Exercise 6.2.23

for all $n\ge N$. But this means that all of them are vectors in ${𝔽}^N$. So it’s an inner product.

And it’s orthonormal since

(a)

If $e_1$ is an element in $W$, we may write

$$e_1=a_1{\sigma }_1+a_2{\sigma }_2+\cdots +a_k{\sigma }_k,$$

where $a_i$ is some scalar. But we may observe that $a_i$ must be zero otherwise the $i$-th entry of $e_1$ is nonzero, which is impossible. So this means that $e_1=0$. It’s also impossible. Hence $e_1$ cannot be an element in $W$.

(b)

If $a$ is a sequence in $W^{\text{⊥}}$, we have $a(1)=-a(n)$ for all $i$ since

$$⟨a,{\sigma }_n⟩=a(1)+a(n)=0$$

for all $i$. This means that if $a$ contains one nonzero entry, then all entries of $a$ are nonzero. This is impossible by our definition of the space $V$. Hence the only element in $W^{\text{⊥}}$ is zero.

On the other hand, we have $W^{\text{⊥}}={\{0\}}^{\text{⊥}}=V$. But by the previous argument, we know that $W\ne V={(W^{\text{⊥}})}^{\text{⊥}}$.