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高等代数教学辅导

4.4 \(\F^m\)的子空间、基与维数

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练习 练习

基础题.

1.
判断下列\(\mathbb{R}^m\)的子集是否为\(\mathbb{R}^m\)的子空间,说明理由。
  1. \(V_1=\{(a_1,\ldots,a_m)^T|a_i\geq 0,i=1,\ldots,m\}\)
  2. \(V_2=\{(a_1,\ldots,a_m)^T|a_1a_2\cdots a_m\geq 0\}\)
  3. \(V_3=\{(a_1,\ldots,a_n,0,\ldots,0)^T|a_1,\ldots,a_n\in\mathbb{R}\}\)
解答.
  1. 不是。因为\(\varepsilon_1=(1,0,\ldots,0)^T\in V_1\),但 \((-1)\varepsilon_1=(-1,0,\ldots,0)^T\not\in V_1\),所以\(V_1\)不是\(\mathbb{R}^m\)的子空间。
  2. 不是。因为\(\alpha=\sum\limits_{i=1}^{m-1}\varepsilon_i\in V_2,\beta=-\varepsilon_m\in V_2\),但 \(\alpha+\beta=(1,\ldots,1,-1)^T\not\in V_2\),所以\(V_2\)不是\(\mathbb{R}^m\)的子空间。
  3. 是。\(V_3\)\(\mathbb{R}^m\)的非空子集,且对任意\(\alpha=(a_1,\ldots,a_n,0,\ldots,0)^T, \beta=(b_1,\ldots,b_n,0,\ldots,0)^T\in V_3\),有
    \begin{equation*} \alpha+\beta=(a_1+b_1,\ldots,a_n+b_n,0,\ldots,0)^T\in V_3, \end{equation*}
    对任意 \(c\in\mathbb{R}\),有
    \begin{equation*} c\alpha=(ca_1,\ldots,ca_n,0,\ldots,0)^T\in V_3, \end{equation*}
    因此\(V_3\)\(\mathbb{R}^m\)的子空间。
2.
判断下列数域\(\mathbb{F}\)\(n\)元方程的解集是否为\(\mathbb{F}^n\)的子空间:
  1. \(\sum\limits_{i=1}^n a_ix_i=0\)
  2. \(\sum\limits_{i=1}^n a_ix_i=1\)
  3. \(\sum\limits_{i=1}^n x_i^2=0\)
解答.
  1. 是。记\(V_1=\{(x_1,\dots,x_n)^T|\sum\limits_{i=1}^n a_ix_i=0\}\),则
    \begin{equation*} (0,\dots,0)^T\in V_1, \end{equation*}
    所以\(V_1\)\(\mathbb{F}^n\)的非空子集。对任意 \(X=(x_1,\dots,x_n)^T,Y=(y_1,\dots,y_n)^T\in V_1\)\(c\in\F\),有
    \begin{equation*} \sum\limits_{i=1}^n a_ix_i=0, \sum\limits_{i=1}^n a_iy_i=0, \end{equation*}
    \begin{equation*} \sum\limits_{i=1}^n a_i(x_i+y_i)=\sum\limits_{i=1}^n a_ix_i+\sum\limits_{i=1}^n a_iy_i=0, \end{equation*}
    \begin{equation*} \sum\limits_{i=1}^n a_i(cx_i)=c\left(\sum\limits_{i=1}^n a_ix_i\right)=0, \end{equation*}
    \begin{equation*} X+Y=(x_1+y_1,\dots,x_n+y_n)^T\in V_1, \end{equation*}
    \begin{equation*} cX=(cx_1,\dots,cx_n)^T\in V_1. \end{equation*}
    因此\(V_1\)\(\mathbb{F}^n\)的子空间。
  2. 不是。
  3. 与数域\(\mathbb{F}\)有关:当\(\mathbb{F}=\C\)时不是;当\(\mathbb{F}\)\(\R\)的子域时是。
3.
