Necessary and Sufficient Conditions for a Basis in Finite-Dimensional Vector Spaces
Theorem1
Let $V$ be a $n$-dimensional vector space. Suppose a subset $S\subset V$ has $n$ elements. A necessary and sufficient condition for $S$ to be a basis of $V$ is that $V = \text{span}(S)$ or $S$ is linearly independent.
Explanation
Vector space, dimension, basis, span, independence - all these fundamental concepts of linear algebra appear here. For a set to be a basis of a vector space, it must be a linearly independent set that spans the vector space. Usually, both conditions must be shown separately, but for sets with a number of elements equal to the dimension, fulfilling one condition implies the fulfilment of the other.
Proof
Since one direction is trivial, proving that $S$ spans $V$ and that $S$ is linearly independent are practically equivalent to the necessary and sufficient condition.
$(\implies)$
This follows trivially from the definition of a basis. If $S$ is a basis of $V$, then $S$ spans $V$ and is linearly independent.
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$(\impliedby)$
If $S$ spans $V$, it is linearly independent
Assume $S$ spans $V$. Now suppose $S$ is not linearly independent. Then some vector $\mathbf{v} \in S$ can be expressed as a linear combination of other vectors in $S$.2 Then, excluding this $\mathbf{v}$ from $S$ still spans the same space. But, a set with fewer elements than the dimension cannot span the vector space. This contradiction implies that the assumption of $S$ being not linearly independent is wrong. Therefore, $S$ is linearly independent.
If $S$ is linearly independent, it spans $V$.
Suppose $S$ is linearly independent. Now suppose $S$ does not span $V$. This means there exists some $\mathbf{v} \in V$ not included in $\text{span}(S)$. Then, adding this vector $\mathbf{v}$ to $S$ still keeps $V$ linearly independent. However, a set with more elements than the dimension is linearly dependent. This contradiction implies that the assumption of $S$ not spanning $V$ is wrong. Therefore, $S$ spans $V$.
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