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Concerning this question I asked (specifically my last comment), is it true that the formula $$L\oplus \ker S =H$$ holds in a finite dimensional vector space, where $S:H\rightarrow H$ is linear operator, $(b_1,\ldots,b_k)$ is a basis of the image of $S$ and $L$ consists of the preimages of the $b_i$ under $S$, i.e. $l_j\in L$ is such that $l_j\in S^{-1} (b_j)$ ?

If we equip $H$ with an inner product, such that $S$ becomes selfadjoint, would then $$\text{im} S\oplus \ker S =H$$ hold ? (Or can you give a counterexample?)

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No, the statement does not hold: $L$ isn't a vector space. If you define $L$ to be the span of (some choice of) preimages of the $b_i$ it will be true: by rank-nullity all you need prove is that $L \cap \ker S = \{ 0 \}$, but this is clear because $f|_L$ is one-to-one.

Suppose $S$ is self-adjoint with respect to some inner product $(,)$. Let $x \in \ker S \cap \operatorname{im} S$. Then $(x,x) = (x,Sy)$ for some $y$, and by self-adjointness, $(x,Sy)=(Sx,y)=(0,y)=0$. So $\ker S \cap \operatorname{im} S = \{0\}$. It follows that $H = \ker S \oplus \operatorname{im} S$.

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I understand, that this argument with selfadjointness also works in the infinite dimensional setting ? –  user36772 Aug 18 '12 at 14:40
    
it shows that $\ker S \cap \operatorname{im} S = \{0\}$ in the infinite-dimensional setting. –  mt_ Aug 18 '12 at 14:42
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If you have $S : H \to H$ a projection, i.e. a linear map such that $S^2 = S$ then it is true that

$$H = \ker S \oplus \textrm{im} S.$$

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It is true that if $L$ is any complementary subspace of $\ker S$, a space that satisfies $H=L\oplus\ker S$, then $S$ maps $L$ isomorphically to $\mathrm{im}\,S$. But you cannot define $L$ as the set of the preimages of the $b_i$ under S, since that is not a vector space but rather the union of $k$ distinct cosets of $\ker S$ (for any preimage, adding anything in $\ker S$ to it gives a preimage of the same element, and these fill one such coset). What you probably intend is to choose a preimage for every $b_i$ and then take the subspace spanned by those vectors; this is indeed a way to obtain a complementary subspace of $\ker S$ (and every choice of preimages gives a different $L$).

As for the other statement, $S$ being selfadjoint is indeed sufficient to have $\text{im}\, S\oplus \ker S =H$, but it is nowhere near necessary. Indeed $S$ diagonalisable is also sufficient, the direct sum then being that of the sum of eigenspaces for all nonzero eigenvalues, and of $0$-eigenspace (kernel).

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