1.
设\(U,V\)是数域\(\mathbb{F}\)上的线性空间，记
\begin{equation*}
U\times V=\{(\alpha,\beta)\ |\ \alpha\in U,\beta\in V\}.
\end{equation*}
\(\forall (\alpha_1,\beta_1),(\alpha_2,\beta_2)\in U\times V,\forall c\in\mathbb{F}\)，规定
\begin{equation*}
(\alpha_1,\beta_1)+(\alpha_2,\beta_2)=(\alpha_1+\alpha_2,\beta_1+\beta_2),\
c(\alpha_1,\beta_1)=(c \alpha_1,c \beta_1).
\end{equation*}
- 证明：\(U\times V\)关于以上运算构成数域\(\mathbb{F}\)上的线性空间；
- 已知\(\dim U=m,\dim V=n\)，求\(\dim (U\times V)\)，并说明理由。
解答.
-
\((0,0)\in U\times V\)，所以\(U\times V\)是非空集合。对任意\((\alpha_1,\beta_1),(\alpha_2,\beta_2)\in U\times V,c\in\mathbb{F}\)，由\(U,V\)是数域\(\mathbb{F}\)上的线性空间得\begin{equation*} (\alpha_1,\beta_1)+(\alpha_2,\beta_2)=(\alpha_1+\alpha_2,\beta_1+\beta_2)\in U\times V, \end{equation*}\begin{equation*} c(\alpha_1,\beta_1)=(c \alpha_1,c \beta_1)\in U\times V, \end{equation*}即\(U\times V\)关于以上运算封闭。对任意\((\alpha_1,\beta_1),(\alpha_2,\beta_2),(\alpha_3,\beta_3)\in U\times V,c,d\in\mathbb{F}\)，有
- \begin{equation*} \begin{array}{ccl} (\alpha_1,\beta_1)+(\alpha_2,\beta_2)&=&(\alpha_1+\alpha_2,\beta_1+\beta_2)\\ &=&(\alpha_2+\alpha_1,\beta_2+\beta_1)\\&=&(\alpha_2,\beta_2)+(\alpha_1,\beta_1); \end{array} \end{equation*}
- \begin{equation*} \begin{array}{ccl} & &\left((\alpha_1,\beta_1)+(\alpha_2,\beta_2)\right)+(\alpha_3,\beta_3)\\ &=&(\alpha_1+\alpha_2,\beta_1+\beta_2)+(\alpha_3,\beta_3)\\ &=&\left((\alpha_1+\alpha_2)+\alpha_3,(\beta_1+\beta_2)+\beta_3\right)\\&=&\left(\alpha_1+(\alpha_2+\alpha_3),\beta_1+(\beta_2+\beta_3)\right)\\ &=&(\alpha_1,\beta_1)+(\alpha_2+\alpha_3,\beta_2+\beta_3)\\ &=&(\alpha_1,\beta_1)+\left((\alpha_2,\beta_2)+(\alpha_3,\beta_3)\right); \end{array} \end{equation*}
- 存在\((0,0)\in U\times V\)，使得对任意\((\alpha,\beta)\in U\times V\)，有\begin{equation*} (\alpha,\beta)+(0,0)=(\alpha,\beta); \end{equation*}
- 对任意\((\alpha,\beta)\in U\times V\)，存在\((-\alpha,-\beta)\in U\times V\)，使得\begin{equation*} (\alpha,\beta)+(-\alpha,-\beta)=(0,0); \end{equation*}
- \begin{equation*} \begin{array}{ccl} c\left((\alpha_1,\beta_1)+(\alpha_2,\beta_2)\right)&=&c(\alpha_1+\alpha_2,\beta_1+\beta_2)\\ &=&\left(c(\alpha_1+\alpha_2),c(\beta_1+\beta_2)\right)\\ &=&(c\alpha_1+c\alpha_2,c\beta_1+c\beta_2)\\ &=&(c\alpha_1,c\beta_1)+(c\alpha_2, c\beta_2)\\ &=&c(\alpha_1,\beta_1)+c(\alpha_2,\beta_2) \end{array} \end{equation*}
