题号:1341    题型:填空题    来源:2003年全国硕士研究生招生考试试题
类型:考研真题
从 $\mathbf{R}^{2}$ 的基 $\boldsymbol{\alpha}_{1}=\left(\begin{array}{l}1 \\ 0\end{array}\right), \boldsymbol{\alpha}_{2}=\left(\begin{array}{c}1 \\ -1\end{array}\right)$ 到基 $\boldsymbol{\beta}_{1}=\left(\begin{array}{l}1 \\ 1\end{array}\right), \boldsymbol{\beta}_{2}=\left(\begin{array}{l}1 \\ 2\end{array}\right)$ 的过渡矩阵为
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答案:
$\left(\begin{array}{cc}2 & 3 \\ -1 & -2\end{array}\right)$

解析:

$n$ 维向量空间中, 从基 $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ 到基 $\beta_{1}, \beta_{2}, \cdots, \beta_{n}$ 的过渡矩阵 $P$ 满足
$$
\left[\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right]=\left[\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right] P,
$$
因此过渡矩阵 $P$ 为:
$$
P=\left[\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right]^{-1}\left[\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right] .
$$
根据定义, 从 $R^{2}$ 的基 $\alpha_{1}=\left(\begin{array}{l}1 \\ 0\end{array}\right), \alpha_{2}=\left(\begin{array}{c}1 \\ -1\end{array}\right)$ 到基 $\beta_{1}=\left(\begin{array}{l}1 \\ 1\end{array}\right), \beta_{2}=\left(\begin{array}{l}1 \\ 2\end{array}\right)$ 的过渡矩阵为
$$
P=\left[\alpha_{1}, \alpha_{2}\right]^{-1}\left[\beta_{1}, \beta_{2}\right]=\left[\begin{array}{cc}
1 & 1 \\
0 & -1
\end{array}\right]^{-1}\left[\begin{array}{cc}
1 & 1 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
1 & 1 \\
0 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & 1 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
-1 & -2
\end{array}\right] .
$$

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