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Let $H$ be a Hilbert Space, and $M$ a closed subspace. Is it true that

$H = M \bigoplus M^{\perp}$

Does this hold if $M$ is not closed? Or only if $H$ is finite/infinite dimensional?

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  • $\begingroup$ How exactly are you defining $\oplus$? $\endgroup$ – Nate Eldredge May 5 '12 at 23:57
  • $\begingroup$ I hadn't considered this, just the usual Cartesian product of the underlying sets I assumed? My lecturer didn't define it. What do you think? $\endgroup$ – rk101 May 6 '12 at 0:44
  • $\begingroup$ Well, some definitions say you take the completion afterwards. If you don't then it certainly won't be true for non-closed $M$. (For example, if $M$ is dense in $H$ then $M^\perp = 0$.) But if you do, then it will be true. $\endgroup$ – Nate Eldredge May 6 '12 at 1:54
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Yes, this is true. Here's how you can go about proving this:

  1. Prove that if $N\leqslant M^\perp$ then $M\oplus N$ is closed.
  2. Prove that if $N\leqslant H$ is closed and $N^\perp=0$ then $N=H$.
  3. Conclude that $M\oplus M^\perp$ is closed and $(M\oplus M^\perp)^\perp=0$.
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  • $\begingroup$ SO this is true in both infinite and finite dimensional Hilbert Spaces? What about just Inner product spaces? $\endgroup$ – rk101 May 5 '12 at 23:39

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