TITLE: When does $V = \ker(f) \oplus \operatorname{im}(f)$?
QUESTION [16 upvotes]: For a vector space $V$ and a linear operator $f : V \to V$, under what conditions does $V = \ker(f) \oplus \operatorname{im}(f)$?
Is it always true, or only in special cases?
Edit: $V$ is finite dimensional.

REPLY [2 votes]: In Hilbert spaces (spaces with scalar product) you have that
$$\ker (B^*) = (\operatorname{ran} B)^\perp,$$
Where $\operatorname{ran}$ is the range, i.e., what you call image;
the $B^*$ is the adjoint operator, defined by the prescription $$(Bx,y)=(x,B^*y);$$
and $U^\perp$ is the orthogonal complement to $U$, i.e.
$$U^\perp=\{v:(\forall u\in U)((u,v)=0)\}.$$
There is a wide class of Hermitean operators, i.e. such that $B=B^*$. All these satisfy $\ker B = (\operatorname{ran} B)^\perp$ therefore $\ker B\oplus \operatorname{ran} B=V$.
In the finite-dimensional case, every Hermitean matrix (i.e., $a_{ij}=\overline{a_{ji}}$) satisfies this property.