Sufficient Conditions for a Set of Vectors to be a Basis
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# Sufficient Conditions for a Set of Vectors to be a Basis

Recall from the Dimension of a Vector Space page that if $V$ is a finite-dimensional vector space then the length of any basis of $V$ is the same and we defined that number to be the dimension of said vector space.

Also recall that if $V$ if a finite-dimensional vector space, then $V$ can be spanned by a finite set of vectors $\{ v_1, v_2, ..., v_n \}$. If this set of vectors is also linearly independent in $V$ then we said that $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$.

Sometimes it can be tedious to have to verify that a certain set of vectors both spans $V$ and is linearly independent in $V$ to show that a set is a basis of $V$. If we happen to instead know the dimension of $V$ and that a specific set has $\mathrm{dim} V$ vectors in it, then as we will see with the following two theorems - determining whether a set of vectors is a basis or not requires that we only need to verify one of the two properties of a basis.

 Theorem 1: If $V$ is a finite-dimensional vector space with $\mathrm{dim} (V) = n$ and if $\{ v_1, v_2, ..., v_n \}$ is linearly independent in $V$ then $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$.
• Proof: Let $\{ v_1, v_2, ..., v_n \}$ be a linearly independent set of vectors in $V$ where $\mathrm{dim} (V) = n$. We already know that we can thus extend this set of vectors to a basis of $V$. In doing so, we note that every basis of $V$ must have $\mathrm{dim} V = n$ vectors though, and so, in the extension of $\{ v_1, v_2, ..., v_n \}$ to form a basis - we end up adding no vectors at all. Therefore $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$. $\blacksquare$
 Theorem 2: If $V$ is a finite-dimensional vector space with $\mathrm{dim} (V) = n$ and if $\{ v_1, v_2, ..., v_n \}$ spans $V$ then $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$.
• Proof: Let $\{ v_1, v_2, ..., v_n \}$ span $V$ where $\mathrm{dim} (V) = n$. We already know that we can reduce this set of vectors to a basis of $V$. In doing so, we note that every basis of $V$ must have $\mathrm{dim} V = n$ vectors though, and so, in the reduction of $\{ v_1, v_2, ..., v_n \}$ to form a basis - we end up deleting no vectors at all. Therefore $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$. $\blacksquare$
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