Theorems Regarding a Basis of a Vector Space

# Theorems Regarding a Basis of a Vector Space

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 $n ≤ m$.

- 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 $V$. $\blacksquare$