2 } }{ 2 ( \varphi_{11} - \varphi_{22} )^2 + 2 ( \varphi_{12} + \varphi_{21} )^2 }.\tag{7}\] (Necessity): Assume now that \(\Phi\) is diagonal with respect to a basis \(\mathcal{B}\) as before. If \(\theta \neq \pi - \xi\), as \(\sin^2 \theta \neq \sin^2 \xi\), there are two different roots of \(P(t)\) which are independent from \(x\) and \(y\). As a consequence, the polynomial has positive discriminant so ?? is satisfied. Calling \(t_1\) and \(t_2\) the result of equation 7 , \(A = t_1 + t_2\) and \(B = (t_1 - t_2)^2\) are constant numbers.
In the case that \(\theta = \pi - \xi\), we have that \(\sin \theta = \sin \xi\) and \(\cos \theta = - \cos \xi\). Thus, \(\cos^2 \theta \cdot \varphi_{21} - \sin^2 \theta \cdot \varphi_{21} = \cos^2 \xi \cdot \varphi_{21} - \sin^2 \xi \cdot \varphi_{2,1}\) and so by equation 5 , \[\label{eq:sin95cos95vphi950} \sin \theta \cdot \cos \theta \cdot (\varphi_{11} - \varphi_{22}) = 0 = \sin \xi \cdot \cos \xi \cdot (\varphi_{11} - \varphi_{22}).\tag{8}\] But \(\sin \theta\) and \(\cos \theta\) cannot be both \(0\), leading to \(\varphi_{11} = \varphi_{22}\). Therefore, formulas of 7 are the same (unique) solution of \(P(t) = 0\). The solution is \(t = \sin^2 \theta = \sin^2 \xi = A\) and \(B = 0\), giving as before independent values from \(x\) and \(y\). Moreover, from \(0 \leq A \leq 1\) it can be derived that \(\varphi_{12} \varphi_{21} \geq - \varphi_{11}^2\) and \(\varphi_{12} \varphi_{21} \geq - \varphi_{11}^2\). As \(\varphi_{12} \neq - \varphi_{21}\), we conclude that \(\varphi_{12} \varphi_{21} \geq 0\) and so ?? is satisfied.
(Sufficiency): Conversely, assume now that ?? is satisfied and \(A\) and \(B\) are constant real numbers for any \(x\) and \(y\) in wich it can be computed. We exclude the case where \(\varphi_{11} = \varphi_{22}\) and \(\varphi_{12} = \varphi_{21} = 0\) for any possible \(x\) and \(y\), since it represents the case of already diagonal expression on the canonical basis, so we can assume that \(A\) and \(B\) are well defined. Note that if we are able to find two different solutions \(\theta, \xi \in [ 0, \pi [\) for 5 , we will have the basis \(\mathcal{B}\) defined in which \(\Phi\) is diagonal by reasoning as in Lemma 2.
As \(B \geq 0\), we can define \[\label{eq:t12} t_1 = \frac{A + \sqrt{B}}{2} \quad \text{and} \quad t_2 = \frac{A - \sqrt{B}}{2}.\tag{9}\] First of all, we have to see that both values belong to \([0,1]\). Completing the square, \[\sqrt{B} \leq \frac{ ( \varphi_{11} - \varphi_{22} )^2 + 2 \varphi_{12} \varphi_{21} }{ ( \varphi_{11} - \varphi_{22} )^2 + ( \varphi_{12} + \varphi_{21} )^2 },\] and hence \(t_1 \geq t_2 \geq 0\), and \(t_2 \leq t_1 \leq 1\).
We start by studying the case in which \(B = 0\). In this case, \(P(t)\) has only one solution \(t_0 = A / 2\), so for the existence of different \(\theta\) and \(\xi\) solutions of the equation as \[\sin^2 \theta = t_0 = \sin^2 \xi,\] we need that \(t_0 < 1\).
