> [!Definition] Definition (Linear Independence) > A subset $S$ of a [[Vector spaces|vector space]] $V$ is said to be _linearly independent_ iff whenever $ \lambda_1\underline{v}_1+\ldots+\lambda_s\underline{v}_s=\underline{0}, $for $\lambda_{1},\dots,\lambda_{s}\in\mathbb{F}$ and $\underline{v}_{1},\dots,\underline{v}_{s}\in S$ we necessarily have that $\lambda_{1}=\dots=\lambda_{s}=0$. > [!NOTE] Definition (Linear dependence) > A subset $S \subset V$ is ***linearly dependent*** if and only if there is an equation$ \lambda_1\underline{v}_1+\ldots+\lambda_s\underline{v}_s=\underline{0} $ for $\underline{v}_{1},\dots,\underline{v}_{s}\in S$ and $\lambda_1,\ldots,\lambda_s\in\mathbb{F}$ which are not all zero. > >This equation is referred to as a ***linear dependence relation ***for the vectors $\underline{v}_1,\ldots,\underline{v}_s.$ # Properties > [!NOTE] Collinear = linear dependent for two vectors > Two vectors $\underline{v}$ and $\underline{w} \in \mathbb{R}^{n}$ are linearly dependent iff they're [[Collinearity of Two Vectors|collinear]]. > *Proof*. Check definitions. >The set $\{ v_{1},\dots,v_{m+1} \}$ is linearly independent by above lemma. $\square$ # Applications - [[Basis of Vector Space]]. # Examples > [!Example] > The vectors $\underline{v}=(1,2)^T$ and $\underline{w}=(-2,3)^T\in\mathbb{R}^2$ are **linearly independent**. Indeed, suppose that $\lambda\underline {v}+ \mu\underline {w}= \underline {0}.$ Then $ \begin{pmatrix}0\\0\end{pmatrix}=\lambda\begin{pmatrix}1\\2\end{pmatrix}+\mu\begin{pmatrix}-2\\3\end{pmatrix}=\begin{pmatrix}\lambda-2\mu\\2\lambda+3\mu\end{pmatrix} $Considering the components separately gives two simultaneous linear equations $ \begin{array}{rcl}\lambda-2\mu&=&0\\2\lambda+3\mu&=&0\end{array} $Solving these shows that $\lambda=\mu=0$. > [!Example] > In contrast, the vectors $\underline{u}_1=(2,-6)$ and $\underline{u}_2=(-3,9)$ are not **linearly dependent**. You may see at once that $3\underline{u}_1+2\underline{u}_2=\underline{0}$, which is a nontrivial linear combination giving the zero vector. > > > [!Example] > Let $\underline{v}_{1}=(2,1,3)^{T}$ and $\underline{v}_{2}=(1,-1,1)^{T}$. > We consider the three different target vectors: > 1. $\underline{b}=\underline{0}$ > 2. $\underline{b}=(1,5,3)^{T}$ > 3. $\underline{b}=\underline{e}_{1}=(1,0,0)^{T}$ > > **Solution** > 1. Solve $\lambda_{1} (2,1,3)^{T}+\lambda_{2}(1,-1,1)^{T}=(0,0,0)^{T}$ for $\lambda_{1},\lambda_{2}\in \mathbb{R}$. > In components $\begin{align}2 \lambda_{1} + \lambda_{2} = 0 \tag{1} \\ \lambda_{1} = \lambda_{2} = 0 \tag{2} \\ 3\lambda_{1} + \lambda_{2} = 0 \tag{3} \end{align}$ $(1)+(2)$ gives $\lambda_{1}=\lambda_{2}=0$. Hence $\underline{v}_{1},\underline{v}_{2}$ are [[Linear Independence|linearly independent]]. > 2. Solve $\lambda_{1} (2,1,3)^{T}+\lambda_{2}(1,-1,1)^{T}= (1,5,3)^{T}$for $\lambda_{1},\lambda_{2}\in\mathbb{R}$. > In components $\begin{align}2 \lambda_{1} + \lambda_{2} = 1 \tag{1} \\ \lambda_{1} = \lambda_{2} = 5 \tag{2} \\ 3\lambda_{1} + \lambda_{2} = 3 \tag{3} \end{align}$ $(1)+(2)$ gives $3\lambda_{1}=6\implies\lambda_{1}=2$. Substituting in $(1)$ gives $(\lambda_{1},\lambda_{2})=(2,-3)$ which indeed satisfies all three equations. > 3. We attempt to solve them as before. In this case there is no solution at all. > > > - The collection of all possible vectors $\lambda_{1} \underline{v}_{1}+\lambda_{2}\underline{v}_{2}$ describes a flat, linear plane $W$ through the origin in $\mathbb{R}^{3}$. > > - In case 2, the vector $\underline{b}=(1,5,3)^{T}$ happens to lie on this plane $W$. The question is then how many different solutions are there, and the answer is the same as in case 1: there was precisely one solution then, and so there is also precisely one solution in this case (necessarily, as we shall prove later). > > - In case 3, the vector $\underline{e}_{1}$ does not lie on $W$. > > > [!Example] > Let $\underline{v}_{1}=(2,-1)^{T},\underline{v}_{2}=(1,1)^{T}$ and $\underline{v}_{3}=(3,1)^{T} \in \mathbb{R}^{2}$. > Consider the target vectors: > 1. $\underline{b}=\underline{0}$ > 2. $\underline{b}=(7,-1)^T$ > > **Solution** > 1. $\lambda_{2}=-(5/3)\lambda_{3}$ and $\lambda_{1}=-\frac{2}{3}\lambda_{3}$ for $\lambda_{3}\in\mathbb{R}$. It follows that $\underline{v}_{1},\underline{v}_{2},\underline{v}_{3}$ are linearly dependent.