Exercise 1.3.8

Question

Determine whether the following sets are subspaces of $\mathrm{R}^{3}$ under the operations of addition and scalar multiplication defined on $\mathrm{R}^{3}$. Justify your answers.
\begin{enumerate}[label=(\alph*)]
\item $\mathrm{W}_{1}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in \mathrm{R}^{3}: a_{1}=3 a_{2}\right.$ and $\left.a_{3}=-a_{2}\right\}$
\item $\mathrm{W}_{2}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in \mathrm{R}^{3}: a_{1}=a_{3}+2\right\}$
\item $\mathrm{W}_{3}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in \mathrm{R}^{3}: 2 a_{1}-7 a_{2}+a_{3}=0\right\}$
\item $\mathrm{W}_{4}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in \mathrm{R}^{3}: a_{1}-4 a_{2}-a_{3}=0\right\}$
\item $\mathrm{W}_{5}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in \mathrm{R}^{3}: a_{1}+2 a_{2}-3 a_{3}=1\right\}$
\item $\mathrm{W}_{6}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in \mathrm{R}^{3}: 5 a_{1}^{2}-3 a_{2}^{2}+6 a_{3}^{2}=0\right\}$
\end{enumerate}

Jephian C.-H. Lin · Accepted Answer

Just check whether it's closed under addition and scalar multiplication and whether it contains $0$. And here $s$ and $t$ are in $\mathbb{R}$.
\begin{enumerate}
\item Yes. It's a line $t(3,1,-1)$.
\item No. It contains no $(0,0,0)$.
\item Yes. It's a plane with normal vector $(2,-7,1)$.
\item Yes. It's a plane with normal vector $(1,-4,-1)$.
\item No. It contains no $(0,0,0)$.
\item No. We have both $(\sqrt{3},\sqrt{5},0)$ and $(0,\sqrt{6},\sqrt{3})$ are elements of $W_6$ but their sum $(\sqrt{3},\sqrt{5}+\sqrt{6},\sqrt{3})$ is not an element of $W_6$.
\end{enumerate}

Exercise 1.3.8

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Exercise 1.3.8

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