\(\alpha_1,\ldots,\alpha_s,\beta_1,\ldots,\beta_t\in\F^m\),证明:
\begin{equation*} \langle\alpha_1,\ldots,\alpha_s\rangle +\langle\beta_1,\ldots,\beta_t\rangle =\langle\alpha_1,\ldots,\alpha_s,\beta_1,\ldots,\beta_t\rangle . \end{equation*}
解答.
\(V_1=\langle\alpha_1,\ldots,\alpha_s\rangle\)\(V_2=\langle\beta_1,\ldots,\beta_t\rangle\)\(V=\langle\alpha_1,\ldots,\alpha_s,\beta_1,\ldots,\beta_t\rangle\)
对任意\(\gamma\in V_1 +V_2\),存在 \(\alpha\in V_1, \beta\in V_2\),使得
\begin{equation*} \gamma=\alpha+\beta. \end{equation*}
\(\alpha\in V_1\)知存在 \(a_1,\dots,a_s\in\F\),使得
\begin{equation*} \alpha=a_1\alpha_1+\cdots+a_s\alpha_s. \end{equation*}
同理,由\(\beta\in V_2\)知存在\(b_1,\dots,b_t\in\mathbb{F}\),使得
\begin{equation*} \beta=b_1\beta_1+\cdots+b_t\beta_t. \end{equation*}
于是
\begin{equation*} \alpha+\beta=a_1\alpha_1+\cdots+a_s\alpha_s+b_1\beta_1+\cdots+b_t\beta_t\in V, \end{equation*}
因此 \(V_1+V_2\subseteq V\)
反之,对任意\(\xi\in V\),存在\(a_1,\dots,a_s,b_1,\dots,b_t\in\F\),使得
\begin{equation*} \xi=a_1\alpha_1+\cdots +a_s\alpha_s+b_1\beta_1+\cdots +b_t\beta_t. \end{equation*}
\(\xi_1=a_1\alpha_1+\cdots +a_s\alpha_s,\xi_2=b_1\beta_1+\cdots +b_t\beta_t\),则\(\xi_1\in V_1,\xi_2\in V_2\)\(\xi=\xi_1+\xi_2\),故\(V\subseteq V_1+V_2\)。从而 \(V=V_1+V_2\)
4.
\(V_1\)\(V_2\)是线性空间\(V\)的子空间且\(V_1\subseteq V_2\)。证明:\(V_1=V_2\)的充分必要条件是\(\dim V_1=\dim V_2\)
解答.
必要性:显然成立。
充分性:设\(\dim V_1= \dim V_2=m\)\(\xi_1,\xi_2,\cdots,\xi_m\)\(V_1\)的一组基。因\(V_1\subseteq V_2\),所以\(\xi_1,\xi_2,\cdots,\xi_m\)也是\(V_2\)\(m\)个线性无关的向量。由\(\dim V_2=m\)知:\(\xi_1,\xi_2,\cdots,\xi_m\)也是\(V_2\)的一个基,因此\(V_1=V_2\)
5.
在线性空间\(\mathbb{F}^m\)中,证明:
  1. 存在\(\mathbb{F}^m\)的子空间\(U\),使得\(U\)中任一非零向量的分量均不为零;
  2. \(\mathbb{F}^m\)的子空间\(U\)中任一非零向量的分量均不为零,则\(\dim U=1\)
解答.
  1. \(U=\langle (1,1,\dots ,1)^T\rangle\),则\(U\)中任一非零向量的分量均不为零。
  2. 假设存在\(\mathbb{F}^m\)的子空间\(U\),满足\(U\)中任一非零向量的分量均不为零且\(\dim U\geq 2\),则存在
    \begin{equation*} \alpha=(a_1,a_2,\dots ,a_m)^T,\beta=(b_1,b_2,\dots ,b_m)^T\in U, \end{equation*}
    使得\(\alpha,\beta\)线性无关。因\(\beta\)\(U\)中非零向量,故\(b_1,b_2,\dots ,b_m\)均不为零。由\(\alpha,\beta\in U\)\(U\)\(\mathbb{F}^m\)的子空间知:
    \begin{equation*} \alpha-\frac{a_1}{b_1}\beta=(0,a_2-\frac{a_1b_2}{b_1},\dots ,a_m-\frac{a_1b_m}{b_1})^T\in U. \end{equation*}
    \(\alpha,\beta\)线性无关,所以\(\alpha-\frac{a_1}{b_1}\beta\neq 0\)。从而\(U\)中存在非零向量\(\alpha-\frac{a_1}{b_1}\beta\),其分量不全非零,与已知矛盾。因此\(\dim U=1\)
6.
