# Proving a Basis The general requirements for proving basis is argued above (not using math), but proof of this in can be seen as the example below. >The example below is adapted from a 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 steps. Suppose we had a set x, and we wanted to prove this was a basis of $\mathbb{R^2}$ $C = \left\{ \begin{bmatrix} -1 \\ 2 \end{bmatrix},\begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}$ We know since were proving it to be in $\mathbb{R^2}$ it must be contained only two vectors and vectors who are [[Linear Independence & Dependence|linearly independent]]. >[!warning] Assumptions for Checking Span Assume the [[Span|vector equation]] to be true: > $\vec{x} = c_1v_1+ \dots +c_kv_k, \quad ,c_1\dots c_k \in \mathbb{R}$ > Where we will let $\vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \mathbb{R^2}$ in order to check if the span $\in \mathbb{R^2}$ $\begin{array}{c} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = c_1 \begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \\ \\ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -c_1 + c_2 \\ 2c_1 + c_2 \end{bmatrix} \\ \\ (1) \quad -c_1 + c_2 = x_1 \\ (2) \quad 2c_1 + c_2 = x_2 \\ \\ \text{Solve for $c_1, c_2$ in terms of $x_1, x_2$ to prove span} \\ \\ (1) - (2) \\ -3c_1 = x_1 - x_2 \\ (3) \quad c_1 = \frac{1}{3} (x_1 + x_2 ) \\ \\ (3) \Rightarrow (1) \\ c_2 = x_1 + \frac{1}{3} (x_1 + x_2 ) \\ c_2 = \frac{1}{3}(2x_1 + x_2) \\ \\ \text{Sub $c_1,c_2$ into out vector equation} \\ \\ \frac{1}{3} (x_1 + x_2 ) \begin{bmatrix} -1 \\ 2 \end{bmatrix} + \frac{1}{3}(2x_1 + x_2) \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \\\\ \text{Therefore this spans $\mathbb{R^2}$} \end{array}$ >[!warning] Assumptions for Linear Independence To check for [[Linear Independence & Dependence|linear independence]], assume the independence equation: > - $c_1\vec{v_1} + \dots + c_k \vec{v_k} = \vec{0}$ If the equation is satisfied only when all scalar must be 0 then it is linear independent. $\begin{array}{c} \begin{bmatrix} 0 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \\ \\ \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -c_1 + c_2 \\ 2c_1 + c_2\end{bmatrix} \\ \\ \text{Expand and solve vector eqations} \\ (1) \quad 0 = -c_1 + c_2 \\ (2) \quad 0 = 2c_1 + c_2 \\ \\ (1) - (2) \\ 0 = -3c_1 \\ 0 = c_1 \\\\ \text{Its implied that if $c_1 = 0$ then $c_2 =0 $ for it to work in equation (1)} \\ \\ \text{Therefore this is linearly independent} \end{array}$