Endomorphism/Trigonalizable/Canonical additive decomposition/Fact/Proof

Due to fact, we have

${\displaystyle {}V=H_{1}\oplus \cdots \oplus H_{m}\,,}$

where the ${\displaystyle {}H_{i}}$ are the generalized eigenspaces for the eigenvalues ${\displaystyle {}\lambda _{i}}$, and we have

${\displaystyle {}\varphi =\varphi _{1}\oplus \cdots \oplus \varphi _{m}\,}$

with ${\displaystyle {}\varphi _{i}=\varphi {|}_{H_{i}}}$. Let

${\displaystyle p_{i}\colon V\longrightarrow V}$

denote the composition ${\displaystyle {}V\rightarrow H_{i}\rightarrow V}$, that is, ${\displaystyle {}p_{i}}$ is in particular a projection. We set

${\displaystyle {}\varphi _{\rm {diag}}:=\lambda _{1}p_{1}+\cdots +\lambda _{m}p_{m}\,.}$

This mapping is obviously diagonalizable, on ${\displaystyle {}H_{i}}$ it is the multiplication with ${\displaystyle {}\lambda _{i}}$. Sei

${\displaystyle {}\varphi _{\rm {nil}}:=\varphi -\varphi _{\rm {diag}}\,.}$

The property of this mapping of being nilpotent can be checked on the ${\displaystyle {}H_{i}}$ separately. There, we have

${\displaystyle {}{\left(\varphi -\varphi _{\rm {diag}}\right)}{|}_{H_{i}}=\varphi _{i}-{\left(\varphi _{\rm {diag}}\right)}{|}_{H_{i}}=\varphi _{i}-\lambda _{i}\operatorname {Id} _{H_{i}}\,,}$

so this is nilpotent. Moreover, ${\displaystyle {}\varphi _{j}}$ and ${\displaystyle {}p_{i}}$ commute, since ${\displaystyle {}p_{i}}$ induces the identity on ${\displaystyle {}H_{i}}$ and on ${\displaystyle {}H_{j}}$, ${\displaystyle {}j\neq i}$, the zero mapping. Therefore, also the direct sums of those commute, and hence also ${\displaystyle {}\varphi }$ and ${\displaystyle {}\varphi _{\rm {diag}}}$ commute. Thus, ${\displaystyle {}\varphi _{\rm {diag}}}$ and ${\displaystyle {}\varphi -\varphi _{\rm {diag}}=\varphi _{\rm {nil}}}$ commute.