Assume by way of contradiction that \(A = 2\) and fix \((x,y)\) such that \[\varphi_{11} \neq \varphi_{22} \quad \text{or} \quad \varphi_{12} \neq 0 \quad \text{or} \quad \varphi_{21} \neq 0.\] Note that \(A = 2\) implies \[( \varphi_{11} - \varphi_{22} )^2 + 2 \varphi_{12}^2 + 2 \varphi_{12} \varphi_{21} = 0.\] As \(B = 0\), \[\varphi_{11} = \varphi_{22} \quad \text{or} \quad ( \varphi_{11} - \varphi_{22} )^2 + 4 \varphi_{12} \varphi_{21} = 0.\] Each case implies that \(\varphi_{12} = - \varphi_{21}\) and \(\varphi_{12} = \varphi_{21}\), respectively, both leading to \(\varphi_{12} = \varphi_{21} = 0\) and hence, to a contradiction.
When \(B \neq 0\), \(t_1 \neq t_2\) are two different solutions of \(P(t) = t\), so different values for \(\theta\) and \(\xi\) can be determined being solutions of 5 . Each direction is given by \[\sin \theta = \sqrt{t_1}, \quad \cos \theta = \pm \sqrt{1 - t_1}, \quad \theta \in [0,\pi[,\] where the sign of the \(\cos \theta\) is determined by 5 . Note also that it is the same sign for any \((x,y)\) by continuity. The same process determines the value of \(\xi\). ◻
Let us present a comprehensive example in which we study the diagonalization problem of a nonlinear function \(\Phi\colon\mathbb{R}^2\to\mathbb{R}^2\) and find the corresponding basis.
Example 3. Consider the \(\mathcal{C}^1\) function \[\label{eq:phiex} \Phi(x,y)=(2e^{-x+y}-x+2y, e^{-x+y}-x+2y),\qquad{(4)}\] which Jacobian matrix is \[D\Phi(x,y)=\begin{pmatrix} \varphi_{11}(x,y) & \varphi_{12}(x,y)\\ \varphi_{21}(x,y) & \varphi_{22}(x,y) \end{pmatrix}=\begin{pmatrix} -2e^{-x+y}-1 & 2e^{-x+y}+2\\ -e^{-x+y} -1 & e^{-x+y} +2 \end{pmatrix}.\] After some elementary calculations we can find that the expression defined in ?? is positive since \((\varphi_{11}-\varphi_{22})^2+4\varphi_{21}\varphi_{12}=e^{-2x}(e^x+e^y)^2\). Besides, we can get that the expressions \(A\) and \(B\) defined in ?? are, respectively, the constants 7/10 and 9/100. Consequently, the function \(\Phi\) is diagonalizable by Theorem 10.
To find a basis in which \(\Phi\) is diagonal, formed by the unitary vectors \((\cos\theta_1,\sin\theta_1)\) and \((\cos\theta_2,\sin\theta_2)\), recall that the values \(t_1\) and \(t_2\) defined as in 9 gives \(t_i=\sin^2\theta_i\). In this case we get \[t_1=\frac{A+\sqrt{B}}{2}=\frac{1}{2},\quad t_2=\frac{A-\sqrt{B}}{2}=\frac{1}{5}.\] Therefore, the desired basis is \(\left\{(1/\sqrt{2}, 1/\sqrt{2}), (2/\sqrt{5}, 1/\sqrt{5})\right\}\).
Some particular cases of the previous theorem include the following examples, illustrating well-known structural properties of certain nonlinear maps.
Example 4. Assume that \(\Phi\colon\Omega\subseteq\mathbb{R}^2\to\mathbb{R}^2\) is a function such that the denominators of \(A\) and \(B\) are \(0\) for any \((x,y)\), that is, \[\label{eq:t95denominator950} \varphi_{11} = \varphi_{22} \quad \text{and} \quad \varphi_{12} = - \varphi_{21}.\qquad{(5)}\] Therefore, the polynomial 6 becomes \(P(t) = \varphi_{21}^2\), which has no solution unless \(\varphi_{21} = 0\). Then, for \(\Phi\) being diagonal must hold \(\varphi_{21}=\varphi_{12}=0\).
