定义
简单来讲（不严谨）若一个变换 $\mathscr A$ 是线性变换，则对于线性空间中的任意元素 $\alpha$ 和 $\beta$ 它必须满足：
- $\mathscr A\left(\alpha + \beta\right) = \mathscr A\left(\alpha\right) + \mathscr A\left(\beta\right)$
- 对于该数域上的任意值 $k$，有 $\mathscr A\left(k\alpha\right) = k\mathscr A\left(\alpha\right)$
莫比乌斯（Möbius）反演
如果对于数论函数 $f\left(n\right)$ 和 $F\left(n\right)$，有
$$F\left(n\right) = \sum\limits_{d \mid n} f\left(d\right)$$
则将其莫比乌斯反演公式定义为：
$$f\left(n\right) = \sum\limits_{d \mid n}\mu\left(d\right)F\left(\frac nd\right)$$
其中，$\mu$ 为 莫比乌斯（Möbius）函数，具体定义为：
$$ \mu\left(n\right) = \begin{cases} 1 & n = 1\\ \left(-1\right)^k & n\ 无平方数因数，且\ n = p_1p_2\cdots p_k\\ 0 & n\ 有大于\ 1\ 的平方数因数 \end{cases} $$
若要证明莫比乌斯反演公式为线性变换，首先需要证明 $\mathscr M\left(\alpha + \beta\right) = \mathscr M\left(\alpha\right) + \mathscr M\left(\beta\right)$：
$$ \begin{aligned} \mathscr M\left(\alpha\right) = \sum\limits_{d \mid n}\mu\left(\frac nd\right)\alpha\left(d\right)\\ \mathscr M\left(\beta\right) = \sum\limits_{d \mid n}\mu\left(\frac nd\right)\beta\left(d\right) \end{aligned} $$
$$ \begin{aligned} \mathscr M\left(\alpha + \beta\right) &= \sum\limits_{d \mid n}\mu\left(\frac nd\right)\left(\alpha\left(d\right) + \beta\left(d\right)\right)\\ \mathscr M\left(\alpha + \beta\right) &= \sum\limits_{d \mid n}\mu\left(\frac nd\right)\alpha\left(d\right) + \sum\limits_{d \mid n}\mu\left(\frac nd\right)\beta\left(d\right)\\ \mathscr M\left(\alpha + \beta\right) &= \mathscr M\left(\alpha\right) + \mathscr M\left(\beta\right) \end{aligned} $$
证毕。
其次，要满足 $\mathscr M\left(k\alpha\right) = k\mathscr M\left(\alpha\right)$
$$ \begin{aligned} \mathscr M\left(k\alpha\right) &= \sum\limits_{d \mid n}k\mu\left(\frac nd\right)\alpha\left(d\right)\\ \mathscr M\left(k\alpha\right) &= k\sum\limits_{d \mid n}\mu\left(\frac nd\right)\alpha\left(d\right)\\ \mathscr M\left(k\alpha\right) &= kM\left(\alpha\right) \end{aligned} $$
证毕。
两条性质都满足。因此，莫比乌斯反演是线性变换。
二项式反演
如果对于函数 $f\left(n\right)$ 和 $g\left(n\right)$ 有
$$ F\left(n\right) = \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni f\left(i\right) $$
那么就有
$$ f\left(n\right) = \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni F\left(i\right) $$
若要证明二项式反演公式为线性变换，首先需要证明 $\mathscr B\left(f_1 + f_2\right) = \mathscr B\left(f_1\right) + \mathscr B\left(f_2\right)$
$$ \begin{aligned} \mathscr B\left(f_1\right) = \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni f_1\left(i\right)\\ \mathscr B\left(f_2\right) = \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni f_2\left(i\right) \end{aligned} $$
$$ \begin{aligned} \mathscr B\left(f_1 + f_2\right) &= \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni \left(f_1\left(i\right) + f_2\left(i\right)\right)\\ \mathscr B\left(f_1 + f_2\right) &= \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni f_1\left(i\right) + \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni f_2\left(i\right)\\ \mathscr B\left(f_1 + f_2\right) &= \mathscr B\left(f_1\right) + \mathscr B\left(f_2\right) \end{aligned} $$
证毕
其次，要满足 $\mathscr B\left(kf\right) = k\mathscr B\left(f\right)$
$$ \begin{aligned} \mathscr B\left(kf\right) &= \sum\limits_{i = 0}^n \left(-1\right)^i\binom ni kf\left(i\right)\\ \mathscr B\left(kf\right) &= k\sum\limits_{i = 0}^n \left(-1\right)^i\binom ni f\left(i\right)\\ \mathscr B\left(kf\right) &= k\mathscr B\left(f\right) \end{aligned} $$
证毕。
两条性质都满足。因此，二项式反演是线性变换。