例题 4.1

Question

设
$β_{1} β_{2} β_{n} = (1, 1, \dots, 1, 1), = (0, 1, \dots, 1, 1), ⋮ = (0, 0, \dots, 0, 1) \in R^{n} .$

则它们线性无关, 对任意向量 $α = (a_{1}, a_{2}, \dots, a_{n}) \in R^{n}$ , 有

$α = a_{1} β_{1} + (a_{2} - a_{1}) β_{2} + \dots + (a_{n} - a_{n - 1}) β_{n} .$
所以 $β_{1}, β_{2}, \dots, β_{n}$ 也是 $R^{n}$ 的一组基, 且 $(a_{1}, a_{2} - a_{1}, \dots, a_{n} - a_{n - 1})$ 就是 $α$ 在基 $β_{1}, β_{2}, \dots, β_{n}$ 下的坐标.

事实上, 若设 $x_{1} β_{1} + x_{2} β_{2} + \dots + x_{n} β_{n} = 0$ , 则有

(x_{1}, x_{1} + x_{2}, \dots, x_{1} + x_{2} + \dots + x_{n}) = (0, 0, \dots, 0) .

由此可得 $x_{1} = 0, x_{2} = 0, \dots, x_{n} = 0$ , 故向量组 $β_{1}, β_{2}, \dots, β_{n}$ 线性无关.

下证任意 $α = (a_{1}, a_{2}, \dots, a_{n}) \in R^{n}$ 可由 $β_{1}, β_{2}, \dots, β_{n}$ 线性表出. 若设 $α = k_{1} β_{1} + k_{2} β_{2} + \dots + k_{n} β_{n}$ ,于是

(a_{1}, a_{2}, \dots, a_{n}) = (k_{1}, k_{1} + k_{2}, \dots, k_{1} + k_{2} + \dots + k_{n}),

则有线性方程组

⎩ ⎨ ⎧ k_{1} k_{1} + k_{2} \dots\dots k_{1} + k_{2} + \dots + k_{n} = a_{1}, = a_{2}, \dots = a_{n} .

解得 $k_{1} = a_{1}, k_{2} = a_{2} - a_{1}, \dots, k_{n} = a_{n} - a_{n - 1}$ ,故

α = a_{1} β_{1} + (a_{2} - a_{1}) β_{2} + \dots + (a_{n} - a_{n - 1}) β_{n} .

所以 $β_{1}, β_{2}, \dots, β_{n}$ 是 $R^{n}$ 的一组基,且 $(a_{1}, a_{2} - a_{1}, \dots, a_{n} - a_{n - 1})$ 就是 $α$ 在这组基下的坐标.

Youliang Zhong

例题 4.1

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