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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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    @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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    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