《线性代数应该这样学》课内验证2-有限维向量空间

CHAPTER 2 Finite-Dimensional Vector Spaces

If some vectors are removed from a linearly independent list, the remaining
list is also linearly independent, as you should verify.

Proof Suppose $v_1,\cdots ,v_m, \cdots ,v_n$ is linearly independent, then the only choice of $a_1, \cdots, a_n\in \mathbb{F}$ that makes $a_1 v_1 +\cdots +a_n v_n=0$ is $a_1=\cdots = a_n=0$ .

Since the order of vectors does not affect linear independence, we put all the vectors that need to be removed on the right.

Now assume the vectors $v_{m+1}, \cdots ,v_n$ are removed, and the remaining $v_1, \cdots, v_m$ is linearly dependent.

Hence there exist $a_1,\cdots ,a_m \in \mathbb{F}$ , not all $0$ , such that $a_1 v_1 +\cdots + a_m v_m=0$ .

Let $a_{m+1},\cdots,a_n$ all are $0$ , hence $a_1 v_1+\cdots+a_m v_m+\cdots+a_nv_n=0$ , at the same time $a_1,\cdots,a_n$ , not all $0$ . We get a contradiction.

Thus the remaining lise is also linearly independent.

《线性代数应该这样学》课内验证2-有限维向量空间

CHAPTER 2 Finite-Dimensional Vector Spaces

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