命题 6.2.1.
设\(T\)为从线性空间\(V\)的基\((\xi_{1}, \ldots, \xi_{n})\)到基\((\eta_{1}, \ldots, \eta_{n})\)的过渡矩阵，则\(T\)为可逆矩阵。
节 6.2 基变换与坐标变换
子节 6.2.1 基础知识回顾
命题 6.2.2.
设\((\xi_{1}, \ldots, \xi_{n})\)是\(n\)维线性空间\(V\)的一个基，
\begin{equation*}
T = \begin{pmatrix}c_{11}&c_{12}&\cdots&c_{1n}\\ c_{21}&c_{22}&\cdots&c_{2n}\\ \vdots&\vdots&\ddots&\vdots \\ c_{n1}&c_{n2}&\cdots&c_{nn}\end{pmatrix}
\end{equation*}
是一个可逆矩阵，则由下列关系
\begin{equation*}
(\eta_{1}, \ldots, \eta_{n}) = (\xi_{1}, \ldots, \xi_{n}) T
\end{equation*}
所定义的\((\eta_{1}, \ldots, \eta_{n})\)也是\(V\)的基。
命题 6.2.3.
设\((\xi_{1}, \ldots, \xi_{n})\)为\(n\)维空间\(V\)的一个基，若\(\varphi: V \to \F^{n}\)是从\(V\)中向量到其在基\((\xi_{1}, \ldots, \xi_{n})\)下的坐标的映射，即
\begin{equation*}
\varphi(\alpha) = (c_{1}, \ldots, c_{n})^T, \quad \forall \alpha \in V,
\end{equation*}
其中\(\alpha =(\xi_{1}, \ldots, \xi_{n}) \begin{pmatrix}c_{1}\\ \vdots\\ c_{n}\end{pmatrix}\)，则坐标映射\(\varphi\)是双射。
定理 6.2.4.
设\((\xi_{1}, \ldots, \xi_{n})\)是数域\(\F\)上\(n\)维线性空间\(V\)的一个基，\(\varphi: V \to \F^{n}\)为从\(V\)中向量到其在\((\xi_{1}, \ldots, \xi_{n})\)下坐标的映射，则\(\varphi\)满足如下性质：
-
保持加法运算：\(\varphi(\alpha + \beta) = \varphi(\alpha) + \varphi(\beta), ~\forall \alpha, \beta \in V\)；
-
保持数乘运算：\(\varphi(c \alpha) = c \varphi(\alpha), ~\forall c \in \F, \alpha \in V\)。
练习 6.2.2 练习
基础题.
1.
给定\(\alpha = (2,1,3)^{T} \in \R^{3}\)，求：
解答.
-
设\(\alpha\)在基\((\xi_{1}, \xi_{2}, \xi_{3})\)下的坐标为\((c_{1}, c_{2}, c_{3})^{T}\)，则\begin{align*} \alpha \amp= c_{1} \xi_{1} + c_{2} \xi_{2} + c_{3} \xi_{3} \\ \amp = c_{1}(1,1,0)^{T} + c_{2}(1,0,1)^{T} + c_{3}(0,1,1)^{T} \\ \amp = (c_{1} + c_{2}, c_{1} + c_{3}, c_{2} + c_{3})^{T}. \end{align*}与\(\alpha = (2,1,3)^{T}\)比较，得方程组\begin{equation*} \begin{cases}c_{1} + c_{2} = 2, \\ c_{1} + c_{3} = 1, \\ c_{2} + c_{3} = 3.\end{cases} \end{equation*}解之，得\(c_{1} = 0\)，\(c_{2} = 2\)，\(c_{3} = 1\)。所以坐标为\((0,2,1)^{T}\)。
2.
设\(f(x) = x^{5} + 4 x^{4} - 2x^{2} -x + 1 \in \F_{5}[x]\)，求：
解答.
