Consider the inner product space $l_0$ consisting of all infinite sequences of complex numbers with only finitely many non-zero terms, with the inner product of $l^2$ (space of square summable sequences): $$<x,y>=x_1\overline y_1 +x_2\overline y_2 + …$$ for $x = (x_1,x_2,…,x_n,…)$ and $y =(y_1,y_2,…,y_n,…)$.
Let $a=(1,\dfrac12,\dfrac1{2^2},\dfrac1{2^3},…)\in l^2$.
$1)$ Show that $W = \{x\in l_0 : <x,a>=0\}$ is a subspace of $l_0$. Is $W$ finite dimensional or infinite dimensional?
$2)$ Show that $W^\perp=\{y\in l_0: <x,y>=0 \text{ for all } x\in W\}=\{0=(0,0,…0,…)\}$.
I have shown that $W$ is a subspace of $l_0$
And I can find a linearly independent set of sequences $\{b_1,b_2,…\}
\subseteq W$ which spans the space of all sequences.
$b_1=(1,-2,0,0,…..)$
$b_2=(0,1,-2,0,0,….)$
$b_3=(0,0,1,-2,0,0,…)$
…
$b_n=(0,0,0,…,1,-2,0,0,…)$
Any sequence $(x_1,x_2,x_3,…)$ can be expressed as linear combination of $\{b_1,b_2,…\}$ hence they also span $l_0$.
Is this enough to conclude that $W$ is infinite dimensional since it is spanned by infinitely many independent elements? Or do I have to show that $\{b_1,b_2,…\}$ form orthogonal basis?
And I have no idea for part $2)$
Best Answer
Let $y \in W^\perp$, then using your definition, $\forall i, <y,b_i>=0$. For example, $<y,b_1>=0,$ or $y_1 = 2y_2$. By induction, it is easy to prove that $y_1 = 2^{n-1} y_n$. But $y$ has finitely many non-zero terms. Therefore, necessarily, $y_1 = 0$ and finally $y = 0.$