例题 4.2 求坐标

例题 4.2.

在 ${\mathbb{R}}^3$ 中,求向量 $\alpha =\left(4,12,6\right)$ 在基

$${\alpha }_1=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right),{\alpha }_2=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right),{\alpha }_3=\left(\begin{array}{c}-2\\ -5\\ -1\end{array}\right).$$

下的坐标.

解 取 ${\mathbb{R}}^3$ 的标准基 ${\epsilon }_1=\left(1,0,0\right)$, ${\epsilon }_2=\left(0,1,0\right)$, ${\epsilon }_3=\left(0,0,1\right)$, 则 $\alpha$ 在标准基下的坐标 $\left(x_1,x_2,x_3\right)=\left(4,12,6\right)$. 又因为

$$\begin{array}{rl}{\alpha }_1& =-2{\epsilon }_1+{\epsilon }_2+3{\epsilon }_3,\\ {\alpha }_2& =-{\epsilon }_1+{\epsilon }_3,\\ {\alpha }_3& =-2{\epsilon }_1-5{\epsilon }_2-{\epsilon }_3.\end{array}$$

所以由标准基 ${\epsilon }_1,{\epsilon }_2,{\epsilon }_3$ 到基 ${\alpha }_1,{\alpha }_2,{\alpha }_3$ 的过渡矩阵为

$$A=\left(\begin{array}{rrr}-2& -1& -2\\ 1& 0& -5\\ 3& 1& -1\end{array}\right)$$

若设 $\alpha$ 在基 ${\alpha }_1,{\alpha }_2,{\alpha }_3$ 下的坐标为 $\left(y_1,y_2,y_3\right)$ ,则

$$\begin{array}{rl}\left(\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right)& ={\mathbf{A}}^{-1}\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)\\ & =\left(\begin{array}{ccc}\frac{5}{2}& -\frac{3}{2}& \frac{5}{2}\\ -7& 4& -6\\ \frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{c}4\\ 12\\ 6\end{array}\right)\\ & =\left(\begin{array}{c}7\\ -16\\ -1\end{array}\right).\end{array}$$

即 $\alpha$ 在基 ${\alpha }_1,{\alpha }_2,{\alpha }_3$ 下的坐标为 $\left(7,-16,-1\right)$ .