Suppose $\{v_1, ....,v_m\}\in V$. Prove that $\{v_1, ....,v_m\}^\perp=(\text{span}\{v_1, ....,v_m\})^\perp$

Question

I intend to answer this question myself. I am reviewing my copy of Sheldon Axler's excellent "Linear Algebra Done Right", and wrote up a solution for this problem, only to have trouble finding it on the net to verify. It certainly isn't on Math.SE that I could find, and I think it is a subtle enough problem that others looking for it will benefit, so I will post the question and give my answer shortly.

Suppose $\{v_1, ....,v_m\}\in V$, where $V$ is a vector space. Prove that $\{v_1, ....,v_m\}^\perp=(\text{span}\{v_1, ....,v_m\})^\perp$.

Answer 1

Let $\langle\cdot,\cdot\rangle$ be the inner product on $V$. First, suppose $v\in(\text{span}\{v_1,...,v_m\})^\perp$. Then $\forall w\in\text{span}\{v_1,...,v_m\}$ we have $\langle v,w\rangle =0$. But since $v_i\in\text{span}\{v_1,...,v_m\}\;\forall i\in[m]$, this implies that $\langle v,v_i\rangle = 0\;\forall i\in[m]$, hence $v\in\{v_1,...,v_m\}^\perp$, so we have that $(\text{span}\{v_1,...,v_m\})^\perp\subseteq\{v_1,...,v_m\}^\perp$.

For the reverse inclusion, let $v\in\{v_1,...,v_m\}^\perp$. Suppose $w\in(\text{span}\{v_1,...,v_m\})$. Then, by the definition of the span, there exist scalars $a_1,...,a_m$ such that $w=a_1v_1+\cdots+a_mv_m$. Then, we have $\langle w,v\rangle = \langle a_1v_1+\cdots+a_mv_m, v\rangle$, and by the linearity of the inner product in both slots, we have that this is equal to:

$$a_1\langle v_1,v\rangle+\cdots+a_m\langle v_m,v\rangle,$$ but since $v\in\{v_1,...,v_m\}^\perp$, we have that $\langle v_i,v\rangle=0\;\forall i\in[m]$, and so $\langle w,v\rangle=0$, showing that $v\in(\text{span}\{v_1,...,v_m\})^\perp$, and that $\{v_1,...,v_m\}^\perp\subseteq(\text{span}\{v_1,...,v_m\})^\perp$.

By double inclusion, the claim is proved.

I intentionally was extremely detailed. Comments are welcome.

Answer 2

If $v\in\{v_1,\ldots,v_m\}^\bot$ then $v\cdot v_j=0$ for every $1\leq j\leq m$. Pick some $w\in\text{span}\{v_1,\ldots,v_m\}$. By definition you can write $w=\lambda_1v_1+\cdots+\lambda_mv_m$ for some scalars $\lambda_1,\ldots,\lambda_m$, but this means that $$ v\cdot w=v\cdot(\lambda_1v_1+\cdots+\lambda_mv_m)=\lambda_1v\cdot v_1+\cdots+\lambda_mv\cdot v_m=0. $$ As $w$ was arbitrary we get that $v\in\left(\text{span}\{v_1,\ldots,v_m\}\right)^\bot$. This proves the inclusion $$\{v_1,\ldots,v_m\}^\bot\subset\left(\text{span}\{v_1,\ldots,v_m\}\right)^\bot.$$ The other one is trivial because $A\subset B\Rightarrow B^\bot\subset A^\bot$ always holds.