Let \(W\) be a subspace of \({\mathbb{R}^n}\), and let \({W^ \bot }\) be the set of all vectors orthogonal to \(W\). Show that \({W^ \bot }\) is a subspace of \({\mathbb{R}^n}\) using the following steps.

1. Take \({\bf{z}}\) in \({W^ \bot }\), and let \({\bf{u}}\) represent any element of \(W\). Then \({\bf{z}} \cdot {\bf{u}} = 0\). Take any scalar \(c\) and show that \(c{\bf{z}}\) is orthogonal to \({\bf{u}}\). (Since \({\bf{u}}\) was an arbitrary element of \(W\), this will show that \(c{\bf{z}}\) is in \({W^ \bot }\).)
2. Take \({{\bf{z}}_1}\) and \({{\bf{z}}_2}\) in \({W^ \bot }\), and let \({\bf{u}}\) be any element of \(W\). Show that \({{\bf{z}}_1} + {{\bf{z}}_2}\) is orthogonal to \({\bf{u}}\). What can you conclude about \({{\bf{z}}_1} + {{\bf{z}}_2}\)? Why?
3. Finish the proof that \({W^ \bot }\) is a subspace of \({\mathbb{R}^n}\).

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if:

\(\begin{aligned}{l}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} &= {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\\{\rm{and}}\\{\bf{u}} \cdot {\bf{v}} &= 0\end{aligned}\)

Since,in step (a), \({\bf{z}}{\rm{ and }}{\bf{u}}\)are in \({W^ \bot }\) for \(c\)as any scalar,

Then, we have:

\(\begin{aligned}{c}\left( {c{\bf{z}}} \right) \cdot {\bf{u}} &= c\left( {{\bf{z}} \cdot {\bf{u}}} \right)\\ &= c\left( 0 \right)\\ &= 0\end{aligned}\)

Thus, \(c{\bf{z}}\)is orthogonal to \({\bf{u}}\).

Similarly, for \({{\bf{z}}_1}{\rm{ and }}{{\bf{z}}_2}\)are in \({W^ \bot }\), which implies,

\(\begin{aligned}{l}{{\bf{z}}_1} \cdot {\bf{u}} = 0\\{{\bf{z}}_2} \cdot {\bf{u}} = 0\end{aligned}\)

Then, we have:

\(\begin{aligned}{c}\left( {{{\bf{z}}_1} + {{\bf{z}}_2}} \right) \cdot {\bf{u}} &= {{\bf{z}}_1} \cdot {\bf{u}} + {{\bf{z}}_2} \cdot {\bf{u}}\\ &= 0 + 0\\ &= 0\end{aligned}\)

Thus, \({{\bf{z}}_1} + {{\bf{z}}_2}\) is orthogonal to \({\bf{u}}\).

As 0 orthogonal to all the vectors. So, \({W^ \bot }\)is a subspace in \({\mathbb{R}^n}\).

Hence these are the required proofs.