- \begin{equation*} \begin{array}{ccl} (c+d)(\alpha_1,\beta_1)&=&\left((c+d)\alpha_1,(c+d)\beta_1\right)\\&=&(c\alpha_1+d\alpha_1,c\beta_1+d\beta_1)\\ &=&(c\alpha_1,c\beta_1)+(d\alpha_1,d\beta_1)\\&=&c(\alpha_1,\beta_1)+d(\alpha_1,\beta_1); \end{array} \end{equation*}
- \begin{equation*} \begin{array}{ccl} (cd)(\alpha_1,\beta_1)&=&\left((cd)\alpha_1,(cd)\beta_1\right)\\&=&\left(c(d\alpha_1),c(d\beta_1)\right)\\ &=&c(d\alpha_1,d\beta_1)\\&=&c\left(d(\alpha_1,\beta_1)\right); \end{array} \end{equation*}
- \(\displaystyle 1(\alpha_1,\beta_1)=(1\alpha_1,1\beta_1)=(\alpha_1,\beta_1).\)
因此\(U\times V\)关于以上运算构成数域\(\mathbb{F}\)上的线性空间。 -
设\(\xi_1,\xi_2,\cdots ,\xi_m\)是\(U\)的一个基，\(\eta_1,\eta_2,\cdots ,\eta_n\)是\(V\)的一个基。我们断言，\begin{equation*} (\xi_1,0),(\xi_2,0),\cdots ,(\xi_m,0),(0,\eta_1),(0,\eta_2),\cdots ,(0,\eta_n) \end{equation*}是\(U\times V\)的一个基。事实上，设\begin{equation*} k_1(\xi_1,0)+\cdots +k_m(\xi_m,0)+l_1(0,\eta_1)+\cdots +l_n(0,\eta_n)=(0,0), \end{equation*}即\((k_1 \xi_1+k_2 \xi_2+\cdots +k_m \xi_m,l_1\eta_1+l_2\eta_2+\cdots +l_n\eta_n)=(0,0),\)则\begin{equation*} k_1 \xi_1+k_2 \xi_2+\cdots +k_m \xi_m=0\mbox{且}l_1\eta_1+l_2\eta_2+\cdots +l_n\eta_n=0. \end{equation*}由\(\xi_1,\xi_2,\cdots ,\xi_m\)及\(\eta_1,\eta_2,\cdots ,\eta_n\)线性无关，得\begin{equation*} k_1=k_2=\cdots =k_m=l_1=l_2=\cdots =l_n=0. \end{equation*}故\((\xi_1,0),(\xi_2,0),\cdots ,(\xi_m,0),(0,\eta_1),(0,\eta_2),\cdots ,(0,\eta_n)\)线性无关。另一方面，对任意\((\alpha,\beta)\in U\times V\)，因\(\alpha\in U\)且\(\xi_1,\xi_2,\cdots ,\xi_m\)是\(U\)的一个基，所以存在\(a_1,a_2,\cdots ,a_m\in\mathbb{F}\)使得\(\alpha=a_1 \xi_1+a_2 \xi_2+\cdots +a_m \xi_m\)。同理，存在\(b_1,b_2,\cdots ,b_n\in\mathbb{F}\)，使得\(\beta=b_1 \eta_1+b_2\eta_2+\cdots +b_n\eta_n\)。于是，存在\(a_1,a_2,\cdots ,a_m,b_1,b_2,\cdots ,b_n\in\mathbb{F}\)，使得\begin{equation*} (\alpha,\beta)=a_1(\xi_1,0)+\cdots +a_m(\xi_m,0)+b_1(0,\eta_1)+\cdots +b_n(0,\eta_n). \end{equation*}因此\((\xi_1,0),\cdots ,(\xi_m,0),(0,\eta_1),\cdots ,(0,\eta_n)\)是\(U\times V\)的一个基，\begin{equation*} \dim (U\times V)=m+n. \end{equation*}