Moreover, \(\frac{\partial \Psi_1}{\partial x}(x) = \frac{\partial \Psi_2}{\partial y}(y)\) for any \((x,y) \in \Omega\) (omitting the irrelevant variables), and so \(\Psi_1(x) = a x + b\) and \(\Psi_2(y) = a y + c\) with \(a, b, c \in \mathbb{R}\) are affine functions with the same slope. Then, \(\Phi(x,y) = a (x,y) + (b,c)\) and, in fact, \(\Phi\) is diagonal with respect to any basis of \(\mathbb{R}^2\).
This example is notable because the equations obtained in ?? are the Cauchy-Riemann equations, characterising the holomorphic functions in \(\mathbb{C}\). Consequently, the only complex differentiable functions that are diagonalizable are the affine functions.
Observe also that the linear case associated to ?? is the classical problem of diagonalization of the matrix \[M = \begin{pmatrix} a & b \\ -b & a \end{pmatrix},\] a well-known example of non-diagonalizable matrix (in real terms) for \(b \neq 0\) and diagonal with respect to any basis when \(b = 0\).
Example 5. Another important case is when \[\label{eq:t95phi95A195B0} \varphi_{11} = \varphi_{22} \quad \text{and} \quad \varphi_{12} = \varphi_{21}.\qquad{(6)}\] This implies that the Jacobian matrix of \(\Phi\) is symmetric and therefore, there exists \(f : \Omega \subseteq \mathbb{R}^2 \to \mathbb{R}\) differentiable such that \(\Phi = \nabla f\). Moreover, as \(\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial y^2}\), we have that there exist \(g \in\mathcal{C}^2(\Omega_1)\) and \(h \in\mathcal{C}^2(\Omega_2)\) such that \[f(x,y) = g(x+y) + h(x-y).\] The mapping \(\Phi\) is then diagonal with respect to the vectors \((1,1)\) and \((1,-1)\).
The same result can be seen as a consequence of Theorem [teo:latlip95car]. As \(A = 1\) and \(B = 1\), the mapping is diagonal with respect to the basis \[\{ ( 1/\sqrt{2}, 1/\sqrt{2} ), (-1/\sqrt{2}, 1/\sqrt{2}) \}.\]
As in the linear case, finding a diagonal form for a mapping by means of a change of basis provides an interesting advantage for explicit analytical computations of certain fundamental processes for applications. In this section, we proceed to demonstrate three such applications.
Diagonalization for a general function \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\Omega\) is a powerful tool to compute high-order compositions, since \(\Phi^{(k)}=P\circ\Psi^{(k)}\circ P^{-1}\) where \(P\) is the matrix of the isomorphism to get the diagonal form \(\Psi\). And since \(\Psi\) is diagonal, the problem is reduced to compute \(k\) high-order composition but from simpler functions of one variable. In this direction, this section provides the first application of our results.
Example 6. Consider \(\Omega=\{(x,y)\in\mathbb{R}^2:y\neq x\}\) and \(\Phi\colon\Omega\to\mathbb{R}^2\) defined as \[\label{eq:ex} \Phi(x,y)= \left( 4x-2y+\frac{1}{y-x},4x-2y+\frac{2}{y-x} \right)\qquad{(7)}\] It can be checked that \(\Phi(\Omega)\subset\Omega\) and this function is diagonalizable by Theorem 10. One basis in which \(\Phi\) is diagonal is given by \(\{(1,1),(1,2)\}\), being the diagonal expression \(\Psi(s,t)=(2s,1/t)\) with variables relation \((s,t)=(2x-y,-x+y)\). If \((x,y)\in\Omega\) then observe that \(t\neq 0\) and hence \(\Psi\) is well defined.