-
多项式 \(f(x)=x^{5}+4x^{4}-2x^{2}-x+1\) 在基 \(((x-1)^{5},(x-1)^{4},(x-1)^{3},(x-1)^{2},x-1,1)\) 下的坐标，即为将 \(f(x)\) 展开成 \((x-1)\) 的幂多项式时的系数（按降幂排列）。利用泰勒公式\begin{equation*} f(x)=\sum_{k=0}^{5}\frac{f^{(k)}(1)}{k!}(x-1)^{k}, \end{equation*}依次计算 \(f(1)\) 及各阶导数在 \(x=1\) 处的值：\begin{align*} f(1)\amp = 1^5+41^4-21^2-1+1 = 3, \\ f’(x) & = 5x^4+16x^3-4x-1, \\ f’(1)& =5+16-4-1=16, \\ f”(x) & = 20x^3+48x^2-4,\\ f”(1)&=20+48-4=64, \\ f^{(3)}(x)& = 60x^2+96x,\\ f^{(3)}(1)& =60+96=156, \\ f^{(4)}(x)& = 120x+96, \\ f^{(4)}(1)&=120+96=216, \\ f^{(5)}(x)& = 120, \\ f^{(5)}(1)&=120. \end{align*}除以对应的阶乘得到各系数：\begin{equation*} a_{5}=\frac{f^{(5)}(1)}{5!}=\frac{120}{120}=1,\quad a_{4}=\frac{f^{(4)}(1)}{4!}=\frac{216}{24}=9, \end{equation*}\begin{equation*} a_{3}=\frac{f^{(3)}(1)}{3!}=\frac{156}{6}=26,\quad a_{2}=\frac{f''(1)}{2!}=\frac{64}{2}=32, \end{equation*}\begin{equation*} a_{1}=\frac{f'(1)}{1!}=16,\quad a_{0}=f(1)=3. \end{equation*}因此，\(f(x)\) 在给定基下的坐标为\((1,\;9,\;26,\;32,\;16,\;3)\)。
3.
在\(\F^{2 \times 2}\)中，设
\begin{equation*}
\xi_{1} = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix},\quad \xi_{2} = \begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix},\quad \xi_{3} = \begin{pmatrix}1 & -1 \\ 1 & -1\end{pmatrix},\quad \xi_{4} = \begin{pmatrix}1 & -1 \\ -1 & 1\end{pmatrix},
\end{equation*}
\begin{equation*}
\eta_{1} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix},\quad \eta_{2} = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix},\quad \eta_{3} = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix},\quad \eta_{4} = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}.
\end{equation*}
求从基\((\xi_{1}, \xi_{2}, \xi_{3}, \xi_{4})\)到\((\eta_{1}, \eta_{2}, \eta_{3}, \eta_{4})\)的过渡矩阵。
解答.
\(\eta_{1}, \eta_{2}, \eta_{3}, \eta_{4}\)在基\((\xi_{1}, \xi_{2}, \xi_{3}, \xi_{4})\)下的坐标向量（作为过渡矩阵的列）为：
\begin{equation*}
\eta_{1}: \begin{pmatrix}1/2 \\ 0 \\ 0 \\ 1/2\end{pmatrix},\quad \eta_{2}: \begin{pmatrix}0 \\ 1/2 \\ 1/2 \\ 0\end{pmatrix},\quad \eta_{3}: \begin{pmatrix}1/2 \\ 0 \\ 0 \\ -1/2\end{pmatrix},\quad \eta_{4}: \begin{pmatrix}0 \\ 1/2 \\ -1/2 \\ 0\end{pmatrix}.
\end{equation*}
因此，过渡矩阵为
\begin{equation*}
T = \frac{1}{2}\begin{pmatrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & -1 & 0\end{pmatrix}.
\end{equation*}
4.
设\((\xi_{1}, \ldots, \xi_{n})\)为\(n\)维线性空间\(V\)的一个基。
-
若向量\(\alpha\)在基\((\xi_{1}, \ldots, \xi_{n})\)下的坐标为\((n,n-1, \ldots, 2,1)^{T}\)，求\(\alpha\)在基\((\xi_{1}, \xi_{1} + \xi_{2}, \ldots, \xi_{1} + \cdots + \xi_{n})\)下的坐标。
解答.
