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Given $n \in \mathbb{Z}^+$ and $M = \{ (x_1,x_2,..,x_n,0,0,...) \mid x_1,x_2,..,x_n \in \mathbb{R} \}$. Find $M^{\perp}$ in $l^2$?

I can show $M$ is a closed subspace of $l^2$ and a Hilbert space.

Let $P_M : l^2 \to M$: $$x= (x_1,x_2,..,x_n,..) \mapsto (x_1,x_2,..,x_n,0,0,..)$$

$P_M$ is a projection from $l^2$ onto $M$.

But how can I find $M^{\perp}$, $\mathrm{Im} P_M$ and $\mathrm{Ker} P_M$ ?

Jonas Dahlbæk
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Desunkid
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1 Answers1

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Let $X$ be a Hilbert space and $P_H$ be the projection onto a subspace $H$ of $X$. We always have $X = P_HX\oplus (P_HX)^{\perp}$. Recall that the direct sum of two subspaces is the space of all sums of elements from the two spaces assuming they only have zero in common. Specifically, if $X = \ell^2$ and $H$ is a finite projection of $x \in \ell^2$ to its first $n$ coordinates, call this $x^{(n)}$, it follows that $y \in (P_H\ell^2)^{\perp}$ if and only if $$ y = x - x^{(n)} = (0,0,\dots,0,x_{n+1},x_{n+1}, \dots). $$

JMJ
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