The \(k\)-fold composition of \(\Psi\) can be easily derived as \[\Psi^{(k)}(s,t)=(2^ks, t^{(-1)^k}),\] and therefore \[\begin{align} \Phi^{(k)}(x,y)&=(P\circ\Psi^{(k)}\circ P^{-1})(x,y)\\ &=(P\circ\Psi^{(k)})(2x-y,-x+y)\\ &=P\circ\begin{pmatrix} 2(2x-y) \\ \tfrac{1}{y-x} \end{pmatrix}^{(k)}\\&=P\circ\begin{pmatrix} 2^k(2x-y)\\ (y-x)^{(-1)^k} \end{pmatrix}\\&=\begin{pmatrix} 2^k(2x-y)+(y-x)^{(-1)^k}\\ 2^k(2x-y)+2(y-x)^{(-1)^k} \end{pmatrix}. \end{align}\]
As a particular case, we find the computation of the exponential function. Given a matrix \(A\in\mathbb{R}^{n\times n}\), recall that its exponential can be defined as the power series \[e^A=\sum_{k=0}^\infty \frac{1}{k!}A^k.\] In this way, the exponential of a linear function can be defined. Now we are interested in defining an equivalent concept for a general function \(\Phi\colon\Omega\subseteq\mathbb{R}^n\to\Omega\). We define \(e^\Phi\colon\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n\) as \[e^\Phi(\cdot)=\sum_{k=0}^\infty \frac{1}{k!}\Phi^{(k)}(\cdot),\] where \(\Phi^{(k)}\) denotes the \(k\)-fold composition. We will also consider the function \(E_\Phi\colon\Omega\times\mathbb{R}\to\mathbb{R}^n\) defined as \[\label{eq:expxi} E_{\Phi}(x,\xi)=\sum_{k=0}^\infty \frac{\xi^k}{k!}\Phi^{(k)}(x).\tag{10}\] Clearly, \(E_\Phi(x,0) = x\) due to the usual convention \(\Phi^{(0)}=Id\), and also yields \(E_\Phi(x,1) = e^\Phi(x)\). Some basic examples of this function are the following, where \(e^t\) denotes the usual exponential of real numbers.
(i) For the identity function \(\Phi(x) = x\), we get \(E_\Phi(x,\xi) = e^\xi x\).
(ii) More in general, when \(\Phi(x) = \mu x\) for \(\mu \in \mathbb{R}\), yields \(E_\Phi(x,\xi) = e^{\mu \xi} x\).
(iii) If \(\Phi(x) = x + x_0\) for \(x_0 \in \mathbb{R}^n\), then \(E_\Phi(x,\xi) = e^{\xi} (x+\xi x_0)\).
(iv) For a linear map \(\Phi(x) = A x\), where \(A\) is a real \(n \times n\) matrix, our definition of the exponential function matches the well-known matrix exponential: \(E_\Phi(x,\xi) = e^{A \xi} x\).
The following results will cover the study of the convergence of this series.
Proposition 11. For \(\Phi\colon\Omega\subset\mathbb{R}^n\to\Omega\) consider the series defined in 10 . Then, for each fixed \(x\in\Omega\), the series converges for every \(\xi\in\mathbb{R}\) such that \(|\xi|<R(x)\), where \(R(x):=\liminf_{k} \left(\frac{k!}{\|\Phi^{(k)}(x)\|}\right)^{\frac{1}{k}}.\)
**Proof.* For a fixed \(x\in\Omega\) consider the real series whose general term is the norm of the general term of 10 , that can be written as \(\sum_{k=0}^\infty a_k|\xi|^k\) with \(a_k=\|\Phi^{(k)}(x)\|/k!\). By the root test for power series, it converges for \(\xi\in\mathbb{R}\) such that \(|\xi|<R\), being \(R\) the radius of convergence that can be computed as \[|\xi|<R:=\liminf_k|a_k|^{-1/k}=\liminf_k\left(\frac{k!}{\|\Phi^{(k)}(x)\|}\right)^{1/k}.\] Hence, for such \(\xi\), the vector series \(E_\Phi(x,\xi)\) converges absolutely in \(\mathbb{R}^n\) and therefore converges. ◻*
Let us now consider the Lipschitz condition for \(\Phi\) as well as \(\Phi(0)=0\) in order to show that, under particular requirements, we can obtain better convergence results. Note, however, that the example at the beginning of this section does not satisfy these properties.
