浙江大学真题 第20995题
📝 题目
8．（20分）已知3维线性空间 $V$ 有两组基：（I）$\left\{\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right\}$ ，（II）$\left\{-\varepsilon_{3},-2 \varepsilon_{2},-3 \varepsilon_{1}\right\}$
（1）写出（I）到（II）的过渡矩阵；
（2）若向量 $\alpha$ 在基（I）下坐标为 $(1,2,3)^{T}$ ，写出 $\alpha$ 在基（II）下的坐标；
（3）定义线性变换 $A$ 为：$A\left(\varepsilon_{1}\right)=\varepsilon_{1}, A\left(\varepsilon_{2}\right)=2 \varepsilon_{2}, A\left(\varepsilon_{3}\right)=3 \varepsilon_{3}-\varepsilon_{1}$ 分别写出 $A$ 关于基（I），
（II）的矩阵；
（4）求 $A(\alpha)$ ．
💡 答案解析
8．解（1）由于
$$
\left(-\varepsilon_{3},-2 \varepsilon_{2},-3 \varepsilon_{1}\right)=\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)\left(\begin{array}{ccc}
0 & 0 & -3 \\
0 & -2 & 0 \\
-1 & 0 & 0
\end{array}\right)
$$
故所求过渡矩阵为
$$
\left(\begin{array}{ccc}
0 & 0 & -3 \\
0 & -2 & 0 \\
-1 & 0 & 0
\end{array}\right)
$$
（2）由 $\alpha=\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ 以及（1），可得
$$
\alpha=\left(-\varepsilon_{3},-2 \varepsilon_{2},-3 \varepsilon_{1}\right)\left(\begin{array}{ccc}
0 & 0 & -3 \\
0 & -2 & 0 \\
-1 & 0 & 0
\end{array}\right)^{-1}\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)=\left(-\varepsilon_{3},-2 \varepsilon_{2},-3 \varepsilon_{1}\right)\left(\begin{array}{l}
-3 \\
-1 \\
-\frac{1}{3}
\end{array}\right)
$$
故 $\alpha$ 在基（II）下的坐标 $\left(-3,-1,-\frac{1}{3}\right)^{T}$ ．
$$
\begin{aligned}
& \text { (3) } \begin{aligned}
& A\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)=\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)\left(\begin{array}{ccc}
1 & 0 & -1 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right) \\
& \begin{aligned}
A\left(-\varepsilon_{3},-2 \varepsilon_{2},-3 \varepsilon_{1}\right) & =A\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)\left(\begin{array}{ccc}
0 & 0 & -3 \\
0 & -2 & 0 \\
-1 & 0 & 0
\end{array}\right) \\
& =\left(-\varepsilon_{3},-2 \varepsilon_{2},-3 \varepsilon_{1}\right)\left(\begin{array}{ccc}
0 & 0 & -3 \\
0 & -2 & 0 \\
-1 & 0 & 0
\end{array}\right)^{-1}\left(\begin{array}{ccc}
1 & 0 & -1 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right)\left(\begin{array}{ccc}
0 & 0 & -3 \\
0 & -2 & 0 \\
-1 & 0 & 0
\end{array}\right)
\end{aligned}
\end{aligned} \text { }
\end{aligned}
$$
$$
=\left(-\varepsilon_{3},-2 \varepsilon_{2},-3 \varepsilon_{1}\right)\left(\begin{array}{ccc}
3 & 0 & 0 \\
0 & 2 & 0 \\
-\frac{1}{3} & 0 & 1
\end{array}\right)
$$
故 $A$ 关于基（I），（II）的矩阵分别为
$$
\begin{gathered}
\left(\begin{array}{ccc}
1 & 0 & -1 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right),\left(\begin{array}{ccc}
3 & 0 & 0 \\
0 & 2 & 0 \\
-\frac{1}{3} & 0 & 1
\end{array}\right) \cdot \\
\text { (4) } A(\alpha)=A\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)=\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)\left(\begin{array}{ccc}
1 & 0 & -1 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right)\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)=\left(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}\right)\left(\begin{array}{c}
-2 \\
4 \\
9
\end{array}\right) \cdot
\end{gathered}
$$
📋 详细解题步骤
步骤 1/1
目标：1. 由基变换定义写出过渡矩阵；2. 利用坐标变换公式计算新坐标；3. 由线性变换定义求其在两组基下的矩阵；4. 通过矩阵乘法计算A(α)。
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