尋找滿足一個函數方程的矩陣

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Find a $4 × 4$, nonsingular, nonconstant matrix function $N (x)$ that satisfies the functional equation$$N(2 x)-(N(x))^{8}=0$$

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maa.org/sites/default/files/Kominers-CMJ0918817.pdf

At first glance, this problem appears to be quite difficult. Beyond the likely difficulty of finding such a matrix $N (x)$, it is not even immediately clear how one would prove that a matrix $N (x)$ is actually a solution without a great deal of matrix algebra. However, this problem is not hard as it seems. In fact, it is one of a large class of problems that can be solved via a surprising method based upon single-variable calculus. In this JOURNAL , Khan [2] used nilpotent matrices and Taylor series to find matrix functions satisfying the exponential functional equation, $f (x + y) = f (x) · f (y)$. His method is an example of a much more general theory of matrix power series due to Weyr [4], which can be used to find matrix functions satisfying a variety of functional equations. (Rinehart [3] gives an excellent survey of Weyr’s approach. Higham [1, ch. 4] gives a more comprehensive account, as well as further applications.)

We say that a set of real-valued functions $\{ f_ i (x)\}^n_{i=1} ⊂{\cal C}^∞(\Bbb R)$ satisfies an analytic functional equation $E$ if there is an analytic function $E$ such that$$ E\left(f_{1}, \ldots, f_{n}\right)(x)=0 $$ identically for all $x \in \mathbb{R}$. For example, the trigonometric functions $f_{1}(x)=\sin (x)$ and $f_{2}(x)=\cos (x)$ satisfy the analytic functional equation $$ E\left(f_{1}, f_{2}\right)=f_{1}^{2}+f_{2}^{2}-1 \equiv 0 $$ Now, for any set of real-valued functions $\left\{f_{i}(x)\right\}_{i=1}^{n} \subset \mathcal{C}^{\infty}(\mathbb{R})$ satisfying the analytic functional equation $E$, we will find a set of associated matrix functions $\left\{A_{i}(x)\right\}_{i=1}^{n}$ satisfying the same functional equation $E$.

Approximating each $f_{i}$ by its Taylor expansion about the origin, we obtain $$ f_{i}(x)=f_{i}(0) x^{0}+\frac{f_{i}^{(1)}(0) x}{1 !}+\frac{f_{i}^{(2)}(0) x^{2}}{2 !}+\cdots+\frac{f_{i}^{(k)}(0) x^{k}}{k !}+\cdots $$ where $f_{i}^{(j)}$ is the $j$ th derivative of the function $f_{i}$. We let $A$ be any nilpotent matrix with index of nilpotence $k$ and then take $$ \begin{aligned} A_{i}(x):=f_{i}(A x) &=f_{i}(0) I+\frac{f_{i}^{(1)}(0) A x}{1 !}+\frac{f_{i}^{(2)}(0) A^{2} x^{2}}{2 !}+\cdots \\ &=f_{i}(0) I+\frac{f_{i}^{(1)}(0) A x}{1 !}+\frac{f_{i}^{(2)}(0) A^{2} x^{2}}{2 !}+\cdots+\frac{f_{i}^{(k-1)}(0) A^{k-1} x^{k-1}}{(k-1) !} \end{aligned} $$ If the functions $\left\{f_{i}(x)\right\}_{i=1}^{n}$ satisfy the analytic functional equation $E$, then the Taylor series of the $f_{i}$ do as well, as $E$ is continuous. Thus, the matrix functions $\left\{A_{i}(x)\right\}_{i=1}^{n}$ found from the Taylor series of the $f_{i}$ must also satisfy the functional equation $E$. We begin with a simple example. For aesthetic reasons, we will work with the nilpotent matrix $$ A=\left(\begin{array}{llll} 0 & 0 & 12 & 0 \\ 0 & 0 & 0 & 12 \\ 6 & 6 & 0 & 0 \\ -6 & -6 & 0 & 0 \end{array}\right) $$

