Vector space/Characterizations of basis/Maximal/Minimal/Fact/Proof

Proof

Proof by ring closure. $(1) \Rightarrow (2)$ . The family is a generating system. Let us remove a vector, say $v_{1}$ , from the family. We have to show that the remaining family, that is $v_{2}, \dots, v_{n}$ , is not a generating system anymore. So suppose that it is still a generating system. Then, in particular, $v_{1}$ can be written as a linear combination of the remaining vectors, and we have

v_{1} = \sum_{i = 2}^{n} s_{i} v_{i} .

But then

v_{1} - \sum_{i = 2}^{n} s_{i} v_{i} = 0

is a nontrivial representation of $0$ , contradicting the linear independence of the family. $(2) \Rightarrow (3)$ . Due to the condition, the family is a generating system, hence every vector can be represented as a linear combination. Suppose that for some $u \in V$ , there is more than one representation, say

u = \sum_{i = 1}^{n} s_{i} v_{i} = \sum_{i = 1}^{n} t_{i} v_{i},

where at least one coefficient is different. Without loss of generality, we may assume $s_{1} \neq t_{1}$ . Then we get the relation

(s_{1} - t_{1}) v_{1} = \sum_{i = 2}^{n} (t_{i} - s_{i}) v_{i}

Because of $s_{1} - t_{1} \neq 0$ , we can divide by this number and obtain a representation of $v_{1}$ using the other vectors. In this situation, due to exercise, also the family without $v_{1}$ is a generating system of $V$ , contradicting the minimality. $(3) \Rightarrow (4)$ . Because of the unique representability, the zero vector has only the trivial representation. This means that the vectors are linearly independent. If we add a vector $u$ , then it has a representation

u = \sum_{i = 1}^{n} s_{i} v_{i},

and, therefore,

0 = u - \sum_{i = 1}^{n} s_{i} v_{i}

is a non-trivial representation of $0$ , so that the extended family $u, v_{1}, \dots, v_{n}$ is not linearly independent. $(4) \Rightarrow (1)$ . The family is linearly independent, we have to show that it is also a generating system. Let $u \in V$ . Due to the condition, the family $u, v_{1}, \dots, v_{n}$ is not linearly independent. This means that there exists a non-trivial representation

0 = s u + \sum_{i = 1}^{n} s_{i} v_{i} .

Here $s \neq 0$ , because otherwise this would be a non-trivial representation of $0$ with the original family $v_{1}, \dots, v_{n}$ . Hence, we can write

u = - \sum_{i = 1}^{n} \frac{s_{i}}{s} v_{i},

yielding a representation for $u$ .

To fact