Proposition 12. Let \(\Phi\colon\Omega\subset\mathbb{R}^n\to\Omega\) be a Lipschitz function such that \(\Phi(0)=0\) and consider the series 10 . Then, for every \(\xi\in\mathbb{R}\) the series converges absolutely and uniformly in the compact subsets of \(\Omega\). Furthermore, \(E_{\Phi}(\cdot,\xi)\) is Lipschitz with \(\|E(\cdot,\xi)\|_{Lip}\leq e^{|\xi|\|\Phi\|_{Lip}}\).
**Proof.* Recall that the Lipschitz seminorm of \(f\colon\Omega\to\mathbb{R}^n\) is \[\|f\|_{Lip}=\sup\left\{\frac{\|f(x)-f(y)\|}{\|x-y\|}:x,y\in\Omega, \, x\neq y\right\},\] and \(f\) is said to be Lipschitz if \(\|f\|_{Lip}<\infty\). This seminorm is submultiplicative under composition, that is, \(\|f\circ g\|_{Lip}\leq\|f\|_{Lip}\|g\|_{Lip}\). By induction this gives \(\|\Phi^{(k)}\|_{Lip}\leq \|\Phi\|_{Lip}^k\) for all \(k\in\mathbb{N}\).*
Fix \(\xi\in\mathbb{R}\) and note that, for \(x\in\Omega\), holds \[\|\Phi^{(k)}(x)\|=\|\Phi^{(k)}(x)-\Phi^{(k)}(0)\|+\|\Phi^{(k)}(0)\|\leq \|\Phi^{(k)}\|_{Lip}\|x\|\leq \|\Phi\|_{Lip}^k\|x\|\] for all \(k=0,1,\ldots\), and hence \[\left\|\frac{\xi^k}{k!}\Phi^{(k)}(x)\right\|\leq \frac{|\xi|^k}{k!}\|\Phi\|_{Lip}^k\|x\|=:M_k.\] Since the series of \(M_k\) is convergent to \(\|x\|e^{|\xi|\|\Phi\|_{Lip}}\), by the Weierstrass \(M\)-criterion the series 10 is absolutely and uniformly convergent in the compact subsets of \(\Omega\). Also, by similar arguments it can be obtained \[\|E_\Phi(\cdot,\xi)\|_{Lip}\leq\sum_{k=0}^\infty \frac{|\xi|^k}{k!}\|\Phi^{(k)}\|_{Lip}\leq\sum_{k=0}^\infty \frac{|\xi|^k}{k!}\|\Phi\|_{Lip}^k=e^{|\xi|\|\Phi\|_{Lip}}.\] Consequently, \(E_\Phi(\cdot,\xi)\) is a Lipschitz function such that \(\|E_\Phi(\cdot,\xi)\|_{Lip}\leq e^{|\xi|\|\Phi\|_{Lip}}\). ◻
Clearly, the condition \(\Phi(0)=0\) is not necessary to get the Lipschitz property for \(E_\Phi(\cdot,\xi).\) However, this requirement is neither necessary to obtain the convergence results. For any \(x_0\in\Omega\) (not necessarily a fixed point) and \(\Phi\) such that \(\|\Phi\|_{Lip}\neq 1,\) it can be deduced that \[\|\Phi^{(k)}(x_0)\|\leq \frac{1-\|\Phi\|_{Lip}^k}{1-\|\Phi\|_{Lip}}\|\Phi(x_0)-x_0\|+\|x_0\|,\] for each \(k\). In this case, the term \(M_k\) becomes \[M_k=\frac{|\xi|^k}{k!}\left(\|\Phi\|_{Lip}^k\|x-x_0\|+\frac{1-\|\Phi\|_{Lip}^k}{1-\|\Phi\|_{Lip}}\|\Phi(x_0)-x_0\|+\|x_0\|\right),\] and \[\sum_{k=0}^\infty M_k=\|x-x_0\|e^{|\xi|\|\Phi\|_{Lip}}+\frac{e^{|\xi|}-e^{|\xi|\|\Phi\|_{Lip}}}{1-\|\Phi\|_{Lip}}\|\Phi(x_0)-x_0\|+\|x_0\|e^{|\xi|}.\] Therefore, by the Weierstrass \(M\)-criterion we deduce again the convergence results. These are also valid for the case \(\|\Phi\|_{Lip}=1\), which we omit due to its straightforwardness. We left assumption \(\Phi(0)=0\) in Proposition 12 because it has been considered throughout the paper and for the sake of simplicity.
