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$A = (a_1,a_1,a_1,a_4,a_5,...)$ is a vector in $l^2(ℕ)$. I need to show that $A = (span(b_1,b_2))^{\perp} = B^{\perp}$, where $b_1=(1,-1,0,0,...)$ and $b_2=(0,1,-1,0,...)$.

I have already shown that A ⊂ $B^{\perp}$ by using the orthogonal complement as well as defining the inner product to be $\langle a,b \rangle$ = $\sum\limits_{n = 1}^\infty{a_n}{\overline{b_n}}$, proven to be zero after expanding.

I am confused on showing $B^{\perp}⊂A$.

My approach so far: $B^{\perp}$ = $(span(b_1,b_2))^{\perp}$ = $(c_1b_1+c_2b_2)^{\perp}$, for $c_1,c_2 \in F$

I would appreciate any help.

1 Answers 1

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If $(x_1,...,x_n)$ in the orthogonal of $B$, $\langle x,(1,-1,0,...)\rangle=0$ and $\langle x,(0,1,-1,0,..)\rangle=0$. This is equivalent to $x_1-x_2=0$ and $x_2-x_3=0$, we conclude that $x_1=x_2=x_3$.