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If $ x_1=(1,2,0,0,...), x_2=(0,1,2,0,...) $ and in general $ x_i=e_i+2e_{i+1} $, find the projection of an element $ y\in\ell_2 $ into the subspace $S=\overline{sp}\{x_i\}_{i=1}^\infty$. Since $ S $ is closed and separable we know this projection will exist and be unique.

The only thing I've thought to do, which I believe is the wrong approach, is find a vector $x$ such that $x\perp S$, which I found to be $x=\sum_{k=1}^\infty\frac{(-1)^{k-1}\lambda}{2^{k-1}}e_k$ for any $\lambda\in\mathbb{C}$, then since I know $y-P_Sy\perp S$ I set up the equation $\sum_k\langle y,e_k\rangle e_k-\sum_k\alpha_kx_k=\sum_k\frac{(-1)^{k-1}\lambda}{2^{k-1}}e_k$ (which is just $y-P_Sy=x)$. I then tried to solve for $\lambda$ and $\alpha_k's$ using these equations, but I'm at a loss. Is this even a good approach?

mechanodroid
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R. Steven
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1 Answers1

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Notice that $S^\perp$ is one-dimensional. Indeed, if $y = (y_n)_{n=1}^\infty \in \ell^2$ satisfies $y \perp S$, then for every $n \in \mathbb{N}$ we have

$$0 = \langle y, e_n+2e_{n+1}\rangle = y_n + 2y_{n+1} \implies y_{n+1} = -\frac12 y_n$$

which implies $y_n = \left(-\frac12\right)^ny_1$ for all $n \in \mathbb{N}$, or $$y = y_1\left(1, -\frac12, \frac14, -\frac18, \frac1{16}, \ldots\right)$$

Hence if we denote $a = \left(1, -\frac12, \frac14, -\frac18, \ldots\right)$, we have $S^\perp = \operatorname{span}\{a\}$. The orthogonal projection on $S^\perp$ is given by $$P_{S^\perp}(x) = \left\langle x, \frac{a}{\|a\|}\right\rangle \frac{a}{\|a\|} = \frac34\left(\sum_{n=1}^\infty \left(-\frac{1}{2}\right)^{n-1}x_n\right)a$$ and the orthogonal projection on $S$ is then $$P_S = I-P_{S^\perp}.$$

mechanodroid
  • 46,490