In Example SSP4 we showed that \begin{equation*} S=\set{x-2,\,x^2-4x+4,\,x^3-6x^2+12x-8,\,x^4-8x^3+24x^2-32x+16} \end{equation*} is a spanning set for $W=\setparts{p(x)}{p\in P_4,\ p(2)=0}$. We will now show that $S$ is also linearly independent in $W$. Begin with a relation of linear dependence, \begin{align*} 0+0x&+0x^2+0x^3+0x^4\\ &=\alpha_1\left(x-2\right)+\alpha_2\left(x^2-4x+4\right)+\alpha_3\left(x^3-6x^2+12x-8\right)\\ &\quad\quad +\alpha_4\left(x^4-8x^3+24x^2-32x+16\right)\\ &=\alpha_4x^4+ \left(\alpha_3-8\alpha_4\right)x^3+ \left(\alpha_2-6\alpha_3+24\alpha_4\right)x^2\\ &\quad\quad + \left(\alpha_1-4\alpha_2+12\alpha_3-32\alpha_4\right)x+ \left(-2\alpha_1+4\alpha_2-8\alpha_3+16\alpha_4\right) \end{align*}
Equating coefficients (vector equality in $P_4$) gives the homogeneous system of five equations in four variables, \begin{align*} \alpha_4&=0\\ \alpha_3-8\alpha_4&=0\\ \alpha_2-6\alpha_3+24\alpha_4&=0\\ \alpha_1-4\alpha_2+12\alpha_3-32\alpha_4&=0\\ -2\alpha_1+4\alpha_2-8\alpha_3+16\alpha_4&=0\\ \end{align*}
We form the coefficient matrix, and row-reduce to obtain a matrix in reduced row-echelon form \begin{equation*} \begin{bmatrix} \leading{1}&0&0&0\\ 0&\leading{1}&0&0\\ 0&0&\leading{1}&0\\ 0&0&0&\leading{1}\\ 0&0&0&0 \end{bmatrix} \end{equation*}
With only the trivial solution to this homogeneous system, we conclude that only scalars that will form a relation of linear dependence are the trivial ones, and therefore the set $S$ is linearly independent (Definition LI). Finally, $S$ has earned the right to be called a basis for $W$ (Definition B).