求由向量\(\alpha_i\)生成的子空间与由向量\(\beta_i\)生成的子空间的交与和空间的基与维数: \(\left\{\begin{array}{c} \alpha_1=(1,2,1,0)^T,\\ \alpha_2=(-1,1,1,1)^T, \end{array} \right.\quad\left\{\begin{array}{c} \beta_1=(2,-1,0,1)^T,\\ \beta_2=(1,-1,3,7)^T; \end{array}\right.\)
解答.
\(V_1=\langle\alpha_1,\alpha_2 \rangle ,V_2=\langle\beta_1,\beta_2\rangle\)。因为
\begin{equation*} V_1+V_2=\langle\alpha_1,\alpha_2 \rangle +\langle\beta_1,\beta_2\rangle=\langle\alpha_1,\alpha_2 ,\beta_1,\beta_2\rangle , \end{equation*}
所以向量组\(\alpha_1,\alpha_2 ,\beta_1,\beta_2\)的一个极大无关组就是\(V_1+V_2\)的一个基。
\(A=(\alpha_1,\alpha_2 ,\beta_1,\beta_2)\),对\(A\)进行初等行变换,化为阶梯形矩阵:
\begin{equation*} A=(\alpha_1,\alpha_2,\beta_1,\beta_2)=\begin{pmatrix} 1 & -1 & 2 & 1\\ 2 & 1 & -1 & -1\\ 1 & 1 & 0 & 3\\ 0 & 1 & 1 & 7 \end{pmatrix}\xrightarrow{rref}\begin{pmatrix} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 4\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 0 \end{pmatrix}, \end{equation*}
由此得出,\(\alpha_1,\alpha_2,\beta_1\)\(\alpha_1,\alpha_2 ,\beta_1,\beta_2\)的一个极大无关组,从而\(\alpha_1,\alpha_2 ,\beta_1\)\(V_1+V_2\)的一个基,因此\(\dim(V_1+V_2) =r (\alpha_1,\alpha_2,\beta_1,\beta_2)=3\)
从上述简化行阶梯形矩阵的前\(2\)列可以看出:\(\alpha_1,\alpha_2\)线性无关;从后\(2\)列看出,\(\beta_1,\beta_2\)线性无关。因此\(\dim V_1=2,\dim V_2=2\)。由维数公式得
\begin{equation*} \dim (V_1\cap V_2)=\dim V_1+\dim V_2-\dim (V_1+V_2)=2+2-3=1, \end{equation*}
\(V_1\cap V_2\)中任一非\(0\)向量就是 \(V_1\cap V_2\)的一个基。 注意到 \(\alpha_1-4\alpha_2-3\beta_1+\beta_2={\bf 0}\),所以
\begin{equation*} \alpha_1-4\alpha_2=3\beta_1-\beta_2=(5,-2,-3,-4)^T\in V_1\cap V_2. \end{equation*}
因此\((5,-2,-3,-4)^T\)\(V_1\cap V_2\)的一个基。
7.
\(V_1,V_2,V_3\)\(V\)的子空间,举例说明:
\begin{equation*} V_1\cap (V_2 +V_3)=(V_1\cap V_2)+(V_1\cap V_3) \end{equation*}
未必成立。
解答.
\(V=\mathbb{R}^2\)
\begin{equation*} V_1=\left\{(x,x)^T|x\in\mathbb{R}\right\}, \end{equation*}
\begin{equation*} V_2=\left\{(x,0)^T|x\in\mathbb{R}\right\}, \end{equation*}
\begin{equation*} V_3=\left\{(0,y)^T|y\in\mathbb{R}\right\}, \end{equation*}
\begin{equation*} V_1\cap (V_2+V_3)=V_1\cap \mathbb{R}^2=V_1, \end{equation*}
\begin{equation*} (V_1\cap V_2)+(V_1\cap V_3)=\{{\bf 0}\}+\{{\bf 0}\}=\{{\bf 0}\}, \end{equation*}
此时\(V_1\cap (V_2 +V_3)=(V_1\cap V_2)+(V_1\cap V_3)\)不成立。
8.
