>[!summary] Linear independence is another way to determine whether the span of a set of vectors is reduced to its lowest size. > If a set is <u>reduced</u> to its fullest, then it is called **linearly independent**. If a set is <u>not reduced</u> to its fullest, then it is called **linearly dependent.** > **Key equations:** > If $c_1\vec{v_1} + \dots + c_k \vec{v_k} = \vec{0}$ is satisfied and some but not all $c$ scalars are zero, then the set is linearly dependent > If $c_1\vec{v_1} + \dots + c_k \vec{v_k} = \vec{0}$ is only satisfied if all $c$ scalars are zero, then the set is linearly independent. >[!info]+ Read Time **⏱ 4 mins** # Definition A [[Sets|set of vectors]] is **linearly independent** if the sum of all vectors points back to the origin ([[Zero Vector|zero vector]]) **only if the coefficients of each vector are 0.** A [[Sets|set of vectors]] is **linearly dependent** if the addition of some or all the vectors and their coefficients points back to the origin. Geometrically, linear dependent sets will always add to the origin ([[Zero Vector|zero vector]]) while linear independent sets of vectors only add to the origin if they are all the vectors are multiplied by 0. > [!note]+ Case Diagrams of Linear Independence ![[li_2.png]] Case 1: The set containing $\vec{a},\vec{b},\vec{c}$ is linearly dependent since the vectors added together with their coefficients add back to the origin. > Case 2: The set containing $\vec{e},\vec{f},\vec{g}$ is linearly independent since the vectors cannot be added together to add back to the origin. The only way this set of vectors could add to the origin is if all the coefficients were 0. # Derivation To derive a general expression for determining linear independence and dependence is through a [[!! Reducing Span|reducing span]] example. A span can be reduced if one of the vectors is linearly dependent. Explained more below > [!info]+ >The example below is adapted from an example in An Introduction To Linear Algebra For Science and Engineering by Norman, D., & Wolczuk, D. The solution is adapted and is my original interpretation of the steps. Take, for example, this span of 3 vectors in $\mathbb{R^3}$ $Z = \left\{\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\right\}$ ![[Li_1.png|300]] >[!warning] Linear Dependence Assumptions > To determine whether a vector is a linear dependent of another, observe wether the addition of vectors and their scalar multiples adds to zero. Where one or more, **but not all,** scalars are zero, so that the addition of the vectors points back to the origin ([[Zero Vector|zero vector]]. > Mathematically, this is described by the [[Span|vector equation]] $\vec{x} = c_1v_1+ \dots +c_kv_k, \quad ,c_1\dots c_k \in \mathbb{R}$, where $\vec{{x}}=\vec{0}$ so that $\vec{0} = c_1v_1+ \dots +c_kv_k, \quad ,c_1\dots c_k \in \mathbb{R}$ > If the equation is satisfied, this means that the addition of vectors points back to the origin ([[Zero Vector|zero vector]]). This means that one or more vectors are a scalar multiple of other vectors in a [[Sets|set]] and the [[!! Reducing Span|span can be reduced]] > If we cannot find any addition of vectors and their scalar that add to zero, or all scalars have to be zero for it to point to the [[Zero Vector|zero vector]], this means that the [[Span|span]] is already reduced to the most. $ \begin{array}{c} \text{We start with the vector equation} \\ \\ \vec{x} = c_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \\ \\ \text{We want to see wether the vector equation ever points to zero} \\\\ \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \\ \text{Use linear combination and solve for $c_1,c_2,c_3$} \\ \\ \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} c_1 + c_3 \\ c_2 + c_3 \\ c_1 + c_3 \end{bmatrix} \\ \\ c_1 + c_3 = 0\space (1), \quad c_2 + c_3 = 0 \space (2), \quad c_1 +c_2 = 0 \space (3)\\ \\ (1) - (3) \\ c_3 - c_2 = 0 \\ c_3 = c_2 \\ \\ (2) \\ 2c_3 = 0 \\ c_3 = 0 \\\\ \text{Implies that $c_2 = 0$ and $c_1 = 0 $ } \end{array}$ Because $c_1 = c_2 = c_3 = 0$ is the only solution so the set is **linearly independent** If $c_1 =0$ and $c_2, c_3 \neq 0$ and the solution reduced to zero, the set would be **linearly** **dependent**, and **we could reduce the set of vectors**. ## Generalization The set is linearly dependent (formally, a non-trivial solution) if there exists a real coefficient $c_1 \dots c_k$ not all zero such that $c_1\vec{v_1} + \dots + c_k \vec{v_k} = \vec{0}$ The set is linearly independent (cannot be reduced the set) if the only solution to $c_1\vec{v_1} + \dots + c_k \vec{v_k} = \vec{0}$ is $c_1 = \cdots = c_k=0$ (Formally **Trivial solution**) --- > 🧠 Enjoy this walkthrough? [Support Math & Matter](https://github.com/rajeevphysics/Obsidan-MathMatter) with a star and help others learn more easily. ---