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Let $H$ be a separable Hilbert space, and $\{e_n\}$ an orthonormal basis for $H$. Is $\{pe_i\}$ an orthogonal basis for $K$, if $K$ is a closed subspace of $H$, and $p:H\to K$ is an orthogonal projection?

I know for every $\xi\in K$, $$\sum c_ie_i =\xi=p\xi=\sum c_ipe_i$$where $\{c_i\}\in \ell^2$. But what about Perpendicular? Is $\langle pe_i,pe_j\rangle = 0$?

niki
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    If $H$ is finite-dimensional and $K$ is strictly smaller dimensional, then obviously ${ p e_i }$ is not going to be linearly independent (and therefore they can't be mutually orthogonal either assuming they're all nonzero). – Daniel Schepler Jun 20 '17 at 21:19
  • @DanielSchepler , Right. What's your suggestion to extract a basis for $K$ of ${e_i}$? – niki Jun 20 '17 at 21:38
  • There's always Gram-Schmidt... – Nate Eldredge Jun 20 '17 at 21:40
  • @NateEldredge Hi. In general, isn't it true that the restriction of any orthonormal basis ${e_n}_{n=1}^\infty$ for $H$ to a closed subspace $S$ (that is, the set ${e_j: e_j \in S}$) an orthonormal basis for $S$? – asrxiiviii Jul 09 '22 at 22:26
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    @asrxiiviii: No. In fact, the set ${e_j : e_j \in S}$ may very well be empty. Consider $H = \mathbb{R}^2$, the usual orthonormal basis ${e_1, e_2}$, and the subspace $S$ spanned by $e_1 + e_2$. – Nate Eldredge Jul 11 '22 at 03:57

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