We obtain from the Taylor series for $f_{1}=\sin (x)$ and $f_{2}=\cos (x)$ the matrices $$ \begin{aligned} &A_{1}(x)=f_{1}(A x)=A x-\frac{A^{3} x^{3}}{6}=\left(\begin{array}{llll} 0 & 0 & 12 x-144 x^{3} & -144 x^{3} \\ 0 & 0 & 144 x^{3} & 144 x^{3}+12 x \\ 6 x & 6 x & 0 & 0 \\ -6 x & -6 x & 0 & 0 \end{array}\right), \\ &A_{2}(x)=f_{2}(A x)=I-\frac{A^{2} x^{2}}{2}=\left(\begin{array}{llll} 1-36 x^{2} & -36 x^{2} & 0 & 0 \\ 36 x^{2} & 36 x^{2}+1 & 0 & 0 \\ 0 & 0 & 1-36 x^{2} & -36 x^{2} \\ 0 & 0 & 36 x^{2} & 36 x^{2}+1 \end{array}\right) . \end{aligned} $$ We have immediately from this construction that $A_{1}(x)$ and $A_{2}(x)$ commute. More interestingly, these matrix functions satisfy the trigonometric functional equations. We therefore find the familiar identity $$ \left(A_{1}(x)\right)^{2}+\left(A_{2}(x)\right)^{2}=I . $$ Similarly, we obtain analogues of the "double-angle" formulas, $$ \begin{aligned} &A_{1}(2 x)=2 A_{1}(x) A_{2}(x) \\ &A_{2}(2 x)=\left(A_{2}(x)\right)^{2}-\left(A_{1}(x)\right)^{2}=2\left(A_{2}(x)\right)^{2}-I \end{aligned} $$ Although the matrices found via this method need not be nonsingular, in general, the matrix $A_{2}(x)$ is, as $\operatorname{det} A_{2}(x)=1 \neq 0$. We can therefore invert $A_{2}(x)$ and obtain an analogue of the secant-tangent trigonometric square identity: $$ \left(A_{2}^{-1}(x)\right)^{2}=I+\left(A_{1}(x) A_{2}^{-1}(x)\right)^{2} . $$

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As a second example of our approach, let us solve the problem stated at the beginning of this Capsule. We seek a $4 \times 4$, nonconstant matrix function $N(x)$ satisfying the functional equation $$ N(2 x)-(N(x))^{8}=0 . $$ As before, we consider the associated functional equation in nonconstant $\mathcal{C}^{\infty}(\mathbb{R})$ functions, $$ g(2 x)-(g(x))^{8}=0 $$ Now, this condition immediately implies that either $g(0)=0$ or $(g(0))^{7}=1$. In the former case, $g$ would vanish to some order $k>0$ at the origin, and we would then have $k=8 k$ from the functional equation-impossible. Thus, $g(0)$ is a seventh root of unity, and we may require $g(0)=1$ by considering the function $g / g(0)$ if necessary.

Then, we write $g(x)=1+g_{n} x^{n}+\cdots$ for some minimal $n>0$ and $g_{n} \neq 0$. By the functional equation, we must have $$ 1+2^{n} g_{n} x^{n}+\cdots=g(2 x)=(g(x))^{8}=\left(1+g_{n} x^{n}+\cdots\right)^{8}=1+8 g_{n} x^{n}+\cdots $$ Equating coefficients on both sides then gives that $n=3$; we may also assume that $g_{3}=1$ by replacing $x$ by $x / \sqrt[3]{g_{n}}$. If we now write $g(x)=1+x^{3}+g_{m} x^{m}+\cdots$, for $m>3$ minimal and $g_{m} \neq 0$, we obtain from a further application of the functional equation that $m=6$. Thus, for some $g_{6} \neq 0$, $$ g(x)=1+x^{3}+g_{6} x^{6}+\cdots $$ Since the matrix $A$ chosen above has index of nilpotence $4<6$, we have found sufficient information to compute the matrix $N(x):=g(A x)$. Indeed, we have \begin{aligned} N(x)=g(A x)=I+A^{3} x^{3}+g_{6} A^{6} x^{6}+\cdots &=I+A^{3} x^{3} \\ &=\left(\begin{array}{llll}1 & 0 & 864 x^{3} & 864 x^{3} \\ 0 & 1 & -864 x^{3} & -864 x^{3} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right) \end{aligned} It is immediate from this construction that $N(x)$ satisfies the desired functional equation.

Unwinding our technique, we obtain a general approach to problems that ask for matrices satisfying analytic functional equations. Specifically, one can approach a problem asking for matrices $\left\{A_{i}(x)\right\}_{i=1}^{n}$ satisfying an analytic equation $$ E\left(A_{1}(x), \ldots, A_{n}(x)\right) \equiv 0 $$ by trying to solve the equation $E$ for real-valued functions $\left\{f_{i}(x)\right\}_{i=1}^{n}$. If any of these solutions $\left\{f_{i}(x)\right\}_{i=1}^{n}$ satisfy $f_{i} \in \mathcal{C}^{\infty}(\mathbb{R})$ for all $i$, then it is possible to find matrix solutions $\left\{A_{i}(x)\right\}_{i=1}^{n}$ of any dimension $n$ by applying our method with an $n \times n$ nilpotent matrix.

This approach demonstrates a surprising connection between calculus and linear algebra. It could also serve as an elementary introduction to the theory of functions of matrices.