-
令\(\eta_{j} = \xi_{1} + \xi_{2} + \cdots + \xi_{j}\)，\(j=1,\ldots,n\)，其中\(\eta_{1}=\xi_{1}\)。则向量组\((\eta_{1},\ldots,\eta_{n})\)可由基\((\xi_{1},\ldots,\xi_{n})\)线性表示：\begin{equation*} (\eta_{1},\eta_{2},\ldots,\eta_{n}) = (\xi_{1},\xi_{2},\ldots,\xi_{n}) A, \end{equation*}其中\begin{equation*} A = \begin{pmatrix}1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 1 & \cdots & 1 \\ 0 & 0 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1\end{pmatrix}. \end{equation*}矩阵\(A\)是上三角矩阵，且主对角线元素全为\(1\)，故\(\det A = 1 \neq 0\)，因此\(A\)可逆。从而\((\eta_{1},\ldots,\eta_{n})\)与\((\xi_{1},\ldots,\xi_{n})\)等价，所以\((\eta_{1},\ldots,\eta_{n})\)也是\(V\)的一个基。
-
设\(\alpha\)在基\((\eta_{1},\ldots,\eta_{n})\)下的坐标为\((y_{1},y_{2},\ldots,y_{n})^{T}\)，其中\(\eta_{j} = \xi_{1}+\cdots+\xi_{j}\)。 由坐标变换公式，有\begin{align*} (\xi_{1},\ldots,\xi_{n}) \begin{pmatrix}n \\ n-1 \\ \vdots \\ 1\end{pmatrix} \amp= \alpha = (\eta_{1},\ldots,\eta_{n}) \begin{pmatrix}y_{1} \\ y_{2} \\ \vdots \\ y_{n}\end{pmatrix} \\ \amp = (\xi_{1},\ldots,\xi_{n}) A \begin{pmatrix}y_{1} \\ y_{2} \\ \vdots \\ y_{n}\end{pmatrix}, \end{align*}所以\begin{equation*} \begin{pmatrix}n \\ n-1 \\ \vdots \\ 1\end{pmatrix} = A \begin{pmatrix}y_{1} \\ y_{2} \\ \vdots \\ y_{n}\end{pmatrix}. \end{equation*}于是\begin{equation*} \begin{pmatrix}y_{1} \\ y_{2} \\ \vdots \\ y_{n}\end{pmatrix} = A^{-1}\begin{pmatrix}n \\ n-1 \\ \vdots \\ 1\end{pmatrix}. \end{equation*}由\(\eta_{j}\)的表达式可解得\(\xi_{j} = \eta_{j} - \eta_{j-1}\)（约定\(\eta_{0}=0\)），故从基\((\eta_{1},\ldots,\eta_{n})\)到基\((\xi_{1},\ldots,\xi_{n})\)的过渡矩阵为\begin{equation*} A^{-1}= \begin{pmatrix}1 & -1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & -1 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -1 \\ 0 & 0 & 0 & \cdots & 0 & 1\end{pmatrix}. \end{equation*}计算得\begin{equation*} \begin{pmatrix}y_{1} \\ y_{2} \\ y_{3} \\ \vdots \\ y_{n-1}\\ y_{n}\end{pmatrix} = \begin{pmatrix}1 & -1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & -1 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -1 \\ 0 & 0 & 0 & \cdots & 0 & 1\end{pmatrix} \begin{pmatrix}n \\ n-1 \\ n-2 \\ \vdots \\ 2 \\ 1\end{pmatrix} = \begin{pmatrix}1 \\ 1 \\ 1 \\ \vdots \\ 1 \\ 1\end{pmatrix}. \end{equation*}所以\(\alpha\)在基\((\eta_{1},\ldots,\eta_{n})\)下的坐标为\((1,1,\ldots,1)^{T}\)。
提高题.
5.
设\(n\)阶方阵\(T\)为从线性空间\(V\)的基\((\xi_{1}, \ldots, \xi_{n})\)到基\((\eta_{1}, \ldots, \eta_{n})\)的过渡矩阵，证明：\(T^{-1}\)是从基\((\eta_{1}, \ldots, \eta_{n})\)到基\((\xi_{1}, \ldots, \xi_{n})\)的过渡矩阵。
解答.
由过渡矩阵的定义，有
\begin{equation*}
(\eta_{1}, \ldots, \eta_{n}) = (\xi_{1}, \ldots, \xi_{n}) T.
\end{equation*}
由命题6.2.2，可逆矩阵\(T^{-1}\)是从基\((\eta_{1}, \ldots, \eta_{n})\)到另一组基\((\zeta_{1}, \ldots, \zeta_{n})\)的过渡矩阵
\begin{equation*}
(\zeta_{1}, \ldots, \zeta_{n}) = (\eta_{1}, \ldots, \eta_{n}) T^{-1}= ((\xi_{1}, \ldots, \xi_{n})T)T^{-1}.
\end{equation*}
由“结合律”得
\begin{align*}
(\zeta_{1}, \ldots, \zeta_{n})\amp = ((\xi_{1}, \ldots, \xi_{n})T)T^{-1} \\
\amp = (\xi_{1}, \ldots, \xi_{n}) (T T^{-1}) = (\xi_{1}, \ldots, \xi_{n}).
\end{align*}
6.
设\((\eta_{1}, \ldots, \eta_{n})\)为\(n\)维线性空间\(V\)的一个基，若\(V\)中向量\(\xi_{1}, \ldots, \xi_{n}\)与\(n\)阶方阵\(A\)满足关系
\begin{equation*}
(\eta_{1}, \ldots, \eta_{n}) = (\xi_{1}, \ldots, \xi_{n}) A,
\end{equation*}
证明：\((\xi_{1}, \ldots, \xi_{n})\)也是\(V\)的一组基。
解答.
