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Can someone tell me if and why $\text{Im}\,f+\ker f=R$ holds for a selfadjoint operator $f:R\rightarrow R$, where $R$ is a finite dimensional inner product space?

Can someone give me an example of an operator in an infinite dimensional space for which that equation is not true ?

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    It is holds for any finite dimensional space and any linear operator.2012-08-19
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    Ok, but could you give me an idea why it holds ?2012-08-19
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    Represent your linear operator as a matrix. Then you immediately get $\mathrm{dim}\;\mathrm{Ker}f+\mathrm{dim}\;\mathrm{Im}f=\mathrm{dim} R$2012-08-19
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    @Norbert, the dimensions work, but is that enough for the spaces to be the same?2012-08-19
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    Here is the idea: The dimensions fit and the kernel is a subspace of the orthogonal of the image (if $Sz=0$, then, by self-adjointness, $\langle Sx,z\rangle=\langle x,Sz\rangle=0$ for every $x$), hence the kernel **is** the orthogonal of the image, QED.2012-08-19
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    But why do the dimensions fit ? I can't see how I would obtain that, if I represent $f$ by a matrix (which I know is hermitian, if the basis is orthogonal).2012-08-19
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    @Norbert: This is wrong, only the dimension holds.2012-08-19
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    My statement is wrong if we consider inner direct sum of this spaces, otherwise this correct.2012-08-19
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    @Norbert: so I guess what you wanted to say is that your statement is correct for *selfadjoint* operators, not for *any* linear operator, didn't you?2012-08-19
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    @AgustíRoig Nevermind2012-08-19

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In general, this equation is not true. Just consider the map $$ f: \ K^2 \rightarrow K^2 $$ induced by the matrix $$ \left(\begin{array} &0 & 0 \\ 1 & 0 \end{array} \right) $$ Here kernel and image are the same one-dimensional subspace. Indeed the dimension always adds up. To prove this, simply apply the homomorphism theorem. Also for inifite dimensional spaces you can apply the same construction as given here. Differentiate a function does not fulfill the equation.

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    But this map doesn't seem to be selfadjoint. Or am I mistaken ?2012-08-19
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    Summing up, I think the result is: (a) the result is true for *selfadjoint* operators (see did's comment), (b) the result is false for *any* linear map (see this example of sebigu).2012-08-19