The diagonalization tools developed in this paper facilitate the calculation of the series \(E_\Phi\). Indeed, for a diagonal function \(\Psi=(\Psi_1,\ldots,\Psi_n)\) observe that \[(\Psi\circ\Psi)(x)=\Psi(\Psi_1(x_1),\ldots,\Psi_n(x_n))=(\Psi_1^{(2)}(x_1),\ldots,\Psi_n^{(2)}(x_n)),\] and in general \(\Psi^{(k)}(x)=(\Psi_1^{(k)}(x_1),\ldots,\Psi_n^{(k)}(x_n))\). So, if the series \(E_{\Psi}\) is convergent for some \(\xi\in\mathbb{R}\), then \[\begin{align} E_{\Psi}(x,\xi)&=\sum_{k=0}^\infty \frac{\xi^k}{k!}\Psi^{(k)}(x)\\&= \sum_{k=0}^\infty \frac{\xi^k}{k!}(\Psi_1^{(k)}(x_1),\ldots,\Psi_n^{(k)}(x_n))\\ &=\left(\sum_{k=0}^\infty\frac{\xi^k}{k!}\Psi_1^{(k)}(x_1),\ldots,\sum_{k=0}^\infty\frac{\xi^k}{k!}\Psi_n^{(k)}(x_n)\right)\\ &=(E_{\Psi_1}(x_1,\xi),\ldots,E_{\Psi_n}(x_n,\xi)) \end{align}\] and hence \(E_\Psi(\cdot,\xi)\) is also diagonal. Furthermore, if \(\Phi=P\circ \Psi\circ P^{-1}\), then \(\Phi^{(k)}=P\circ\Psi^{(k)}\circ P^{-1}\) and therefore \[\begin{align} E_\Phi(x,\xi)&=\sum_{k=0}^\infty \frac{\xi^k}{k!}\Phi^{(k)}(x)\\ & =\sum_{k=0}^\infty \frac{\xi^k}{k!}(P\circ\Psi^{(k)}\circ P^{-1})(x)\\ & = P\sum_{k=0}^\infty \frac{\xi^k}{k!}\Psi^{(k)}(P^{-1}x)=PE_\Psi(P^{-1}x,\xi). \end{align}\]
An important feature of the introduced exponential expressions is that they are solutions to a wide range of differential equations characterised by functions \(\Phi\) that may be not linear. Therefore, these differential equations can be easily solved as long as such functions \(\Phi\) can be diagonalized.
Proposition 13. Consider \(\Phi\colon\Omega\subset\mathbb{R}^n\to\Omega\) and \(\Gamma\colon\Omega\times I\to\mathbb{R}^n\), where \(I\subset \mathbb{R}\) is an open interval such that \(\Gamma(x,\cdot)\) is derivable in \(I\) for any \(x\in\Omega\). If \(E_{\Phi}\) is uniformly convergent in \(\Omega\) (for example when \(\Phi\) is Lipschitz), one solution for the differential equation \[\label{eq:deexp} \frac{\partial\,\Gamma(x,\xi)}{\partial\,\xi}=\Gamma(\Phi(x),\xi).\qquad{(8)}\] is given by \(\Gamma(x,\xi)=E_{\Phi}(x,\xi).\)
**Proof.* Fix \(x\in\Omega\) and note that \(\xi\mapsto c_k(\xi):=\frac{\xi^k}{k!}\Phi^{(k)}(x)\) is differentiable in \(I\) with continuous derivatives for each \(k=0,1,\ldots\), since \[\frac{\partial c_k(\xi)}{\partial\xi}=\frac{\xi^{k-1}}{(k-1)!}\Phi^{(k)}(x)\] for \(k\ge 1\) and \[\frac{\partial c_0(\xi)}{\partial\xi}=\frac{\partial x}{\partial \xi}=0.\] Moreover, note that \[\sum_{k=0}^\infty \frac{\partial c_k(\xi)}{\partial \xi}=\sum_{k=1}^\infty \frac{\xi^{k-1}}{(k-1)!