Let G be an abelian group, let $$M_{2}(\mathbb {C})$$ be the algebra of complex $$2\times 2$$ matrices, and let $$\varphi :G\rightarrow G$$ be an endomorphism that need not be involutive. We determine the solutions $$\Phi :G\rightarrow M_{2}(\mathbb {C})$$ of the matrix functional equation $$\begin{aligned} \frac{\Phi (x+y)+\Phi (x+\varphi (y))}{2}=\Phi (x)\Phi (y),\quad x,y\in G. \end{aligned}$$ This enables us to characterize the solutions $$g:G\rightarrow \mathbb {C}^{2}$$ and $$\Phi :G\rightarrow M_{2}(\mathbb {C})$$ of the following functional equation $$\begin{aligned} g(x+y)+g(x+\varphi (y))=2\Phi (y)g(x),\quad x,y\in G, \end{aligned}$$ under the invariant condition $$\Phi \circ \varphi =\Phi $$ .

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达朗贝尔矩阵泛函方程与阿贝尔群上的内同态

令 G 为阿贝尔群，令 $$M_{2}(\mathbb {C})$$ 为复数 $$2\times 2$$ 矩阵的代数，令 $$\varphi :G\rightarrow G$$是一个不需要对合的内同态。我们确定矩阵函数方程 $$\begin{aligned} \frac{\Phi (x+y)+\Phi 的解 $$\Phi :G\rightarrow M_{2}(\mathbb {C})$$ (x+\varphi (y))}{2}=\Phi (x)\Phi (y),\quad x,y\in G. \end{aligned}$$ 这使我们能够表征解决方案 $$g :G\rightarrow \mathbb {C}^{2}$$ 和 $$\Phi :G\rightarrow M_{2}(\mathbb {C})$$ 以下函数方程 $$\begin{aligned} g (x+y)+g(x+\varphi (y))=2\Phi (y)g(x),\quad x,y\in G, \end{aligned}$$ 不变条件下$$\ Phi \circ \varphi =\Phi $$ 。