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• Notes

a) $U_{0}=\left\{\lambda \cdot\left(\begin{array}{l}{1} \\ {0}\end{array}\right) | \lambda \epsilon R\right\} \subset R^{2}$

(1) $\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \in U_{0} \Rightarrow \lambda=0 \Rightarrow 0\left(\begin{array}{l}{1} \\ {0}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \Rightarrow$(1)

(2) Let $u_{1}=\lambda_{1}\left(\begin{array}{l}{1} \\ {0}\end{array}\right) \in U_{0}$

$u_{2}=\lambda_{2}\left(\begin{array}{l}{1} \\ {0}\end{array}\right) \in U_{0}$

$u_{1}+u_{2}=\lambda_{1}\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\lambda_{2}\left(\begin{array}{l}{1} \\ {0}\end{array}\right)=\left(\begin{array}{c}{\lambda_{1}+\lambda_{2}} \\ {0}\end{array}\right)$

$=\left(\lambda_{1}+\lambda_{2}\right)\left(\begin{array}{l}{1} \\ {0}\end{array}\right) \in u_{0} \Rightarrow(2)$

$c u_{0}=c \lambda_{1}\left(\begin{array}{l}{1} \\ {0}\end{array}\right)=\left(\begin{array}{c}{c \lambda_{1}} \\ {0}\end{array}\right)=c \lambda_{1}\left(\begin{array}{l}{1} \\ {0}\end{array}\right) \in U_{0} \Rightarrow (3)$

from $1,2$ and $3 U_{0}$ is a subspace

b) $U_{1}=\left\{\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\lambda .\left(\begin{array}{l}{1} \\ {0}\end{array}\right) | \lambda \in R\right\} \subset R^{2}$

(1) $\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\lambda\left(\begin{array}{l}{1} \\ {0}\end{array}\right)=\left(\begin{array}{c}{\lambda+1} \\ {0}\end{array}\right)$ for $\lambda=-1 \Rightarrow\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \in U_{1} \Rightarrow (1)$

(2) Let $u_{1}=\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\lambda_{1}\left(\begin{array}{l}{1} \\ {0}\end{array}\right)$

$u_{2}=\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\lambda_{2}\left(\begin{array}{l}{1} \\ {0}\end{array}\right)$

$u_{1}+u_{2}=\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\lambda_{1}\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\lambda_{2}\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\left(\begin{array}{l}{1} \\ {0}\end{array}\right)=\left(\begin{array}{c}{\lambda_{1}+\lambda_{2}+2} \\ {0}\end{array}\right)$

$=\left(\begin{array}{c}{1} \\ {0}\end{array}\right)+\left(\begin{array}{c}{\lambda_{1}+\lambda_{2}+1} \\ {0}\end{array}\right)=\left(\begin{array}{c}{1} \\ {0}\end{array}\right)+\left(\lambda_{1}+\lambda_{2}+1\right)\left(\begin{array}{c}{1} \\ {0}\end{array}\right) \in U_{1} \Rightarrow (2)$

$c u_{1}=c\left(\begin{array}{l} {1} \\ {0} \end{array}\right)+c \lambda_{1}\left(\begin{array}{l} {1} \\ {0} \end{array}\right)=\left(\begin{array}{c}{c+c \lambda_{1}} \\ {0}\end{array}\right)$

$=\left(\begin{array}{c}{c+c \lambda_{1}+1-1} \\ {0}\end{array}\right)=\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\left(\begin{array}{c}{c+c \lambda_{1}-1} \\ {0}\end{array}\right)$

$=\left(\begin{array}{l}{1} \\ {0}\end{array}\right)+\left(c+c \lambda_{1}-1\right)\left(\begin{array}{c}{1} \\ {0}\end{array}\right) \in U_{1} \Rightarrow(3)$

from $1,2,$ and $3\ U_{1}$, is a vector space

C) $U_{3}=\left\{\left(\begin{array}{l}{x} \\ {y}\end{array}\right) | 2 x-y=0\right\} \subset R^{2}$

(1) $\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \in U_{3} \Rightarrow\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \Rightarrow 2(0)-(0)=0 \quad \Rightarrow (1)$

(2) Let $u=\left(\begin{array}{ll}{u_{1}} \\ {2 u_{1}}\end{array}\right) \in U_{3}$

$v=\left(\begin{array}{l}{v_{1}} \\ {2 v_{1}}\end{array}\right) \in U_{3}$

$u+v=\left(\begin{array}{l}{u_{1}+v_{1}} \\ {2 u_{1}+2 v_{1}}\end{array}\right)=\left(\begin{array}{l}{u_{1}+v_{1}} \\ {2\left(u_{1}+v_{1}\right)}\end{array}\right) \in U_{3}=(2)$

(3) $c u=c\left(\begin{array}{c}{u_{1}} \\ {2 u_{1}}\end{array}\right)=\left(\begin{array}{c}{c u_{1}} \\ {c 2 u_{1}}\end{array}\right)=\left(\begin{array}{c}{cu_{1}} \\ {2 c u_{1}}\end{array}\right) \in U_{3} \Rightarrow(3)$

d) $U_{4}=\left\{\left(\begin{array}{l}{x} \\ {y}\end{array}\right) | 2 x-y=1\right\} \subset R^{2}$

$\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \Rightarrow 2(0)-(0)=0 \neq 1 \Rightarrow\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \notin U_{4}$

$U_{4}$ is not a subspace

e) $U_{2}=\left\{x \in R^{3} |\|x\| \leq 1\right\} \subset R^{3}$

(1) $\left(\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right) \in U_{2}$

(2) Let $u_{1}=\left(\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right), u_{2}=\left(\begin{array}{l}{0} \\ {0} \\ {1}\end{array}\right) \in U_{2}$

$u_{1}+u_{2}=\left(\begin{array}{l}{1} \\ {0} \\ {1}\end{array}\right) \Rightarrow\left\|u_{1}+u_{2}\right\|=\sqrt{1^{2}+0+1^{2}}=\sqrt{2}>1, \ u_{1}+u_{2} \notin U_{2}$

\ $U_{2}$ is not a subspace