Sufficient Conditions for Basis of Finite Dimensional Vector Space

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Theorem

Let $K$ be a division ring.

Let $n \ge 0$ be a natural number.

Let $E$ be an $n$-dimensional vector space over $K$.

Let $B \subseteq E$ be a subset such that $\card B = n$.


The following are equivalent:

$(1): \quad$ $B$ is a basis of $E$.
$(2): \quad$ $B$ is linearly independent.
$(3): \quad$ $B$ is a generator for $E$.


Proof

1 implies 2 and 3

Let $B$ be a basis of $E$.

Then conditions $(2)$ and $(3)$ follow directly by the definition of basis.

$\Box$


2 implies 1

Let $B$ be linearly independent.

Suppose $B$ does not generate $E$.

Then, because $\card B = n$, by Linearly Independent Subset also Independent in Generated Subspace there would be a linearly independent subset of $n + 1$ vectors of $E$.

But this would contradict Size of Linearly Independent Subset is at Most Size of Finite Generator.

Thus condition $(2)$ implies $(1)$.

$\Box$


3 implies 1

Let $B$ be a generator for $E$.

By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, $B$ contains a basis $B'$ of $E$.

But $B'$ has $n$ elements and hence $B' = B$ by Bases of Finitely Generated Vector Space have Equal Cardinality.

Thus condition $(3)$ implies $(1)$.

$\blacksquare$


Examples

Example: $2$ Dimensions

Let $V$ be a vector space of $2$ dimensions.

Let $\mathbf u, \mathbf v \in V$ be vectors

Let $\mathbf u$ and $\mathbf v$ be such that neither $\mathbf u$ nor $\mathbf v$ is a scalar multiple of the other..

Let $\mathbf w \in V$.


Then there exist scalars such that:

$\mathbf w = a \mathbf u + b \mathbf v$


Sources