}\Phi^{(k)}(x)=\sum_{k=0}^\infty \frac{\xi^k}{k!}\Phi^{(k)}(\Phi(x))= E_{\Phi}(\Phi(x),\xi),\] which is uniformly convergent by hypotesis. Consequently, we can derive \(E_{\Phi}(x,\cdot)\) term by term to conclude \[\begin{align} \frac{\partial}{\partial\xi} E_\Phi(x,\xi)=\frac{\partial}{\partial\xi}\sum_{k=0}^\infty c_k(\xi) =\sum_{k=0}^\infty \frac{\partial c_k(\xi)}{\partial \xi}= E_{\Phi}(\Phi(x),\xi). \end{align}\] ◻*
Example 7. For the diagonal function \(\Psi(s,t)=(\Psi_1(s),\Psi_2(t))=(2s,1/t)\) considered in the Example 6 we consider the differential equation defined in ?? . The solution given by Proposition 13 is \(E_{\Psi}(s,t,\xi)\), which can be computed as \((E_{\Psi_1}(s,\xi),E_{\Psi_2}(t,\xi))\), where \[\begin{align} E_{\Psi_1}(s,\xi)&=\sum_{k=0}^\infty \frac{\xi^k}{k!}\Psi_1^{(k)}(s)=\sum_{k=0}^\infty \frac{\xi^k}{k!}2^ks=e^{2\xi}s,\\ E_{\Psi_2}(t,\xi)&=\sum_{k=0}^\infty \frac{\xi^k}{k!}\Psi_2^{(k)}(t)=\sum_{k=0}^\infty \frac{\xi^k}{k!}t^{(-1)^k}\\ &=\sum_{k=0}^\infty \frac{\xi^{2k}}{2k!}t+\sum_{k=0}^\infty \frac{\xi^{2k+1}}{(2k+1)!}\frac{1}{t}\\ &=t\cosh(\xi)+\tfrac{1}{t}\sinh(\xi). \end{align}\] Let us check that we have indeed found a solution. First, we observe that \[\frac{\partial E_\Psi(s,t,\xi)}{\partial\xi}=(2e^{2\xi}s, t\sinh(\xi)+\tfrac{1}{t}\cosh(\xi)).\] On the other hand, \[E_\Psi(\Psi(s,t),\xi)=E_\Psi(2s,1/t,\xi)=(2se^{2\xi},\tfrac{1}{t}\cosh(\xi)+t\sinh(\xi)),\] and therefore \[\frac{\partial E_\Psi(s,t,\xi)}{\partial\xi}=E_\Psi(s,t,\xi).\]
Example 8. Consider the differential equation ?? given by the function defined in ?? . To compute the solution given by Proposition 13, recall that \(\Phi\) is diagonalizable in the basis \(\{(1,1),(1,2)\}\) with diagonal expression \(\Psi(s,t)=(\Psi_1(s),\Psi_2(t))=(2s,1/t)\). Then, the desired solution is \[\begin{align} E_\Phi(x,y,\xi)&=PE_{\Psi}(2x-y,-x+y,\xi)\\ &=P\begin{pmatrix} e^{2\xi} (2x-y)\\ (-x+y)\cosh(\xi)+\tfrac{1}{-x+y}\sinh(\xi) \end{pmatrix}\\ &=\begin{pmatrix} e^{2\xi}(2x-y) + (-x+y)\cosh(\xi)+\tfrac{1}{-x+y}\sinh(\xi)\\ e^{2\xi}(2x-y)+2(-x+y)\cosh(\xi)+\tfrac{2}{-x+y}\sinh(\xi) \end{pmatrix}. \end{align}\]
As we have shown in the previous, the tools we have developed can be useful in the field of differential equations. The following examples illustrate how to diagonalize nonlinear functions that model classical systems of nonlinear differential equations, and how to derive solutions to these problems.