\((\xi_1,\ldots,\xi_n)\)是列向量空间\(V\)的一个基,证明:对任意\(1\leq i\leq n-1\),有
\begin{equation*} V=\langle\xi_1,\ldots,\xi_i\rangle\oplus\langle\xi_{i+1},\ldots,\xi_n\rangle . \end{equation*}
解答.
根据练习 4.4.3
\begin{equation*} \langle\xi_1,\ldots,\xi_i\rangle + \langle\xi_{i+1},\ldots,\xi_n\rangle=\langle \xi_1,\dots,\xi_n\rangle . \end{equation*}
\((\xi_1,\ldots,\xi_n)\)是列向量空间\(V\)的一个基知\(\langle \xi_1,\dots,\xi_n\rangle =V\),故
\begin{equation*} V=\langle\xi_1,\ldots,\xi_i\rangle + \langle\xi_{i+1},\ldots,\xi_n\rangle . \end{equation*}
对任意\(\alpha\in \langle\xi_1,\ldots,\xi_i\rangle \cap \langle\xi_{i+1},\ldots,\xi_n\rangle\),存在 \(a_1,\dots,a_n\in\mathbb{F}\),使得
\begin{equation*} \alpha=a_1\xi_1+\cdots+a_i\xi_i, \end{equation*}
\begin{equation*} \alpha=a_{i+1}\xi_{i+1}+\cdots+a_n\xi_n. \end{equation*}
两式相减得
\begin{equation*} a_1\xi_1+\cdots+a_i\xi_i-a_{i+1}\xi_{i+1}-\cdots-a_n\xi_n={\bf 0}, \end{equation*}
\(\xi_1,\dots,\xi_n\)线性无关知\(a_1=\cdots=a_n=0\),故\(\alpha={\bf 0}\),因此\(\langle\xi_1,\ldots,\xi_i\rangle + \langle\xi_{i+1},\ldots,\xi_n\rangle\)是直和。从而
\begin{equation*} V=\langle\xi_1,\ldots,\xi_i\rangle\oplus\langle\xi_{i+1},\ldots,\xi_n\rangle . \end{equation*}

提高题.

9.
\(V_1\)\(V_2\)是线性空间\(V\)的子空间。证明:\(V_1\bigcup V_2\)\(V\)的子空间的充要条件是\(V_1\subseteq V_2\)\(V_2\subseteq V_1\)
解答.
充分性:当\(V_1\subseteq V_2\)\(V_2\subseteq V_1\)时,\(V_1\bigcup V_2=V_2\)\(V_1\),此时\(V_1\bigcup V_2\)\(V\)的子空间。
必要性:设\(V_1\not\subseteq V_2\),则存在\(\alpha\in V_1\)\(\alpha\not\in V_2\)。对任意\(\beta\in V_2\),由\(V_1\bigcup V_2\)\(V\)的子空间、\(\alpha,\beta\in V_1\bigcup V_2\)可知\(\alpha +\beta\in V_1\bigcup V_2\),则\(\alpha +\beta\in V_1\)\(V_2\)
  • \(\alpha+\beta\in V_2\),由\(\beta\in V_2\)\(V_2\)\(V\)的子空间可知\(\alpha=(\alpha +\beta)-\beta\in V_2\),与假设矛盾。
\(\alpha +\beta\in V_1\)。注意到\(\alpha\in V_1\)\(V_1\)\(V\)的子空间,所以\(\beta =(\alpha +\beta )-\alpha\in V_1\)。因此,\(V_2\subseteq V_1\)
10.
