Let \(\mathfrak{B} = \{\alpha_1, \alpha_2, \alpha_3\}\) be the ordered basis for \(\RR^3\) consisting of \[ \alpha_1 = (1, 0, -1), \quad \alpha_2 = (1, 1, 1), \quad \alpha_3 = (1, 0, 0). \] What are the coordinates of the vector \((a, b, c)\) in the ordered basis \(\mathfrak{B}\)?
Solution
Write \[ (a,b,c)=x\alpha_1+y\alpha_2+z\alpha_3. \] Comparing coordinates gives \[ x+y+z=a, \qquad y=b, \qquad -x+y=c. \] Thus \[ y=b, \qquad x=b-c, \qquad z=a-2b+c. \] Therefore \[ [(a,b,c)]_{\mathfrak B} = \begin{bmatrix} b-c\\ b\\ a-2b+c \end{bmatrix}. \]