Consider derivable functions \(x(t),y(t)\) that satisfy the autonomous system \[\begin{cases} \frac{dx}{dt}=2e^{-x+y}-x+2y,\\ \frac{dy}{dt}=e^{-x+y}-x+2y. \end{cases}\] It can be written as \(\frac{dX}{dt}=\Phi(X)\), being \(X(t):=(x(t),y(t))\) and \(\Phi\) the function defined in ?? . As we have seen in Example 3, this function can be diagonalized in the basis \(\{(2,1),(1,1)\}\). In particular, its diagonal expression is \(\Psi(p,q):=(e^{-p},q)\), with \(p=x-y\) and \(q=-x+2y\). Thus, the original systems is transformed to \[\begin{cases} \frac{dp}{dt}=e^{-p}\\ \frac{dq}{dt}=q, \end{cases}\] than can be easily solved as two separated ODEs. For the second one, we obtain \(q(t)=K_2e^t\). The first is also a separable equation with solution \(p(t)=\log(t+K_1)\). So, the solutions for \(x\) and \(y\) are \(x(t)=2\log(t+K_1)+K_2e^t\) and \(y(t)=\log(t+K_1)+K_2e^t\). Indeed, note that \[\begin{align} \frac{dx}{dt}&=\frac{2}{t+K_1}+K_2e^t=2e^{-\log(t+K_1)}+K_2e^t=2e^{-x+y}-x+2y,\\ \frac{dy}{dy}&=\frac{1}{t+K_1}+K_2e^t=e^{-\log(t+K_1)}+K_2e^t=e^{-x+y}-x+2y. \end{align}\]
On the other hand, in some contexts is useful to invert the function \(\Phi\), see [14]. In our example the diagonal function \(\Psi\) can be easily inverted as \(\Psi^{-1}(p,q)=(-\log(p),q)\), and so \[\begin{align} \Phi^{-1}(x,y)&=(P\circ \Psi^{-1}\circ P^{-1})(x,y)\\ &=(P\circ\Psi^{-1})(x-y,-x+2y)\\ &=P\begin{pmatrix} -\log(x-y)\\ -x+2y \end{pmatrix}=\begin{pmatrix} -2\log(x-y)-x+2y\\ -\log(x-y)-x+2y \end{pmatrix}. \end{align}\]
To finish, let us show how diagonalization can be applied to study the equilibrium points of dynamical systems. Consider the following system: \[\begin{cases} \frac{dx}{dt}=2(2x-y)^2+\sin(-3x+2y),\\ \frac{dy}{dt}= 3(2x-y)^2+2\sin(-3x+2y). \end{cases}\]
We can diagonalize the function \(\Phi\) when defined in the basis \(\{(2,3),(1,2)\}\) as \(\Psi(p,q)=(p^2,\sin(q))\), where \(p=2x-y\) and \(q=-3x+2y\). In this new form, the system becomes \[\begin{cases} \frac{dp}{dt}=p^2,\\ \frac{dq}{dt}= \sin(q), \end{cases}\] and so its equilibrium points are of the form \((p,q)=(0,k\pi)\) with \(k\in\mathbb{Z}\). Consequently, \((x,y)=(k\pi, 2k\pi)\) are the equilibrium points of the original system.
The first author thanks the support of Generalitat Valenciana (Spain), grant number PROMETEO 2024 CIPROM/2023/32. The third author was supported by a contract of the Programa de Ayudas de Investigación y Desarrollo (PAID-01-24), Universitat Politècnica de València. The fourth author was supported by the R&D&I project/grant PID2022-138342NB-I00 funded by MCIN/ AEI/10.13039/501100011033/ (Spain)
The first author thanks the support of Generalitat Valenciana (Spain), grant number PROMETEO 2024 CIPROM/2023/32. The third author was supported by a contract of the Programa de Ayudas de Investigación y Desarrollo (PAID-01-24), Universitat Politècnica de València. The third author was supported by the R&D&I project/grant PID2022-138342NB-I00 funded by MCIN/ AEI/10.13039/501100011033/ (Spain).↩︎