设有矩阵\(B \in \F^{n \times n}\)，其中\(B\)的第\(i\)个列向量是\(\xi_{i}\)在基\((\eta_{1}, \ldots, \eta_{n})\)下的坐标向量。结合已知条件有
\begin{equation*}
(\xi_{1}, \ldots, \xi_{n}) = (\eta_{1}, \ldots, \eta_{n}) B = ((\xi_{1}, \ldots, \xi_{n}) A) B.
\end{equation*}
由“结合律”得
\begin{equation*}
(\xi_{1}, \ldots, \xi_{n}) = (\xi_{1}, \ldots, \xi_{n}) (AB).
\end{equation*}
由形式表示法的定义有\(AB = E_{n}\)，即\(A\)与\(B\)都可逆，且互为逆矩阵。又因为
\begin{equation*}
(\xi_{1}, \ldots, \xi_{n}) = (\eta_{1}, \ldots, \eta_{n}) B,
\end{equation*}
7.
设\((\alpha_{1}, \alpha_{2}, \alpha_{3})\)是\(\F\)上线性空间\(V\)的一个基，\(k \in \F\)，且\(\beta_{1} = 2\alpha_{1} + 2k \alpha_{3}, \beta_{2} = 2\alpha_{2}, \beta_{3} = \alpha_{1} + (k+1)\alpha_{3}\)。
-
当\(k\)为何值时，存在非零向量\(\xi\)，使得它在基\((\alpha_{1}, \alpha_{2}, \alpha_{3})\)和基\((\beta_{1}, \beta_{2}, \beta_{3})\)下的坐标相同？并求所有满足上述条件的\(\xi\)的集合。
解答.
-
向量组\((\beta_{1}, \beta_{2}, \beta_{3})\)在基\((\alpha_{1}, \alpha_{2}, \alpha_{3})\)下的矩阵表示为\begin{equation*} (\beta_{1}, \beta_{2}, \beta_{3}) = (\alpha_{1}, \alpha_{2}, \alpha_{3}) A, \end{equation*}其中\begin{equation*} A = \begin{pmatrix}2&0&1 \\ 0&2&0 \\ 2k&0&k+1\end{pmatrix}. \end{equation*}因为\((\alpha_{1}, \alpha_{2}, \alpha_{3})\)是基，所以\((\beta_{1}, \beta_{2}, \beta_{3})\)是基当且仅当\(A\)可逆，即\(\det A \neq 0\)。计算\begin{align*} \det A \amp = \begin{vmatrix}2&0&1 \\ 0&2&0 \\ 2k&0&k+1\end{vmatrix} \\ \amp = 2 \cdot \begin{vmatrix}2&1 \\ 2k&k+1\end{vmatrix}\\ \amp2 \big( 2(k+1) - 2k \big) = 4 \neq 0. \end{align*}所以\(A\)可逆，从而\((\beta_{1}, \beta_{2}, \beta_{3})\)是\(V\)的一个基。
-
设向量\(\xi\)在基\((\alpha_{1}, \alpha_{2}, \alpha_{3})\)下的坐标为\((x_{1}, x_{2}, x_{3})^{\top}\)，则\(\xi = (\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix}\)。 由\((\beta_{1}, \beta_{2}, \beta_{3}) = (\alpha_{1}, \alpha_{2}, \alpha_{3}) A\)，其中\(A\)如上。\(\xi\)在基\((\beta_{1}, \beta_{2}, \beta_{3})\)下的坐标若也为\((x_{1}, x_{2}, x_{3})^{\top}\)，则\begin{align*} \xi\amp= (\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix} \\ \amp= (\beta_{1}, \beta_{2}, \beta_{3}) \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix} \\ \amp = (\alpha_{1}, \alpha_{2}, \alpha_{3}) A \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix}. \end{align*}由于\((\alpha_{1}, \alpha_{2}, \alpha_{3})\)是基，坐标表示唯一，所以\begin{equation*} \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix} = A \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix}, \quad \text{即}\quad (A - E_{3}) \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix} = 0. \end{equation*}存在非零向量\(\xi\)当且仅当齐次线性方程组\((A - E_{3})X = 0\)有非零解，即\(\det(A - E_{3}) = 0\)。 计算行列式\begin{equation*} \det(A - E_{3}) = \begin{vmatrix}1&0&1 \\ 0&1&0 \\ 2k&0&k\end{vmatrix} = 1 \cdot \begin{vmatrix}1&1 \\ 2k&k\end{vmatrix} = k - 2k = -k. \end{equation*}所以\(\det(A - E_{3}) = 0\)当且仅当\(k = 0\)。因此，当\(k=0\)时，存在非零向量\(\xi\)满足条件。此时\begin{equation*} A - E_{3} = \begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}. \end{equation*}解方程组\((A-E_{3})X = 0\)得非零通解为\((x_{1}, x_{2}, x_{3})^{\top} = t(-1, 0, 1)^{\top}, t \neq 0 \in \F\)。 于是所有满足条件的非零向量为\begin{equation*} \xi = (\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{pmatrix}-t \\ 0 \\ t\end{pmatrix} = t(-\alpha_{1} + \alpha_{3}), \quad t \in \F, t \neq 0. \end{equation*}即集合为\(\{ t(-\alpha_{1} + \alpha_{3}) \mid t \in \F, t \neq 0 \}\)。
8.
考虑次数不超过\(n-1\)的多项式构成的线性空间\(\F_{n-1}[x]\)，设有\(n\)个两两不同的数\(c_{1}, \ldots, c_{n} \in \F\)，且设
\begin{equation*}
f_{i}(x) = \prod_{j \neq i}(x- c_{j}), \quad i = 1, \ldots, n.
\end{equation*}
解答.
-
又因为\(\F_{n-1}[x]\)是数域\(\F\)上的\(n\)维线性空间，例如\((1, x, x^{2}, \ldots, x^{n-1})\)是一个基。要证明\((f_{1}(x), \ldots, f_{n}(x))\)是基，只需证明它们线性无关。设存在\(a_{1}, \ldots, a_{n} \in \F\)使得\begin{equation*} a_{1} f_{1}(x) + a_{2} f_{2}(x) + \cdots + a_{n} f_{n}(x) = 0. \end{equation*}对任意\(k = 1, \ldots, n\)，将\(x = c_{k}\)代入上式。注意到当\(i \neq k\)时，\(f_{i}(c_{k}) = \prod_{j \neq i}(c_{k} - c_{j})\)，其中因子包含\((c_{k} - c_{k})=0\)，所以\(f_{i}(c_{k})=0\)；而当\(i=k\)时，\(f_{k}(c_{k}) = \prod_{j \neq k}(c_{k} - c_{j}) \neq 0\)，因为\(c_{1}, \ldots, c_{n}\)两两不同。因此\begin{equation*} a_{1} f_{1}(c_{k}) + \cdots + a_{n} f_{n}(c_{k}) = a_{k} f_{k}(c_{k}) = 0. \end{equation*}由于\(f_{k}(c_{k}) \neq 0\)，得\(a_{k} = 0\)。这对每个\(k\)都成立，所以\(a_{1} = a_{2} = \cdots = a_{n} = 0\)。因此\(f_{1}(x), \ldots, f_{n}(x)\)线性无关。又因为\(\dim \F_{n-1}[x] = n\)，所以\((f_{1}(x), \ldots, f_{n}(x))\)是\(\F_{n-1}[x]\)的一组基。
-
设 \(f(x)\) 在该基下的坐标为 \((\lambda_{1}, \ldots, \lambda_{n})\)，即\begin{equation*} f(x) = \sum_{i=1}^{n} \lambda_{i} f_{i}(x). \end{equation*}对每个 \(k = 1, \ldots, n\)，代入 \(x = c_{k}\)。由 \(f_{i}(c_{k})\) 的性质（当 \(i \neq k\) 时 \(f_{i}(c_{k})=0\)，\(f_{k}(c_{k}) \neq 0\)）得\begin{equation*} f(c_{k}) = \lambda_{k} f_{k}(c_{k}), \end{equation*}故\begin{equation*} \lambda_{k} = \frac{f(c_{k})}{f_{k}(c_{k})}= \frac{1 + c_{k} + \cdots + c_{k}^{n-1}}{\prod_{j \neq k}(c_{k} - c_{j})}. \end{equation*}因此，所求坐标为\begin{equation*} \left( \frac{1 + c_{1} + \cdots + c_{1}^{n-1}}{\prod_{j \neq 1}(c_{1} - c_{j})},\; \frac{1 + c_{2} + \cdots + c_{2}^{n-1}}{\prod_{j \neq 2}(c_{2} - c_{j})},\; \ldots,\; \frac{1 + c_{n} + \cdots + c_{n}^{n-1}}{\prod_{j \neq n}(c_{n} - c_{j})}\right). \end{equation*}若记 \(g(x) = \prod_{j=1}^{n} (x - c_{j})\)，则 \(g'(c_{k}) = \prod_{j \neq k}(c_{k} - c_{j})\)，于是亦可写为 \(\lambda_{k} = \dfrac{f(c_k)}{g'(c_k)}\)。
