TITLE: Given a linear map $T:V\to V$, is it true that $V=\ker(T) \oplus \mathrm{im}(T)$?
QUESTION [5 upvotes]: I was wondering if $V=\ker(T) \oplus \mathrm{im}(T)$ if $T:V \to V$. I know the theorem that if $T:V\to W$ is linear then $\dim(V) = \dim(\ker(T)) + \dim(\mathrm{im}(T))$. This should imply $V=\ker(T) \oplus \mathrm{im}(T)$ because if $\dim(V) = \dim(\ker(T)) + \dim(\mathrm{im}(T))$ it mean that a basis of $\ker(T)$ plus a basis of $\mathrm{im}(T)$ forms a basis of $V$ which is the same like  $V=\ker(T) \oplus \mathrm{im}(T)$. Is it correct?

REPLY [4 votes]: If $V$ is a finite dimensional vector space it is true up to isomoprhism. Let $T:V\rightarrow V$ be a linear map. Then there exists the short exact sequence 
$$0\longrightarrow \ker(T)\xrightarrow{\;\mathrm{id}\;} V\xrightarrow{\;\pi\;} V/\ker(T)\longrightarrow 0$$
The sequence splits because $V/\ker(T)$ is a free module. So $V\cong \ker(T)\oplus V/\ker(T)$. Finally by first isomorphism theorem $V/\ker(T)\cong\mathrm{im}(T)$. So the statement is true if you replace the $=$ with $\cong$.