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Is it true that in an infinitdimensional Hilbert space the formula $$\text{im} S\oplus \ker S =H$$holds, where $S:H\rightarrow H$ ? I know it is true for finitely many dimensions but I'm not so sure about infinitely many. Would it be true under some additional assumption, like assuming that the rank of $S$ is finite ?

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There are two problems with your question. 1) $S:H\to S$ isn't what you mean. $H \to H$ maybe? 2) this statement is not true in the finite dimensional case. $\dim \operatorname{im} S+ \dim \ker S = \dim H$ is true for $H$ finite dimensional and $S:H \to H$ a linear map, but $H = \operatorname{im} S \oplus \ker S$ is usually false. –  mt_ Aug 18 '12 at 12:53
    
Here's something you might mean: If $S:U\to V$ is a linear map between vector spaces $U$ and $V$, then $\operatorname{im} S\oplus\ker S\simeq H$? –  joriki Aug 18 '12 at 13:01
    
Umm...I'm quite confused since I thought the formula $$\text{im} S\oplus \ker S =H$$ were true in finitely many dimensions...Are there at least conditions like selfadjointness under which it might be true in finitely many dimensions ? –  user36772 Aug 18 '12 at 13:04
    
@joriki I'm not sure yet. Would would $H$ be ? –  user36772 Aug 18 '12 at 13:05
    
@user36772: There may be a misunderstanding regarding the use of $=$. If you really want $S$ to map from $H$ to $H$ and $=$ to mean equality, then the statement is clearly false -- for instance, if $S$ is rotation by $\pi/2$ in the plane followed by projection onto a line, then that line is both the image and the kernel of $S$. However, if by $=$ you mean isomorphism, the statement is true. If the domain and codomain were different, one would usually implicitly assume that you mean isomorphism, but with domain and codomain coinciding, equality would also make sense, so you should use $\simeq$. –  joriki Aug 18 '12 at 13:12

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Assuming you intended to ask what I proposed in a comment, the answer is yes, this also holds for infinite-dimensional vector spaces, assuming the axiom of choice. Take a basis of $\ker S$ and extend it to all of $S$. The additional basis vectors induce a basis of $S/\ker S$, which by the first isomorphism theorem is isomorphic to $\operatorname{im} S$. This established a linear bijection between $S$ and $\operatorname{im} S\oplus\ker S$. Note that there is no canonical choice for the basis of $S/\ker S$, and you need the axiom of choice to get one.

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Consider $l_2$, and the translation operator $T:~e_n\mapsto e_{n+1}$, it's injective but not surjective. So that $ker~T=0,im~T\neq l_2,ker~T\oplus im~T\neq l_2$.

If $rank~im T$ is finite, i remember i have learned that the equality holds in some book. (but it's vague to me now, so take care)

You can try to consider the coordinate function on the finite dimensional space $im~T$ in this case.

Other kinds of additional assumptions may concern with

  1. Projection operator

  2. Compact operator

  3. Fredholm operator

Try your best

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