Theorems Regarding a Basis of a Vector Space

Table of Contents

Recall from the Basis of a Vector Space page the definition of a basis:

Definition: A set of vectors

\{ v_1, v_2, ..., v_n \}

is said to be a Basis of the

\mathbb{F}

-vector space

$V$

if both

V = \mathrm{span} (v_1, v_2, ..., v_n)

and

\{v_1, v_2, ..., v_n \}

is a linearly independent set.

We will now look at some important theorems regarding a basis of a vector space.

Theorem 1: Every spanning set

B = \{ v_1, v_2, ..., v_n \}

of a vector space

$V$

can be reduced (by throwing away certain vectors in

$B$

) to a basis of

$V$

Proof: We will prove this by successively deleting vectors from $$B$$ until $$B$$ becomes linearly independent which preserving the fact that $V = \mathrm{span} B$ .

Let $\{ v_1, v_2, ..., v_n \}$ be a spanning set of vectors of $$V$$ , that is for every $v \in V$ we have that $$v = a_1v_1 + a_2v_2 + ... + a_nv_n$$ .

Step 1: First, if $$v_1 = 0$$ , then delete $$v_1$$ from $$B$$ . If $v_1 \neq 0$ , then keep $$v_1$$ . In either case, $$B$$ will still be a spanning set of $$V$$ .

Step 2: Secondly, if $$v_2$$ is a linear combination of $$v_1$$ , that is if $$v_2$$ is a scalar multiple of $$v_1$$ so that $$v_2 = av_1$$ for some $a \in \mathbb{F}$ , then delete $$v_2$$ from $$B$$ . If $v_2 \neq av_1$ then keep $$v_2$$ . Once again, $$B$$ is still a spanning set of $$V$$ .

Step j: If $$v_j$$ is a linear combination of the vectors $v_1, v_2, ..., v_{j-1}$ , then delete $$v_j$$ from $$B$$ . If $$v_j$$ is not a linear combination of the vectors $v_1, v_2, ..., v_{j-1}$ then keep $$v_j$$ .

After $$n$$ many steps, $$B$$ will have become a linearly independent set that spans $$V$$ and so $$B$$ will have been reduced to a basis of $$V$$ . $\blacksquare$

Corollary 1: Every finite dimensional vector space

$V$

has a basis.

Proof: Since $$V$$ is a finite dimensional vector space, it follows that for some finite set of vectors $\{ v_1, v_2, ..., v_n \}$ that $V = \mathrm{span} (v_1, v_2, ..., v_n)$ . This set of vectors $\{ v_1, v_2, ..., v_n \}$ be can reduced to a linearly independent set $$B$$ from theorem 2 to become a basis, and so $$V$$ has a basis. $\blacksquare$

Theorem 2: Every linearly independent set of vectors

B = \{ v_1, v_2, ..., v_n \}

from a finite dimensional vector space

$V$

can be extended to a basis of

$V$

Proof: To turn the linearly independent set of vectors $$B$$ into a basis for the finite dimensional vector space $$V$$ we must show that by adding vectors to $$B$$ we can ensure that $$B$$ spans $$V$$ while preserving the linear independence of the set of vectors in $$B$$ .

Recall since $$V$$ is finite dimensional, $$V$$ can be spanned by a finite set of vectors $\{ w_1, w_2, ..., w_m \}$ , and we should recall that the size of any linearly independent set of vectors is less than or equal to the size of any spanning set of $$V$$ , that is $$n \leq m$$ .

Step 1: If $$w_1$$ is a linear combination of the vectors $$v_1, v_2, ..., v_n$$ then $$B$$ stays the same. If $$w_1$$ is not a linear combination of the vectors $$v_1, v_2, ..., v_n$$ then add $$w_1$$ to $$B$$ so that $B = \{ v_1, v_2, ..., v_n, w_1 \}$ .

Step j: If $$w_j$$ is a linear combination of the vectors $$v_1, v_2, ..., v_n$$ and the set of vectors $$w_i$$ where $i \in I$ ( $$I$$ being the indices of the vectors added to $$B$$ ), then keep $$B$$ the same. Otherwise add $$w_j$$ to $$v_1, v_2, ..., v_n, w_i, ...$$ so that $B = \{ v_1, v_2, ..., v_n, ..., w_j \}$ .

After at most $$m$$ many steps, $$B$$ will still be a linearly independent set and will also span $$V$$ , and so $$B$$ will be a basis of $$V$$ . $\blacksquare$

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