## You are here

Homeproperties of linear independence

## Primary tabs

# properties of linear independence

Let $V$ be a vector space over a field $k$. Below are some basic properties of linear independence.

1. $S\subseteq V$ is never linearly independent if $0\in S$.

###### Proof.

Since $1\cdot 0=0$. ∎

2. If $S$ is linearly independent, so is any subset of $S$. As a result, if $S$ and $T$ are linearly independent, so is $S\cap T$. In addition, $\varnothing$ is linearly independent, its spanning set being the singleton consisting of the zero vector $0$.

###### Proof.

If $r_{1}v_{1}+\cdots r_{n}v_{n}=0$, where $v_{i}\in T$, then $v_{i}\in S$, so $r_{i}=0$ for all $i=1,\ldots,n$. ∎

3. If $S_{1}\subseteq S_{2}\subseteq\cdots$ is a chain of linearly independent subsets of $V$, so is their union.

###### Proof.

Let $S$ be the union. If $r_{1}v_{1}+\cdots r_{n}v_{n}=0$, then $v_{i}\in S_{{a(i)}}$, for each $i$. Pick the largest $S_{{a(i)}}$ so that all $v_{i}$’s are in it. Since this set is linearly independent, $r_{i}=0$ for all $i$. ∎

4. $S$ is a basis for $V$ iff $S$ is a maximal linear independent subset of $V$. Here, maximal means that any proper superset of $S$ is linearly dependent.

###### Proof.

If $S$ is a basis for $V$, then it is linearly independent and spans $V$. If we take any vector $v\notin S$, then $v$ can be expressed as a linear combination of elements in $S$, so that $S\cup\{v\}$ is no longer linearly independent, for the coefficient in front of $v$ is non-zero. Therefore, $S$ is maximal.

Conversely, suppose $S$ is a maximal linearly independent set in $V$. Let $W$ be the span of $S$. If $W\neq V$, pick an element $v\in V-W$. Suppose $0=r_{1}v_{1}+\cdots r_{n}v_{n}+rv$, where $v_{i}\in S$, then $-rv=r_{1}v_{1}+\cdots+r_{n}v_{n}$. If $r\neq 0$, then $v$ would be in the span of $S$, contradicting the assumption. So $r=0$, and as a result, $r_{i}=0$, since $S$ is linearly independent. This shows that $S\cup\{v\}$ is linearly independent, which is impossible since $S$ is assumed to be maximal. Therefore, $W=V$. ∎

Remark. All of the properties above can be generalized to modules over rings, except the last one, where the implication is only one-sided: basis implying maximal linear independence.

## Mathematics Subject Classification

15A03*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections