Sufficient Conditions for a Set of Vectors to be a Basis

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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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Sufficient Conditions for a Set of Vectors to be a Basis