\(A\in\mathbb{F}^{m\times n}\)\(A=(\alpha_1,\dots ,\alpha_n)\),其中\(\alpha_i\in\mathbb{F}^m(1\leq i\leq n)\)。记
\begin{equation*} V=\{\ AX\ |\ X\in\mathbb{F}^n\ \}. \end{equation*}
求证:
  1. \(V\)\(\mathbb{F}^m\)的子空间;
  2. \(V=\langle \alpha_1,\dots ,\alpha_n\rangle\)
  3. \(\dim V=r(A)\)
解答.
  1. 显然\(0\in V\),所以\(V\)\(\mathbb{F}^m\)的非空子集。
    对任意\(\beta,\gamma\in V, c \in\mathbb{F}\),存在\(X,Y\in \mathbb{F}^n\),使得\(\beta=AX,\gamma=AY\),则
    \begin{equation*} \beta+\gamma=AX+AY=A(X+Y)\in V, \end{equation*}
    \begin{equation*} c\beta=c(AX)=A(cX)\in V. \end{equation*}
    因此\(V\)\(\mathbb{F}^m\)的子空间。
  2. 对任意\(\beta\in V\),存在\(X\in\mathbb{F}^n\)使得\(\beta=AX\)。记\(X=(x_1,\dots ,x_n)^T\),则
    \begin{equation*} AX = x_1\alpha_1+\cdots +x_n\alpha_n\in \langle \alpha_1,\dots ,\alpha_n\rangle, \end{equation*}
    所以\(V\subseteq \langle \alpha_1,\dots ,\alpha_n\rangle\)。反之,对任意\(\beta\in \langle \alpha_1,\dots,\alpha_n\rangle\),有\(\beta=a_1\alpha_1+\cdots +a_n\alpha_n\)。令\(X=(a_1,\dots,a_n)^T\),则\(X\in\mathbb{F}^n\)\(\beta=AX\),故\(\beta\in V\),因此\(\langle \alpha_1,\cdots,\alpha_n\rangle\subseteq V\)。从而\(V=\langle \alpha_1,\dots ,\alpha_n\rangle\)
  3. 由上题结论知\(\dim V= r(\alpha_1,\dots ,\alpha_n)=r(A)\)
11.
写出\(\F^m\)\(s(s\geq 2)\)个子空间\(V_1,\dots ,V_s\)相应的维数公式,并予以证明。
解答.
推广的维数公式为
\begin{equation*} \dim \left(\sum\limits_{i=1}^{s} V_i\right)=\sum\limits_{i=1}^{s} \dim V_i-\sum\limits_{k=2}^{s}\dim\left[(\sum\limits_{j=1}^{k-1}V_j)\cap V_k\right]. \end{equation*}
下面用数学归纳法证明。
  1. \(s=2\)时,由维数公式知结论成立。
  2. 假设对\(s-1\)个子空间结论成立,则
    \begin{equation*} \dim \left(\sum\limits_{i=1}^{s-1} V_i\right)=\sum\limits_{i=1}^{s-1} \dim V_i-\sum\limits_{k=2}^{s-1}\dim\left[(\sum\limits_{j=1}^{k-1}V_j)\cap V_k\right]. \end{equation*}
    因为\(\sum\limits_{i=1}^{s} V_i=\left(\sum\limits_{i=1}^{s-1} V_i\right)+V_s\),所以根据维数公式
    \begin{equation*} \begin{array}{ll} & \dim\left(V_1+\cdots +V_s\right)\\ = & \dim\left(\sum\limits_{i=1}^{s-1} V_i\right)+\dim V_s-\dim\left[\left(\sum\limits_{i=1}^{s-1} V_i\right)\cap V_s\right].\end{array} \end{equation*}
    于是由归纳假设得
    \begin{equation*} \dim \left(\sum\limits_{i=1}^{s} V_i\right)=\sum\limits_{i=1}^{s} \dim V_i-\sum\limits_{k=2}^{s}\dim\left[(\sum\limits_{j=1}^{k-1}V_j)\cap V_k\right]. \end{equation*}
12.
\begin{equation*} U=\{(a_1,\ldots,a_m)^T\in\F^m|a_1=\cdots=a_m\}, \end{equation*}
\begin{equation*} V=\{(a_1,\ldots,a_m)^T\in\F^m|a_1+\cdots +a_m=0\}, \end{equation*}
证明:
  1. \(U,V\)\(\F^m\)的子空间;
  2. \(\mathbb{F}^m=U\oplus V\)
解答.
