Exercise 6.C.1

Proof. Suppose $w\in {\{v_1,\dots ,v_m\}}^{\text{⊥}}$. Let $v=\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m)$. We have that

for some $a_1,\dots ,a_m\in 𝔽$. Moreover

Thus $w\in {(\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m))}^{\text{⊥}}$ and so ${\{v_1,\dots ,v_m\}}^{\text{⊥}}\subset {(\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m))}^{\text{⊥}}$.

Now suppose $w\in {(\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m))}^{\text{⊥}}$. Since each $v_j$ is in $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m)$, it follows that $w$ is orthogonal to each $v_j$. Therefore $w\in {\{v_1,\dots ,v_m\}}^{\text{⊥}}$ and thus ${(\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m))}^{\text{⊥}}\subset {\{v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m\}}^{\text{⊥}}$. □