Proving that a basis exists for a vector Subspace

Question

let $P=span${$\vec{a_1}, \vec{a_2}$} be a plane in $\Bbb{R^4}$ and let $S$ be a subspace such that $S=$ {$\vec{x} \in\Bbb{R^4}|\vec{x} \cdot \vec{a_1}=0, and, \vec{x} \cdot \vec{a_2}=0 $} how would you prove that there exist $\vec{a_3}, \vec{a_4} \in \Bbb{R^4}$ such that {$\vec{a_3}, \vec{a_4}$} is a basis for $S$

Answer 1

You cannot choose $\vec{a_3}=\vec{a_1}$ because $\vec{a_1}\not\in S$, because $\vec{v_1}\cdot\vec{v_1}\ne 0$. Similar for $\vec{v_4}$.

How it is actually proven depends on what you can assume is already known. One way is to:

• Take the basis $\{\vec{v_1}, \vec{v_2}\}$,
• Extend it to a basis $\{\vec{v_1}, \vec{v_2}, \vec{v_3}, \vec{v_4}\}$ for $\mathbb R^4$,
• Apply the Gramm-Schmidt algorithm to this basis, and this will give you an orthonormal basis $\{\vec{v'_1}, \vec{v'_2}, \vec{v'_3}, \vec{v'_4}\}$.

Now it is easy to prove that $\operatorname{span}(\vec{v'_1}, \vec{v'_2})=\operatorname{span}(\vec{v_1}, \vec{v_2}) = P$, and $\operatorname{span}(\vec{v'_3}, \vec{v'_4})=S$, so as the basis of $S$ you can take $\{\vec{v'_3}, \vec{v'_4}\}$.

My problem here is whether I can assume you are already familiar with Gramm-Schmidt algorithm and with this construction, or is this problem a sort of introduction to that material.