  1. 由于\((0,\dots ,0)^T\in U\),所以\(U\)\(\F^m\)的非空子集。对任意 \(\alpha=(a_1,\ldots,a_m)^T,\beta=(b_1,\ldots,b_m)^T\in U, c\in\F\),有 \(a_1=\cdots=a_m, b_1=\cdots=b_m\), 则
    \begin{equation*} a_1+b_1=\cdots=a_m+b_m,\ ca_1=\cdots=ca_m, \end{equation*}
    \begin{equation*} \alpha+\beta=(a_1+b_1,\dots,a_m+b_m)^T\in U, \end{equation*}
    \begin{equation*} c\alpha=(ca_1,\dots,ca_m)^T\in U. \end{equation*}
    因此\(U\)\(\F^m\)的子空间。
    由于\((0,\dots ,0)^T\in V\),所以\(V\)\(\F^m\)的非空子集。对任意 \(\alpha=(a_1,\ldots,a_m)^T,\beta=(b_1,\ldots,b_m)^T\in V, c\in\F\),有 \(a_1+\cdots+a_m=0, b_1+\cdots+b_m=0\), 则
    \begin{equation*} (a_1+b_1)+\cdots+(a_m+b_m)=\sum\limits_{i=1}^ma_i+\sum\limits_{i=1}^mb_i=0, \end{equation*}
    \begin{equation*} ca_1+\cdots+ca_m=c(a_1+\cdots+a_m)=0, \end{equation*}
    \begin{equation*} \alpha+\beta=(a_1+b_1,\dots,a_m+b_m)^T\in V, \end{equation*}
    \begin{equation*} c\alpha=(ca_1,\dots,ca_m)^T\in V. \end{equation*}
    因此\(V\)\(\F^m\)的子空间。
  2. 对任意\(\alpha=(a_1,\dots ,a_m)^T\in U\cap V\),有
    \begin{equation*} a_1=\cdots =a_m,\ a_1+\cdots +a_m=0, \end{equation*}
    \(a_1=\cdots =a_m=0\),即\(\alpha={\bf 0}\),故\(U+V\)是直和。
    对任意\(\alpha=(a_1,\dots ,a_m)^n\in\mathbb{F}^m\),记\(a=\frac{a_1+\cdots +a_m}{m}\),则\(\beta=(a,\dots ,a)^T\in U,\gamma=(a_1-a,\dots ,a_m-a)^T\in V\),且\(\alpha=\beta+\gamma\),故\(\mathbb{F}^m=U+V\)。综上,\(\mathbb{F}^m=U\oplus W\)
13.
\(V_1\)\(V_2\)是列向量空间\(V\)的子空间,且\(\dim V_1=\dim V_2\)。证明:存在\(V\)的子空间\(U\),使得\(V=U\oplus V_1=U\oplus V_2\)
解答 1.
