Proposition 18

— Let $E$ be a finite dimensional vector space over a commutative field $K$, let $u$ be an endomorphism of $E$ and let $E'$ be a subspace of $E$ closed under $u$. Let $u'$ (resp. $u''$) be the endomorphism of $E'$ (resp. $E/E'$) induced by $u$. Then $\chi_u = \chi_{u'} \cdot \chi_{u''}$.

For $u$ to have a Jordan decomposition, it is necessary and sufficient that $u'$ and $u''$ have ; moreover, if this holds then the absolutely semi-simple (resp. nilpotent) components of $u'$ and $u''$ are the endomorphisms of $E'$ and $E/E'$ induced by the absolutely semi-simple (resp. nilpotent) component of $u$.

Let $B$ be a basis of $E$ containing a basis $B'$ of $E'$, and let $B''$ be the basis of

$E'' = E/E'$ which is the image of $B - B'$. Let $U, U', U''$ be the matrices of $u, u', u''$ with respect to $B, B', B''$ respectively. Then $U$ has the form

and $\chi_u = \chi_U = \chi_{U'}\chi_{U''} = \chi_{u'}\chi_{u''}$ (cf. III, p. 533, Ex. 2). We deduce that the set of eigenvalues of $u$ is the union of the sets of eigenvalues of $u'$ and $u''$. If $u'$ and $u''$ have Jordan decompositions then the eigenvalues of $u'$ and $u''$ are separable over K, hence so are the eigenvalues of $u$, and $u$ has a Jordan decomposition (VII, p. 43, Th. 1). Conversely, if $u$ has a Jordan decomposition $(s, n)$ then $s$ and $n$ leave $E'$ invariant because they are polynomials in $u$; let $s', n', s'', n''$ denote the endomorphisms of $E', E', E'', E''$ induced by $s, n, s, n$ respectively. Then $s'$ and $s''$ are absolutely semi-simple, since their minimal polynomials divide that of $s$ (VII, p. 42, Prop. 15); also $n'$ and $n''$ are nilpotent. Finally $u' = s' + n', u'' = s'' + n'', s'n' = n's', \text{ and } s''n'' = n''s''$, which completes the proof.