TITLE: When  image + kernel span the whole space
QUESTION [5 upvotes]: 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 ?

REPLY [2 votes]: 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.