\[ \newcommand{R}{\mathbb{R}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad} \]

Linear dependence

Let \(X\) be a vector space, and \(S \subset X\) any subset of \(X\).


A linear combination of vectors \(u_1, \ldots, u_n\) is a finite sum \[ \sum_{j=1}^n a_j u_j, \] where \(a_1, \ldots, a_n\) are scalars. The (linear) span of \(S \subset X\) is the set of all linear combinations of vectors in \(S\): \[ \mathrm {span}(S) \stackrel{\text{def.}}{=} \big\{ \sum_\text{finite} a_j x_j \colon x_j \in S, a_j \text{ scalars}\big\}. \] For convenience, we define \(\mathrm{span}(\emptyset) \stackrel{\text{def.}}{=} \{\mathbf 0\}\). If \(V = \mathrm{span} (S)\) we say that \(S\) generates \(V\).

› The linear span of a set S is the smallest subspace containing S

For any \(S \subset X\), \(\mathrm{span}(S)\) is a subspace of \(X\), and \[\mathrm{span}(S) = \bigcap_{S \subset V} \{ V \colon V \text{ is a subspace of } X\}.\]

  • Let \(x = (1,0)\), \(y = (2,0)\) and \(z = (1,1)\) be vectors in \(\R^2\). Then \[\mathrm{span}\{x\} = \mathrm{span}\{y\} = \mathrm{span}\{x,y\} = \{(\lambda,0) \colon \lambda \in \R\},\] \[\mathrm{span} \{z\} = \{ (\lambda,\lambda) \colon \lambda \in \R\}, \] \[ \mathrm{span}\{x,z\} = \mathrm{span}\{y,z\} = \mathrm{span}\{x,y,z\} = \R^2.\]
  • The vectors \(e_1 = (1,0, \ldots)\), \(e_2 = (0,1,0, \ldots)\), \(\ldots\), \(e_n = (0,\ldots,0,1)\) generate \(\R^n\).
  • In general, the span of a set differs between real and complex vector spaces: \[ \mathrm{span}_{\R}\{1\} = \R \quad\text{ but }\quad \mathrm{span}_{\mathbb C}\{1\} = \mathbb C. \]

Linear dependence

A family of vectors \(u_1, u_2, \ldots \) is called linearly dependent if one of them is linear combination of some of the others: \[ \{ u_1, u_2, \ldots \} \text{ linearly dependent} \defarrow \sum_{j=1}^n a_j u_j = \mathbf 0 \quad\text{ for some } n \in \mathbb N \:\text{ and at least one }\: a_j \neq 0. \] Else, the family is linearly independent: \[ \{ u_1, u_2, \ldots \} \text{ linearly independent} \defarrow \big[ \text{for any } n \in \mathbb N: \quad \sum_{j=1}^n a_j u_j = \mathbf 0 \: \Longrightarrow\: a_j = 0 \:\forall\: j\big]. \] More generally, a set \(S\) is linearly independent if all finite subsets of it are linearly independent.

  • The vectors \(x= (1,0)\), \(y= (2,0)\) and \(z=(1,1)\) are linearly dependent in \(\R^2\), since \[ \begin{bmatrix}2\\0\end{bmatrix} = 2\begin{bmatrix}1\\0\end{bmatrix}.\] But both the sets \(\{x,z\}\) and \(\{y,z\}\) are linearly independent, since \[a_1 x + a_2 z = \mathbf 0 \quad\Longleftrightarrow\quad a_1 \begin{bmatrix}1\\0\end{bmatrix} + a_2 \begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \quad\Longleftrightarrow\quad \begin{bmatrix}a_1 + a_2\\a_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} \quad\Longleftrightarrow\quad a_1 = a_2 = 0, \] and similarly for \(\{y,z\}\).
  • If \(\mathbf 0 \in S\), then \(S\) is linearly dependent.
  • \(\{1,x,x^2, \ldots\}\) is linearly independent in \(P(\R)\).
  • \(\{1,\cos(x), \sin(x), \cos(2x), \sin(2x), \ldots\}\) is linearly independent in \(C(I,\R)\).
2017-03-24, Hallvard Norheim Bø