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.
[Math] When does $V = \ker(f) \oplus \operatorname{im}(f)$
linear algebra
Best Answer
$\def\im{\operatorname{im}}$As from the rank-nullity theorem as Matt remarked we always have $$ \dim\ker f + \dim\im f = \dim V $$ the only thing that is missing in the finite-dimensional case, is $\ker f \cap \im f = 0$, that is $f|_{\im f}\colon \im f \to V$ has to be one-to-one.
This is sometimes true, for example for projections (that is idempotent) $f$ where you have $f^2 = f$, and hence a $y =f(x)\in \im f$ with $f(y) = 0$ has $$0 = f(y) = f^2(x) = f(x) = y.$$ Sometimes it is not true, consider for example $$ f = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} \colon \mathbb R^2 \to \mathbb R^2 $$ Here you have $\ker f = \im f = \operatorname{span} e_1$, so $f|_{\im f} = 0$ is not one-to-one.
I'm not quite sure if something general can be said in the infinite-dimensional-case.