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 ${}v_{2},\ldots ,v_{n}$, 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 non-trivial 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 presented 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

${}{\left(s_{1}-t_{1}\right)}v_{1}=\sum _{i=2}^{n}{\left(t_{i}-s_{i}\right)}v_{i}\,$

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 ${}u,v_{1},\ldots ,v_{n}$ 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 ${}u,v_{1},\ldots ,v_{n}$ is not linearly independent. This means that there exists a non-trivial representation

${}0=su+\sum _{i=1}^{n}s_{i}v_{i}\,.$

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

${}u=-\sum _{i=1}^{n}{\frac {s_{i}}{s}}v_{i}\,,$

yielding a representation for ${}u$.

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