Nilpotent endomorphism/Jordan normal form/Mapping lemma/Fact/Proof

Proof

Let ${}\varphi ^{s}=0$ and suppose that ${}s$ is minimal with this property. We consider the linear subspaces

${}V_{i}:=\operatorname {kern} \varphi ^{i}\,.$

Let ${}U_{s}$ be a direct complement for ${}V_{s-1}$, therefore,

${}V=V_{s-1}\oplus U_{s}\,.$

Because of fact, we have

${}\varphi ^{-1}(V_{s-2})=V_{s-1}\,$

and

${}\varphi (U_{s})\cap V_{s-2}=0\,.$

Therefore, there exists a linear subspace ${}U_{s-1}$ of ${}V_{s-1}$ with

${}V_{s-1}=V_{s-2}\oplus U_{s-1}\,$

and with

${}\varphi (U_{s})\subseteq U_{s-1}\,.$

In this way, we obtain linear subspaces ${}U_{i}\subseteq V_{i}$ such that

${}V_{i}=V_{i-1}\oplus U_{i}\,$

and

${}\varphi (U_{i})\subseteq U_{i-1}\,.$

Morover,

${}V=U_{1}\oplus \cdots \oplus U_{s}\,,$

since we refine the preceding direct sum decomposition in every step. Also, ${}\varphi$, restricted to ${}U_{i}$ with ${}i\geq 2$ is injective. For ${}v\in U_{i}\cap \operatorname {kern} \varphi =U_{i}\cap V_{1}\subseteq U_{i}\cap V_{i-1}$, it follows

${}v=0\,$

by the directness of the composition. We construct now a basis with the claimed properties. For this, we choose a basis ${}{\mathfrak {u}}_{s}$ of ${}U_{s}$. We can extend the (linearly independent) image ${}\varphi ({\mathfrak {u}}_{s})$ to get a basis ${}{\mathfrak {u}}_{s-1}$ of ${}U_{s-1}$, and so forth. The union of these bases is then a basis of ${}V$. The basis element of ${}{\mathfrak {u}}_{i}$ for ${}i\geq 2$ are sent by construction to other basis elements, and the basis elements of ${}{\mathfrak {u}}_{1}$ are sent to ${}0$. To get an ordering, we choose a basis element from ${}{\mathfrak {u}}_{s}$, together with all its successive images, then we choose another basis element of ${}{\mathfrak {u}}_{s}$, together with all its successive images, until the ${}{\mathfrak {u}}_{s}$ is exhausted. Then we work with ${}{\mathfrak {u}}_{s-1}$ in the same way. In the last step, we swap the ordering of the basis elements just constructed.

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