Consider the vector $\vect{u}=15+10x-6x^2\in P_2$ from the vector space of polynomials with degree at most 2 (Example VSP). A nice basis for $P_2$ is \begin{equation*} B=\set{1,\,x,\,x^2} \end{equation*} so that \begin{equation*} \vect{u}=15+10x-6x^2=15(1)+10(x)+(-6)(x^2) \end{equation*} so by Definition VR \begin{equation*} \vectrep{B}{\vect{u}}=\colvector{15\\10\\-6} \end{equation*}
Another nice basis for $P_2$ is \begin{equation*} C=\set{1,\,1+x,\,1+x+x^2} \end{equation*} so that now it takes a bit of computation to determine the scalars for the representation. We want $a_1,\,a_2,\,a_3$ so that \begin{equation*} 15+10x-6x^2=a_1(1)+a_2(1+x)+a_3(1+x+x^2) \end{equation*}
Performing the operations in $P_2$ on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars, \begin{align*} 15&=a_1+a_2+a_3\\ 10&=a_2+a_3\\ -6&=a_3 \end{align*}
The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB), \begin{align*} a_1&=5&a_2&=16&a_3&=-6 \end{align*} so by Definition VR \begin{equation*} \vectrep{C}{\vect{u}}=\colvector{5\\16\\-6} \end{equation*}
While we often form vector representations relative to “nice” bases, nothing prevents us from forming representations relative to “nasty” bases. For example, the set \begin{equation*} D=\set{ -2-x+3x^2,\, 1-2x^2,\, 5+4x+x^2 } \end{equation*} can be verified as a basis of $P_2$ by checking linear independence with Definition LI and then arguing that 3 vectors from $P_2$, a vector space of dimension 3 (Theorem DP), must also be a spanning set (Theorem G).
Now we desire scalars $a_1,\,a_2,\,a_3$ so that \begin{equation*} 15+10x-6x^2=a_1(-2-x+3x^2)+a_2(1-2x^2)+a_3(5+4x+x^2) \end{equation*}
Performing the operations in $P_2$ on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars, \begin{align*} 15&=-2a_1+a_2+5a_3\\ 10&=-a_1+4a_3\\ -6&=3a_1-2a_2+a_3 \end{align*}
The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB), \begin{align*} a_1&=-2&a_2&=1&a_3&=2 \end{align*} so by Definition VR \begin{equation*} \vectrep{D}{\vect{u}}=\colvector{-2\\1\\2} \end{equation*}