设 [tex=0.857x1.286]ZpwhzmyivskaH5M1X7ozaQ==[/tex] 为 [tex=0.786x1.286]dSWbQCTjdbLxKy7q0ps2gg==[/tex] 上 [tex=0.643x0.786]/he/ol8BkDuTTL9yMPtH4Q==[/tex] 维线性空间. [tex=4.5x1.071]d9y+KLOQpwgiIZSOmy+NwuKjcx8pDZqu3h/Q7GEiRUWsitTO5mMLH8wF4N4pZRWx[/tex]  [tex=1.0x1.0]XyzqB7VduQKuYOSYbtY4TQ==[/tex] 为 [tex=0.857x1.286]ZpwhzmyivskaH5M1X7ozaQ==[/tex] 的子空间. [tex=9.143x1.357]N9dOzoEP62XohFERsEMpa2KyZcTDUOKkxV5al4c4zXfioI+38ZHntN3Pf7e2CD2sm8trZqWKDvESKlRpunPzeQ==[/tex] 证明 [tex=15.857x1.357]NovbxKl63Ey/milqTcbe/77BIJFM7Bwv6Jpu9w6yQOABriEkIESbhU5mAczCl7oMD+WEWFyQOwQdyq+Z1W8FB4lvgYHEQQPEhMmak8OGud90OalRo/QhPADZxVhN2wEObF+/N1LAVZ4Vhj8aQCAc5m7mMHeHlS1rpzgJSni4Asw=[/tex]证 $\quad$ 因为 $\mathcal{A}$ 限制在 $W$ 上是 $W$ 到 $\mathcal{A}(W)$ 的线性映射，而此限制的核为 $\operatorname{ker} \mathcal{A} \cap$ $W$，于是 $\operatorname{dim} \mathcal{A}(W)+\operatorname{dim}(\operatorname{ker} \mathcal{A} \cap W)=\operatorname{dim} W$..[br][/br]

$\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) A: $(\frac{1}{2},\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ B: $(1,\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ C: $(\frac{1}{2},1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ D: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$