If $f:\ V\longrightarrow V$ be linear then how the fact $V=ker\,f+im\,f$ relate to the diagonalization / Jordanization of $f$?

Question

Let $V$ be a n-dimensions vector space and $f$ be a linear map on $V$. Then is there any relevant between $V=ker\,f+im\,f$ and the diagonalization/Jordanization of $f$ ? Thanks

Answer 1

Given the rank-nullity theorem, if $V = \ker T + \operatorname{im} T$, then the sum is direct, and hence $\ker T \cap \operatorname{im} T = \{0\}$. Hence, $$T^2 v = 0 \implies Tv \in \ker T \cap \operatorname{im} T = \{0\} \implies Tv = 0,$$ and so $\ker T^2 = \ker T$. Thus, the generalised eigenspace and eigenspace corresponding to $0$ agree (if $0$ is an eigenvalue), which is equivalent to all Jordan blocks with $0$ on the diagonal are $1 \times 1$.

The converse holds true too. If $\ker T^2 = \ker T$, then any $v \in \ker T \cap \operatorname{im} T$ must satisfy $v = Tw$ for some $w$ and $Tv = T^2w = 0$. Thus, $Tw = 0$, hence $v = 0$, i.e. $\ker T \cap \operatorname{im} T = \{0\}$. Thus, the two spaces sum directly, so by the rank-nullity theorem, their sum must be all of $V$.