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Let $H$ be a Hilbert space and $T: H \rightarrow H$ a bounded linear operator. Let $N = ker(T)$. It is well-known that $H = N \oplus N^{\perp}$.

Is it also true that $H = N \oplus M$, where $M$ is a closed linear subspace and $M$ is not necessarily $N^{\perp}$? What would such $M$ be?

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Given a Hilbert space $H$ with inner $\langle\cdot,\cdot\rangle$, there are many topologically equivalent inner products $\langle \cdot,\cdot\rangle'$ that are not the same as the original. For any closed subspace $N$, there are corresponding orthogonal complements $N^{\perp}$ and $N^{\perp'}$, potentially leading to different direct sum decompositions that are not the same: $$ H=N\oplus N^{\perp} = N\oplus N^{\perp'}. $$ Every such inner product $\langle \cdot,\cdot\rangle'$ can be written as $$ \langle x,y\rangle' = \langle Ax,y\rangle $$ for a unique bounded selfadjoint $A$ on $H$ that is continuously invertible. Any such $A$ can be used to generate a topologically equivalent inner product, and some $A$ can lead to different decompositions.

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