Basis Properties and Results

Every spanning list contains a basis

A spanning list in a vector space may not be a basis because it is not linearly independent. Our next result says that given any spanning list, some (possibly none) vectors in it can be discarded so that the remaining list is linearly independent and still spans the vector space.

Every spanning list contains a basis.

Every spanning list in a vector space can be reduced to a basis of the vector space.

Proof. Suppose $v_1,\dots ,v_n$ spans $V$. We want to remove some of the vectors from $v_1,\dots ,v_n$ so that the remaining vectors form a basis of $V$. We do this through the process described below.

Start with $B$ equal to the list $v_1,\dots ,v_m$.

Step 1: If $v_1=0$, then delete $v_1$ from $B$. If $v_1\ne 0$, then leave $B$ unchanged.

Step k: If $v_k$ is in $\text{span}(v_1,\dots ,v_{k-1})$, then delete $v_k$ from the list $B$. If $v_k$ is not in $\text{span}(v_1,\dots ,v_{k-1})$, then leave $B$ unchanged.

Stop the process after step $n$, getting a list $B$. This list $B$ spans $V$ because our original list spanned $V$ and we have discarded only vectors that were in the span of the previous vectors. The process ensures that no vector in $B$ is in the span of the previous ones. Thus $B$ is linearly independent, by the linear dependence lemma. Hence $B$ is a basis of $V$. $■$

Basis of Finite-Dimensional Vector Space

The above leads us to this important result:

Basis of Finite-Dimensional Vector Space

Every finite-dimensional vector space has a basis.

Proof. By definition, a finite-dimensional vector space has a spanning list. We just saw that each spanning list can be reduced to a basis.

Every linearly independent list extends to a basis

Every linearly independent list extends to a basis

Every linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space.

Proof. Suppose $u_1,\dots ,u_m$ is linearly independent in a finite-dimensional vector space $V$. Let $w_1,\dots ,w_n$ be a list of vectors in $V$ that spans $V$. Thus the list

$$u_1,\dots ,u_m,w_1,\dots ,w_n$$

spans $V$. Applying the procedure from the above result every spanning list contains a basis to reduce this list to a basis of $V$ produces a basis consisting of the vectors $u_1,\dots ,u_m$ and some $w$‘s. (None of the $u$‘s get deleted in this procedure since $u_1,\dots ,u_m$ is linearly dependent).

As an example in ${\mathbb{F}}^3$, suppose we start with the linearly independent list $(2,3,4),(9,6,8)$. If we take $w_1,w_2,w_3$ to be the standard basis of ${\mathbb{F}}^3$, then applying the produces in the proof above produces the list

$$(2,3,4),(9,6,8),(0,1,0)$$

which is a basis of ${\mathbb{F}}^3$.

Every Subspace of $V$ is a part of a direct sum equal to $V$

As an application of the result above, we now show that every subspace of a finite-dimensional vector space can be paired with another subspace to form a direct sum of the whole space.

Every subspace of $V$ is part of a new direct sum equal to $V$

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then, there is a subspace $W$ of $V$ such that $V=U\oplus W$.

Proof. Because $V$ is finite-dimensional, so is $U$. Thus there is a basis $u_1,\dots ,u_m$ of $U$. Of course $u_1,\dots ,u_m$ is a linearly independent list of vectors in $V$. Hence this list can be extended to a basis $u_1,\dots ,u_m,w_1,\dots ,w_n$ of $V$. Let $W=\text{span}(w_1,\dots ,w_n)$.

To prove that $V=U\oplus W$, by the condition for direct sum of two subspaces,we only need to show that

$$V=U+W\text{and}U\cap W=\{0\}$$

To prove the first equation, suppose $v\in V$. Then, because the list $u_1,\dots u_m,w_1,\dots ,w_n$ spans $V$, there exist $a_1,\dots ,a_m,b_1,\dots ,b_n\in \mathbb{F}$ such that

$$v=\underset{u}{\underbrace{a_1{u}_1+\cdots +a_m{u}_m}}+\underset{w}{\underbrace{b_1{w}_1+\cdots +b_n{w}_n}}$$

We have $v=u+w$, where $u\in U$ and $w\in W$ are defined as above. Thus $v\in U+W$, completing the proof that $V=U+W$.

To show that $U\cap W=\{0\}$, suppose $v\in U\cap W$. Then there exist scalars $a_1,\dots ,a_m,b_1,\dots ,b_m\in \mathbb{F}$ such that

$$\begin{array}{rl}v& =a_1{u}_1+\cdots +a_m{u}_m\\ v& =b_1{w}_1+\cdots +b_n{w}_n\end{array}$$

Subtracting the two equations from each other:

$$a_1{u}_1+\cdots +a_m{u}_m-b_1{w}_1-\cdots -b_n{w}_n=0$$

Because the list $u_1,\dots ,u_m,w_1,\dots ,w_n$ is linearly independent, this implies that

$$a_1=\cdots =a_m=b_1=\cdots =b_n=0$$

Thus $v=0$, completing the proof that $U\cap W=\{0\}$.