Linear Independence and Bases

In this article, I cover linear independence and the Linear Independence Lemma. After that, I discuss bases and finite-dimensional vector spaces. Several examples and corollaries are presented.

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 $\mathop{span}(S)=P_3.$

David A. Smith at Dave4Math

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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