Trigonalizable endomorphism/Jordan normal form/Fact/Proof

Proof

Since ${}\varphi$ is trigonalizable, we can apply fact. Hence, there exists a direct sum decomposition

${}V=\operatorname {GeEig} _{\lambda _{1}}(\varphi )\oplus \cdots \oplus \operatorname {GeEig} _{\lambda _{m}}(\varphi )\,,$

where the generalized eigenspaces are ${}\varphi$-invariant. Looking at the situation for each generalized eigenspace, we may assume that ${}\varphi$ has only one eigenvalue ${}\lambda$, and that

${}V=\operatorname {GeEig} _{\lambda }(\varphi )\,$

holds. Then,

${}\psi =\varphi -\lambda \operatorname {Id} _{V}\,$

is nilpotent. Therefore, because of fact, there exists a basis such that ${}\psi$ is described by a matrix of the form

${\begin{pmatrix}0&c_{1}&0&\cdots &\cdots &0\\0&0&c_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&0&c_{n-2}&0\\0&\cdots &\cdots &0&0&c_{n-1}\\0&\cdots &\cdots &\cdots &0&0\end{pmatrix}},$

where the ${}c_{i}$ equal ${}0$ or equal ${}1$. With respect to this basis,

${}\varphi =\psi +\lambda \operatorname {Id} _{V}\,$

has the form

${\begin{pmatrix}\lambda &c_{1}&0&\cdots &\cdots &0\\0&\lambda &c_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda &c_{n-2}&0\\0&\cdots &\cdots &0&\lambda &c_{n-1}\\0&\cdots &\cdots &\cdots &0&\lambda \end{pmatrix}}.$

To fact