In this article, I explain what linear independence is, what spanning sets are, and explain what bases are. In the end, I discuss finite-dimensional spaces.

## Linear Independence

Let $\mathbb{V}$ be a linear save over a field $\mathbb{F}.$ A subset $S$ of $\mathbb{V}$ is said to be ** linearly dependent** if there exist distinct $v_1, v_2, \ldots, v_n$ in $S$ and scalars $a_1, a_2, \ldots, a_n$ in $\mathbb{F}$, not all zero, such that $a_1 v_1 + a_2 v_2 + \cdots + a_n v_n {a}{v} = 0.$ If a set $S$ contains only finitely many $v_i$ we sometimes say that $v_1, v_2, \ldots, v_n$ are dependent.

**Definition**. A list of vectors $(v_1, v_2, \ldots, v_m)$ in $\mathbb{V}$ is called ** linearly independent** if the only choice of $a_1, a_2, \ldots, a_m\in k$ for $a_1 v_1 + a_2 v_2 + \cdots + a_n v_n +a_m v_m$ is $a_1=\cdots=a_m=0.$ A list of vectors that are not linearly independent is called

**.**

*linearly dependent*Notice from the definition we can conclude that any set which contains a linearly dependent set is linearly dependent. Any subset of a linearly independent set is linearly independent. Any set which contains the zero vector is linearly dependent. A set $S$ of vectors is linearly independent if and only if each finite subset of $S$ is linearly independent.

**Lemma**. (** Linear Dependence**) If $(v_1, v_2, \ldots, v_m)$ is linearly dependent in $\mathbb{V}$ and $v_1\neq 0$ then there exists $j\in {2,\ldots,m}$ such that the following hold: $v_j\in\mathop{span}(v_1,\ldots,v_{j-1})$ and if the $j^{th}$ term is removed from $(v_1, v_2, \ldots, v_m)$, the span of the remaining list equals $\mathop{span}(v_1, v_2, \ldots, v_m).$

**Example**. Show that the following subset of $M_{2\times 2}$ is a linear dependent set. \begin{equation} \label{lindeexample} \left\{ \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix}, \begin{bmatrix} -11 & 3 \\ -2 & 2 \end{bmatrix}, \begin{bmatrix} 6 & -1 \\ 3 & 4 \end{bmatrix}, \begin{bmatrix} -1 & 0\\ 2 & 2 \end{bmatrix} \right\} \end{equation}

**Example**. Suppose that $S$ is the subset $$ S=\{2x^3-x+3, 3x^3+2x-2, x^3-4x+8, 4x^3+5x-7\} $$ of $P_3.$

1) Show that $S$ is linear dependent.

2) Show that every three element subset of $S$ is linear dependent.

3) Show that every two element subset of $S$ is linear independent.

**Example**. Show that no linear independent set can contain zero.

**Lemma**. (** Linear Independence Lemma**) Let $S$ be a linearly independent subset of a vector space $\mathbb{V}.$ Suppose $v$ is a vector in $\mathbb{V}$ which is not in the subspace spanned by $S.$ Then the set obtained by adjoining $v$ to $S$ is linearly independent.

**Proof**. Suppose $u_1,\ldots,u_n$ are distinct vectors in $S$ and that $a_1 u_1+\cdots + a_n u_n+a v =0.$ Then $a=0;$ for otherwise, $$ v=\left (-\frac{a_1}{a}\right ) u_1+\cdots + \left (-\frac{a_n}{a} \right )u_n $$ and $v$ is in the subspace spanned by $S.$ Thus $a_1 u_1+\cdots + a_n u_n=0$, and since $S$ is a linearly independent set each $a_i=0.$

**Theorem**. If $S$ is a set of vectors in a vector space $\mathbb{V}$ over a field $F$ then the following are equivalent.

1) $S$ is linearly independent and spans $\mathbb{V}$

2) For every vector $v\in V$, there is a unique set of vectors $v_1,\ldots,v_n$ in $S$, along with a unique set of scalars in $\mathbb{F}$ for which $v=a_1 v_1+\cdots+a_n v_n$

3) $S$ is a minimal spanning set in the sense that $S$ spans $\mathbb{V}$, and any proper subset of $S$ does not span $\mathbb{V}$

4) $S$ is a maximal linearly independent set in the sense that $S$ is linearly independent, but any proper superset of $S$ is not linearly independent.

**Proof**. Then $S$ is a spanning set. If some proper subset $S’$ of $S$ also spanned $\mathbb{V}$, than any vector in $S-S;$ would be a linear combination of the vectors in $S’.$ contradicting the fact that the vectors in $S$ are linearly independent. Conversely, if $S$ is a minimal spanning set, then it must be linearly independent. For if not, some vector $s\in S$ would be a linear of the other vectors in $S$, and so $S-S’$ would be a proper spanning subset of $S$, which is not possible.

Then $S$ is linearly independent. If $S$ were not maximal, there would be a vector $v\in V-S$ for which the set $S\cup {v}$ is linear independent. But then $\mathbb{V}$ is not in the span of $S$, contradicting the fact that $S$ is a spanning set. Hence, $S$ is a maximal linearly independent set. Conversely, if $S$ is a maximal independent set,. then it must span $\mathbb{V}$, for if not, we could find a vector $v\in S-S’$ that is not a linear combination of the vectors in $S.$ Hence, $S\cup {v}$ would be a contradiction.

**Definition**. A ** basis** of $\mathbb{V}$ is a list of vectors in $\mathbb{V}$ that is linearly independent and spans $\mathbb{V}.$

**Example**. Find a basis for the space of all $2\times 2$ matrices $S$ such that $$ \begin{bmatrix} 1 & 1 \ 1 & 1\end{bmatrix}S=S. $$ Let $S=\begin{bmatrix} a & b \ c & d \end{bmatrix}.$ Then $\begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix}=\begin{bmatrix} a & b \ c & d \end{bmatrix} $, meaning $$ \begin{bmatrix} a+c & b+d \ a+c & b+d \end{bmatrix}=\begin{bmatrix} a & b \ c & d \end{bmatrix}.$$ So $a+c=a, b+d=b, a+c=c,$ and $b+d=d.$ These imply, respectively that $a=b=c=d=0.$

**Corollary**. A list $(v_1,\ldots,v_m)$ of vectors in $\mathbb{V}$ is a basis of $\mathbb{V}$ if and only if every $v\in V$ can be written uniquely in the form $v=a_1 v_1+\cdots+a_m v_m$ where $a_1, a_2, \ldots, a_m \in k.$

**Corollary**. Any $n$ linearly independent vectors in a linear space $\mathbb{V}$ of dimension $n$ constitute a basis for $\mathbb{V}.$

**Example**. Show that the space of all $m\times n$ matrices over a field $\mathbb{F}$ has dimension $mn.$

**Corollary**. Any finitely generated linear space, generated by a asset of nonzero vectors, has a basis.

The zero linear space has no basis, because any subset contains the zero vector and must be linearly dependent.

**Example**. Show that $$ B=\{[2,3,0,-1],[-1,1,1,-1],[3,2,-1,0],[2,3,0,-1]\} $$ is a maximal linearly independent subset of $$ S=\{[1,4,1,-2],[-1,1,1,-1],[3,2,-1,0],[2,3,0,-1]\}. $$ Determine $\dim\mathop{span}(S)$ and determine whether or not $\mathop{span}(S)=P_3.$