Vector space/Basis/Characterization/Existence/Section

The following theorem gives an important characterization for a family of vectors to be basis.

Theorem

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. Let ${}v_{1},\ldots ,v_{n}\in V$

be a family of vectors. Then the following statements are equivalent.

1. The family is a basis of ${}V$.
2. The family is a minimal generating system; that is, as soon as we remove one vector ${}v_{i}$, the remaining family is not a generating system any more.
3. For every vector ${}u\in V$, there is exactly one representation

   ${}u=s_{1}v_{1}+\cdots +s_{n}v_{n}\,.$

4. The family is maximally linearly independent; that is, as soon as some vector is added, the family is not linearly independent any more.

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 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

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

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 otherwise 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 ${}u$.

$\Box$

Remark

Let a basis ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ of a ${}K$-vector space ${}V$ be given. Due to fact  (3), this means that for every vector ${}u\in V$, there exists a unique representation (a linear combination)

${}u=s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}\,.$

Here, the uniquely determined elements ${}s_{i}\in K$ (scalars) are called the coordinates of ${}u$ with respect to the given basis. This means that for a given basis, there is a correspondence between vectors and coordinate tuples ${}(s_{1},s_{2},\ldots ,s_{n})\in K^{n}$. We say that a basis determines a linear coordinate system of ${}V$. To paraphrase, a basis gives, in particular, a bijective mapping

$\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\longmapsto s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}.$

The inverse mapping

${\left(\Psi _{\mathfrak {v}}\right)}^{-1}\colon V\longrightarrow K^{n}$

is also called the coordinate mapping.

Theorem

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space with a finite generating system. Then ${}V$ has a finite

basis.

Proof

Let ${}v_{i}$, ${}i\in I$, be a finite generating system of ${}V$ with a finite index set ${}I$. We argue with the characterization from fact  (2). If the family is minimal, then we have a basis. If not, then there exists some ${}k\in I$ such that the remaining family, where ${}v_{k}$ is removed, that is, ${}v_{i}$, ${}i\in I\setminus \{k\}$, is also a generating system. In this case, we can go on with this smaller index set. With this method, we arrive at a subset ${}J\subseteq I$ such that ${}v_{i}$, ${}i\in J$, is a minimal generating set, hence a basis.

$\Box$

Remark

In general, the Theorem of Hamel says that every vector space has a basis. The proof of this theorem uses strong set-theoretical methods, in particular the axiom of choice and the Lemma of Zorn. This is the reason why many statements about finite-dimensional vector spaces pass over to vector spaces of infinite dimension. In this course, we will concentrate on the finite-dimensional case.