Set inclusion question Suppose I want to show ${\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}=span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}}\}$
If I do this with set inclusions, for the forward direction I would start with $x\in \{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$, and by definition,only  $\begin{bmatrix} 0 \\ 0 \\ a \end{bmatrix}$ and $\begin{bmatrix} 2b \\ -b \\ 0 \end{bmatrix}$ would be in $\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$ because their inner product with $\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}$ is $0$, and so it is the span of of the right hand side. My question is, how do I do the inclusion formally? Like starting from $x\in \{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$ and ending at $x \in span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}\}$? I have no clue when dealing the above with an  "$x$"
 A: If $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \left\{ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \right\}^\perp$$ this means $x_1+2x_2=0$, that is, $x_1=-2x_2$. Thus $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2x_2 \\ x_2 \\ x_3 \end{bmatrix} = -x_2 \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \in \operatorname{span} \left\{ \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}.$$
A: Let $a= \begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}$, $b= \begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix}$ and  $c=\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}$
$V:=\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\} ^{\perp} $ and $ U:={span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}}\}$

Now,take any $x\in V$ then $x= m\cdot a$ so: $$\langle x,b\rangle  = m\langle a,b\rangle =0$$ and $$\langle x,c\rangle  = m\langle a,c\rangle =0$$ so $x\in U$ and thus $\boxed{V\subseteq U}$.

For converse, take any $x\in U$, then $x= m\cdot b+n\cdot c$ and we have $$\langle a,x\rangle  = m\langle a,b\rangle +n\langle a,c\rangle  = 0$$  so $x\in V$ and thus $\boxed {U\subseteq V}$ and so $U=V$.