\(\xi_1,\cdots ,\xi_r\)\(V_1\bigcap V_2\)的基。由于\(\dim V_1=\dim V_2\),所以将其扩充为\(V_1\)的基
\begin{equation*} \xi_1,\cdots ,\xi_r,\alpha_1,\cdots ,\alpha_s, \end{equation*}
也可将其扩充为\(V_2\)的基
\begin{equation*} \xi_1,\cdots ,\xi_r,\beta_1,\cdots ,\beta_s. \end{equation*}
此时,\(\xi_1,\cdots ,\xi_r,\alpha_1,\cdots ,\alpha_s,\beta_1,\cdots ,\beta_s\)是子空间\(V_1+V_2\)的基。因此,可将其扩充为\(V\)的基\(\xi_1,\cdots ,\xi_r,\alpha_1,\cdots ,\alpha_s,\beta_1,\cdots ,\beta_s,\gamma_1,\cdots ,\gamma_t\)。令
\begin{equation*} W=\langle \alpha_1+\beta_1,\cdots ,\alpha_s+\beta_s,\gamma_1,\cdots ,\gamma_t\rangle, \end{equation*}
我们断言,\(V=V_1\oplus W=V_2\oplus W\)。事实上,对任意\(\alpha\in V_1\bigcap W\),有
\begin{equation*} \alpha=\sum\limits_{i=1}^rk_i\xi_i+\sum\limits_{j=1}^sl_j\alpha_j=\sum\limits_{j=1}^sp_j(\alpha_j+\beta_j)+\sum\limits_{m=1}^tq_m\gamma_m, \end{equation*}
\begin{equation*} \sum\limits_{i=1}^rk_i\xi_i+\sum\limits_{j=1}^s(l_j-p_j)\alpha_j-\sum\limits_{j=1}^sp_j\beta_j-\sum\limits_{m=1}^tq_m\gamma_m=0. \end{equation*}
\(\xi_1,\cdots ,\xi_r,\alpha_1,\cdots ,\alpha_s,\beta_1,\cdots ,\beta_s,\gamma_1,\cdots ,\gamma_t\)线性无关可知:
\begin{equation*} k_1=\cdots=k_r=l_1=\cdots=k_s=p_1=\cdots=p_s=q_1=\cdots=q_t=0, \end{equation*}
\(\alpha=0\),故\(V_1\bigcap W=0\)。对任意\(\beta\in V\),有
\begin{equation*} \beta=\sum\limits_{i=1}^ra_i\xi_i+\sum\limits_{j=1}^sb_j\alpha_j+\sum\limits_{j=1}^sc_j\beta_j+\sum\limits_{k=1}^td_k\gamma_k, \end{equation*}
\begin{equation*} \beta=\left[\sum\limits_{i=1}^ra_i\xi_i+\sum\limits_{j=1}^s(b_j-c_j)\alpha_j\right]+\left[\sum\limits_{j=1}^sc_j(\alpha_j+\beta_j)+\sum\limits_{k=1}^td_k\gamma_k\right]\in V_1+W, \end{equation*}
所以\(V=V_1+W\),从而\(V=V_1\oplus W\)。同理可证,\(V=V_2\oplus W\)
解答 2.
\(\dim V_1=\dim V_2=m\),对\(n-m\)作数学归纳法。
  1. \(n-m=1\)时,则\(n=m+1\)。因为\(V_1,V_2\)\(V\)的真子空间,所以存在\(\alpha\in V\text{,}\)\(\alpha\not\in V_1\cup V_2\)。令\(W=\langle\alpha\rangle\),则\(V=V_1\oplus W=V_2\oplus W\)
  2. 假设命题对\(n-m=k\)时成立,以下考虑\(n-m=k+1\)的情形。取\(\alpha\in V\),但\(\alpha\not\in V_1\cup V_2\)。令\(V_1'=V_1\oplus\langle\alpha\rangle,V_2'=V_2\oplus\langle\alpha\rangle\),则
    \begin{equation*} \dim V_1'=\dim V_2'=m+1. \end{equation*}
    此时\(n-(m+1)=(n-m)-1=k\)。由归纳假设,存在\(V\)的子空间\(W'\),使得\(V=V_1'\oplus W'=V_2'\oplus W'\)。令\(W=\langle\alpha\rangle\oplus W'\),则
    \begin{equation*} V=V_1\oplus W=V_2\oplus W. \end{equation*}

挑战题.

14.
\(V\)\(\F^m\)的子空间,且\(V\)中每个非零向量的零分量个数不超过\(r\),证明:\(\dim V\leq r+1\)
解答.
\(U=\langle \varepsilon_1,\dots,\varepsilon_{m-r-1}\rangle\),则\(U\)中向量的的非零分量个数不超过\(m-r-1\),零分量个数不少于\(r+1\)。而\(V\)中每个非零向量的零分量个数不超过\(r\),因此\(V\cap U=\{{\bf 0}\}\)。于是由维数公式得
\begin{equation*} \dim V+\dim U=\dim (U+V)\leq m. \end{equation*}
从而 \(\dim V\leq m-\dim U=r+1\)